Munkres - Topology - Chapter 3 Solutions: Section 24

 

Munkres - Topology - Chapter 3 Solutions Section 24 Problem 24.3. f    and   f (x) − x  where f (x) − iR (x) =  f ( identity function. Since  Since   f  and   iR  where   iR   is the identity Solution:   Define  Define   g   :   X   →   R  where  where   g (x) =   f ( are continuous, g continuous,  g  is continuous by Theorems 18.2(e) and 21.5. Since X  Since  X  is  is connected for all three possibilities given in this problem and   R  is ordered, the intermediate-value theorem applies.  X  =  g (1) ≤  1 − 1 = 0. Clearly if either g  g(1) For X  For  = [0 [0,, 1] 1],, observe that g that  g(0) (0) ≥  0 − 0 = 0  and  g(1) either  g(0) (0) or  or g (1) equals  equals 0  0,, then there is an  an   x0   ∈   X   X   where  where   f  f ((x0 ) − x0   = 0, so so   f ( f (x0 ) =   x0 . Oth Otherw erwise ise,,   g (0) and   g (1) (0)   >   0   and  (1)   <   0, so by the intermediate-value theorem there is some x some  x 1  ∈  X  where  X  where g  g((x1 ) = 0, so so f   f ((x1 ) =  x 1 . The proposition is not necessarily true if  X  =  X  = [0,  X  = Let  f ((x) = (1+ x)/2, which is obviously continuous. [0, 1) or 1)  or X   = (0, (0, 1). 1). Let f  It follows that f  that  f ((x) =  x if  x  if and only if  x  x  = 1, which not in X  in  X ..

Problem 24.4. Solution:   If   X  has X  has only one element, it is trivially a linear continuum, so we will assume  X   X  has  has at least two elements. Let x, y   ∈  Y   Y    where  X  is connected,  where   x < y . Sinc Sincee   X  is connected,   (−∞, y )   and and   (x, ∞)   cannot cannot be a sepa separation ration of the space. Since the two two open sets are clearly non-empty, it must be that they are not disjoint. Therefore there is some  z  ∈  ( −∞, y ) ∩ (x, ∞), from y. which it follows that z that  z < y   and x and  x < z . We infer that x that  x < z < y. X  that is bounded above, and suppose  Y   Y  has Next, let Y  let  Y  be  be a non-empty subset of   X  that  has no supremum. Define M  Define  M    =  {m  ∈ X   X   :  m  ≥  y  for all   y   ∈  Y  }  (in other words, the set of upper bounds of   Y ). Y    is bounded above,  M  is non-empty. of   Y  ). Since Since   Y  above,   M  is  B  = m∈M (m, ∞). It follows that A  B  are a separation of   X , X , contradicting that X  Then let A let  A =  = y∈Y  (−∞, y)  and  and B that  A and  and B that  X  is connected. connected. Given x Given x 0  ∈  X ,  X , if  x  x 0  < y  for some y some  y  ∈  Y ,  Y  , then x then  x 0  ∈  ( −∞, y )  ⊂  A  A.. On the other hand, if   x x 0   ≥  y  for all Y  all  Y ,, then there is some m some  m  ∈  M   M  such  such that m that  m < x 0 , so so x  x 0  ∈  (m,  B . Hence A Hence  A  and  B  partition  partition X   X .. Further, if   x x 1  ∈  A ∩ B ,  ( m, ∞)  ⊂  B. then x then  x 1  < y  for some y some  y  ∈  Y    Y   and and x  x 1  > y  for all y all  y  ∈  Y ,  Y  , which is impossible. Therefore A Therefore  A  and  B  are disjoint. Since A Since  A  and B  form an impossible separation of  X  of  X ,, we conclude that Y  that  Y    must have a supremum. Accordingly, X  Accordingly,  X  is  is a linear continuum.

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Problem 24.4. Solution:   Given x, Given  x, y  ∈  X  ×  × [0,  x < y , we have x have  x  = x and  y  = y Since  [0,, 1)  is a linear continuum, [0, 1) where 1)  where x  =  x 0 × x1   and y  =  y 0  × y1 . Since [0 1) is  x 0  < y0 , let z x1 , 1)  x 0  =  y0 , let z  z  = x y. if  x let  z 1  ∈  (  (x 1);; if  x let  z 1  ∈  (x  ( x1 , y1 ). Hence if  z  =  x 0 × z1 , then x then  x < z < y. U   be a non-empty subset of   X   × M   =   {m   ∈   X   × Now let  let   U   × [0, [0, 1) tha 1)  thatt is bounded bounded above. above. Define Define   M   × [0 [0,, 1) :   m   ≥   a   for M   is non-empty. all   a   ∈   A}, which is the set of all upper bounds of   of   A. Sinc Sincee   A   is bounded above, we’re assured that   M   X    is well-ordered, there is a least U   =   {π1 (m) :   m   ∈   M }, which must be a non-empty subset of   X . Since Since   X  Designate   U   Designate x  × [0  A . Further, if   x x  < x , then any element of   x U . If   x ∈ /  π (A), then x  x  of   U . [0,, 1) 1) must  must all  a  ∈  A. then  x  × 0  > a  for all a element x element   1  A be less than some element of   of   because no upper bound of   of   A  has a first coordinate less than   x . He Henc ncee   x × 0  is the  supremum of   A. On the other other hand, hand, if   x ∈  π 1 (A), the set  set   V   V   =  { π2 (a) :  a  ∈  A }  is a non-empty subset of   [0  that is [0,, 1) 1) that some  d > c  such that x that  x  × d  ∈  A  A,, which bounded above by 1 by  1,, so it has a supremum  y  . Accordingly, if  c  c  ∈  [0,  [0 , y  ), there is some d is greater than x than  x  × c. Therefore Therefore x  x  × y  is the supremum of  A,  A , and X  and  X  × [0  has the supremum property. We conclude [0,, 1) 1) has that X  that  X  × [0  is a linear continuum. [0,, 1) 1) is Problem 24.9. Solution:   Designate  Designate   X   X   =   R2 \A, and let  let   x, y   ∈  X   X  be  be given. If there there is no ele elemen mentt of   A  on the straight-line path in   R2 from x from  x to  to y  y,, then there is obviously a path between between the two points by exercise exercise 24.8(a). In the non-trivial case where there is an element of  A  A  on the straight-line path between x between  x  and  and y  y,, designate D designate  D 0  =  { θ  ∈  [0 π2 (a) − π2 (x))  [0,, 2π ) : tan−1 [( [(π ))//(π1 (a) − π1 (x))]   θ for A  on the straight line passing =  θ  for all a all  a  ∈  A }, which are all the angles around x around  x for  for which there is no element of   A on  x at through x through  at that angle. We will show that D that  D 0  is not empty. Assume the contrary is true and D and  D 0  is empty. It would follow A  on the line at that angle. Consequentl that for every element of the interval [0 interval  [0,, 2π ), there is some element of   A Consequentlyy, there is  A . Becaus an injection from  from   [0 [0,, 2π)  →  A. Becausee   A  is countable, there would be an injection from   [0 [0,, 2π )  →  N , establishing that  D 0 .  D 0  is not empty. Choose an arbitrary θ  from D arbitrary  θ 0  from [0, [0, 2π )  is countable, countable, a contradict contradiction. ion. Therefore Therefore D

 

 θ for =  θ  for all a all  a  ∈  A}, which are all the angles ))/(π1 (a) − π1 (y ))]   [0 , 2π ) : tan−1 [(π [(π2 (a) − π2 (y ))/ Next, designate D designate  D 1  =  {θ  ∈  [0,  y  for which there is no element of  A around y around of  A  on the straight line passing through y through  y  at that angle. By the same argument, argument,  D  θ D1  is not empty. Choose an arbitrary θ  from D 1 . arbitrary 1  from Now let   z   ∈   X   X   be the point at which the two lines from   x   and   y  at the angles   θ0   and   θ1 , respectively, intersect. There is no element of   A  on the line segments from  x  to  z   and and z  z   to y to  y . Therefore Therefore we can construct the contin continuous uous maps k0   : [0, ,   →   X   X    and   k ,   →   X   X    where where    k   x ,   k   k   z , and and    k   y . By the pastin pas ting g lemma, lemma , we [0 1] [1 2] = = 1 (1) =  = = 1   : [1, 0 (0) =  0 (1) =  1 (2) =  can generate the continuous map   k   : [0, X   from  from   k0   and and   k1   that forms a path from   x  and to   z . Accordingly Accordingly,,   X  X    is [0, 2]   →   X   path-connected.

Problem 24.10. Solution:   Suppose U  Suppose  U  is  is an open connected subspace of  R  R 2 . Let x Let  x 0  ∈  U   U  be  be given and V  and  V  be  be the set of points in U  in  U  to  to which 2 2 there is a path from x from  x 0   in U  in  U .. Since U  Since  U  is  is open in   R , there is some non-empty basis element B element  B 0  ⊆  U   U    of   R that contains x0 . By exercise 24.8(a), there is a path from x from  x 0  to every point in B in  B 0 , so so B  B 0  is contained in V  in  V .. Hence V  Hence  V  is  is non-empty. 2  v  ∈  V   V ,, there is a basis element B  v that If  v element B1  containing  containing v  that is contained in U  in U .. Since B Since  B1  is a subset of R , it is path-connected.  Since there is a path from  x 0   to Consequently, to   v  and a path from  from   v  to any point v point  v ∈  B 1 , there is a path from   x0   to v to  v  . Consequently, B1  is contained in V  U , so in  V .. Because this is true for every element of  V  of  V ,, it follows that V  that  V  is  is the union of basis elements of   U , V  V  is  is open in U  in  U .. U .. Since W    =   U \V  V .. If   W ,, there is a basis element   B2   containing  containing   w  that is contained in   U  Since   B2   is pathNow let  let   W  If    w   ∈  W   W ;; otherwise, there would be a path from  to   w , from   x0   to  element   w ∈  W  connected, there cannot be any path from   x0  to any element  then   w to then to   w, contradicting that there is no path from   x0   to to   w. There Therefore fore   B2  is contained in  in   W  W ,, from which it follows that B that  B 2  is the union of basis elements of   U . U . Hence W  Hence  W  is  is open in U  in  U .. Since   V  Since V    =  U \W  W ,, the set  set   V  is V  is closed in   U . U . Sin Since ce   V   V   is both open and closed in the connected subspace   U  U ,, it either  U  or  U .. We conclude that U  equals U  equals  or is empty. We know V  know  V    is non-empty, so V  so  V    =  U  that  U    is path-connected.

Section 25 Problem 25.1. Solution:   Suppose  Suppose   A  is a connected subspace of   R   containing a given pair of points   x   and  and   y   where  where   x < y. It follo follows ws that   A   = ((∞, y ) ∩ A) ∪ ([y, that ([y, ∞) ∩ A), where the two rays are open in   A  and therefore form a separation of the space. Accordingly, for any two distinct points in  in   A  there is no connected subspace of   R   containing containing both points. We conclude conclude that the components of   A  consist of the individual elements of   R . Accordingl Accordinglyy, the only connected connected subspace subspacess of   R   are single-elemen single -elementt spaces. It follows that the path compone components nts are also the individual elements elements of  R  because a path-connected space must be connected. The only continuous functions  f    f   : R → R  are constant functions. By Theorem 23.5,  f   f ((R)  is a connected subspace of  R . From the previous result,  f (  f (R)  must be a single element of   R , implying that f  that  f ((x) =  a0  for some a some  a 0  ∈ R. Problem 25.2. Solution: Part Part (a)  From exercise 24.8(a), since   R  is path-connected   Rω is path-connected, so the only path component is   Rω . Further, this implies that   Rω is connected, so the only component is   Rω . Part (b) Part (c) Problem 25.3. Solution:   Let U   I 02  is a linear continuum, it is connected, as is any interval Let  U  be  be any neighborhood of a given x given  x  ∈  I 02 . Since Since I  U ,, which must 2be an interval of   I 02 ;  I 02   containing x containing  x  that is contained in  in   U  contained in it. There is some basis element contained element of  I   U    contains x , so  I 0  is locally connected consequently, the basis element is connected. Hence  U  contains a conne connected cted neighbo neighborhood rhood of   x, so I  (in addition to be connected in general!)

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U    be some neighborhood of   x. Any The space  space   I 02   is not locally path-connecte path-connected. d. Let Let   x  =  x 0  × 1   for for   x0   ∈   [0 [0,, 1) 1)   and  and   U   x  contained in U  x0  × β, x1  × 0)  β  ∈  [0 neighborhood neighbo rhood of  x in  U  must  must contain some basis element B element  B  of the form  (  (x 0) where  where β   [0,, 1) 1) and  and x1  ∈  (x a, b]  →  I 02  ( x0 , 1] 1].. Following the argument of example 24.6, B 24.6,  B  is not path connected: if we assume a continuous map [ map  [a, for a path between any two points with different first coordinates in   B , we derive a contradiction that an interval of real  x 0  ∈  [0,  [0 , 1) is 1)  is not locally path-connected. An analogous argument shows that  for x numbers is countable. Hence any x any  x 0  × 1  for  x 0  ∈  (0  x 0  × 0  for  (0,, 1] 1] is  is not locally path-connected.  for x any x any The components of   I  I 02 are the vertical intervals (i.e.,  { x0 } × I  for   for any x any  x0  ∈  I   I ). ). Obviously if points x points  x and  and y  y are  are such that π1 (x) =  π 1 (y), they are contained in an interval of  I  of  I  and  and thus are path-connecte path-connected. d. On the other hand, if   π π 1 (x)   =  π 1 (y), then by the arguments above they are not path-connected.

Problem 25.4. Solution:   Let X  Let  X  be  be a locally path-connected space. Given a non-empty, connected, open set U  set  U    of   X , X , for each x each  x  ∈  U   U  there  there is a path-connected neighborhood V  neighborhood  V x   of  x that  x  that is contained in U  in  U .. By Theorem 25.4, V  25.4,  V x  is open in X  in  X ,, as well as open in U  in  U  since it is contained in U  in  U .. Hence  Hence   U  is U  is the union of a collection  V   =  { V x  :  x  ∈  U }  of open path-connected path-connected sets. Further, Further,  V x  is closed in U  each V  each in  U  as  as well. Observe Observe that the sets in  V  are pairwise disjoint because, by Theorem 25.1, each must be contained in one path component, which are all disjoint. We infer that: 0

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U \V x   = 0

V x

V  x ∈V\ V  x0

is open in  in   U . U . By the definition definition of a conne connected cted space, space,   V x  is either empty or equal to  to   U  U .. Since V  Since  V x  must at least contain x0 , it follows that U  that  U    equals V  equals  V x   and is path-connected. 0

0

0

Problem 25.6. Solution:  Suppose space X  space  X  is  is weakly locally connected connected at each of its points. Given x Given x  ∈  X   X ,, let U  let  U x  be a neighborhood of  x  and  C  a  C  a component of   U . U . Since X   x,, there is a subspace W  Since  X  is  is weakly locally connected at  x subspace  W x  containing a neighborhood neighborhood V x   of  x.  x . By Theorem Theorem 25. 25.1, 1,   W x  is contained in C  in  C ,, so so   V x  is also contained in  in   C . Accordingly, Accordingly,   C   = x∈C  V x , from which it follows that C  that  C  is  is open. Therefore every component in any open set of  X  of  X  is  is open. Applying Theorem 25.3, X  25.3,  X  is  is locally connected.

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Problem 25.8. Solution:   Let  Let   p  :  X   →  Y   Y    be a quotient map where  where   X   X   is locally connected. connected. Given an open set  set   U  U    of   Y  Y ,, the set  set   p−1 (U  U )) U ))  is open. is open because  because   p   is quotient quotient map (and therefore therefore continuous) continuous).. By Theorem Theorem 25.3, every component component of   of   p−1 (U  Now let  let   C  C  be  be a component of   of   U . U . If   If   y   ∈   C , then since  since   p  is surjective   p−1 ({y})   is non-empty non-empty. Fo Forr each   x   ∈   p−1 ({y}),   −1 U )  contains  x.. By Theorem 23.5, p  contains x 23.5,  p((C x )  is connected because p some component C  component  C x  of  p because  p is  is continuous. Again applying  of  p (U ) Theorem 25.3, p 25.3,  p((C x )  must be contained in C  in  C  because  because it intersects C  intersects  C    at y at  y.. Since this is the case for each y each  y  ∈  C , it follows −1   C  =  V    in X   p((V  V ))  ⊆  C , then V   p((C x ). Observe that if there is any  V   that C  that  = x∈ p (C ) p in  X  where  where p then  V   ⊆ p (C )  and each element of   −1  V  V  is so  C  open,   p−1 (C )  is open, so C  each   C x  is open,  Hence ce we infer infer that  that   p (C ) = x∈ p (C ) C x . Since each   is contained in some   C x . Hen  Y    is open. By Theorem 25.3, Y  is open because p because  p  is a quotient map. To summarize, each component of any open set in  Y  25.3,  Y  is locally connected. connected.

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Problem 25.9.  G  containing the identity element e x  ∈  G  G,, then x  e · x  = x Solution:   Suppose C  Suppose  C  is  is a component of  G containing element  e.. If   x then  x · e  = x  =  x is  is in xC  in  xC  and  a nd e  =  x Cx.. Since the is in  in   Cx the operator operator  x · y  for any  any   y  is a continuous map, it follows that   xC   and  and   Cx  are connected connected.. By the hint,  xC  = they are components, as well. Since the components of  G of  G  must be disjoint, we infer that  xC   =  C x  because they intersect  x.. Therefore  C  is  G . at at x Therefore C   is a normal subgroup of  G.

Problem 25.10.

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 X ,, there are no disjoint proper subsets Solution: Part (a)  The reflexive and symmetry properties properties are trivial. Given  Given  x  ∈  X  A  a  B   of   X  whose X  whose union is X   x.. Given x,  X    where x  y,, it follows that  and nd B is  X    where x where  x  is in both  both   A  and  B . Thus x Thus  x  ∼  x Given  x, y  ∈  X  where  x  ∼  y  x  ∈  A  and  y  ∈  B  B.. Clearly then, y there are disjoint open sets A sets  A  and  B  whose union is X  is  X  where  where x then,  y  ∼ x. Establishing Establ ishing transitivity transitivity requires a little more work. Given x Given x,, y,z  ∈  X   X    where x where  x  ∼  y   and y and  y   ∼  z  z,, there are disjoint open sets A sets  A  and  B  whose union is X  is  X    and and x  x  ∈  A  and  y  ∈  B  as well as disjoint open sets  C   C    and D and  D  whose union is X  is  X    and y and  y  ∈  C  and   z   ∈  D. and  D . Let  Let   E   =  A ∪ D   and and   F   F   =  B  ∩ C , each of which must be open because their constituent constituent sets sets are open. We will show that E  that  E    and F  and  F  are  are disjoint and have a union equal to X  to  X .. Clearly E  Clearly  E   ∪ ∪ F  F  is  is a subset of   X . X . Let w Let  w  ∈  X   X .. If   w w  is in A  or   D, then   E .. Moreover, it cannot be that   w  is in both   C  because  D ; hence then   w  ∈  E  both   B   and and C   because they are disjoint with  with   A  and  D; w∈ /  F .  F . On the other hand, if  w  w  is in neither A  D,, then w  w  ∈  F   F    and w /  E . Thus E  neither  A  nor  nor D then  w  is in both B both  B   and C  and  C ,, so so w and  w ∈ Thus  E    and F  and  F   X . Since x  y  is in F  are disjoint and have a union equal to  X . Since  x  and  z  are in E  in  E    and and y in  F ,, we conclude that  x ∼ z . C  be X  and fix an  X  is C    =   X , so clearly   x   ∼   y   for all Part (b)   Let   C   be a component of   of   X  and an   x   in   C . If   If   X   is connected, then  then   C  x, y  ∈  X   X  because  X ; therefore X   C  is  X    is  because there is no separation of  X ; therefore  X  is  is a quasicomponent, of which  C   is a subset. subset. Otherwise, Otherwise, X   x,, it follows that  x  is in some quasicomponent  Q  Q.. If   yy  ∈  C   C ,, then there is a connected subspace disconnect discon nected. ed. Since x Since x  ∼  x V  V    of   X   X   that contains x contains  x  and  y . Since X  Since  X  is  is disconnected, there are disjoint open set   A   and  and   B  whose union equals  equals   X . By Lemma 23.2, we may assume without loss of generality that that  V  is  V   is contained in A in  A;; accordingly, both x both  x  and  y  are in A in  A.. Consequently, there is no separation of  X    X   where  where   x   and and   y  are in different sets of the separation, it follows that  x   ∼  y , so y  ∈  Q.  Q. Therefore C  Therefore  C  is  is contained in Q in  Q.. X   is locally connected. W    be the union of components of   X  X  not C  that Now suppose  suppose   X   connected. Let  Let   W   not equal to   C   that intersect  intersect   Q. By  X  is contained in some quasicomponent. It follows that  W   W  is the previous result, every component of  X  is  is contained in Q in  Q,, and W .. Next, let V  since there is some component contained in W  in  W  for  for every point in Q in  Q,, we infer that  Q  = C   =  C   ∪ ∪ W  let  V    be the union W ,, which of all quasicomponents other than Q than  Q.. Thus V  Thus  V  is  is open since every quasicomponent is open. Then let  U 0  =  V   ∪ W  W )) ∪ C  =  X ,, it follows that they form a separation of   X . X . Since we know C   =  X  know  C    is also must be open. Since U  Since  U 0  ∪ C   = (V   ∪ W  not empty, W  empty,  W  must  must be empty because no two points in  Q  can be in separate disjoint, open sets that form a separation of  X . Therefore C  Therefore  C  =  Q,, and the components and quasicomponents of  X  of  X  are  are equal.  =  Q

Part (c)  I only respond for A for  A here.  here. This set is not connected or path-connected, generally or locally. We can separate  A by the line  {1} × [0  and then the union of all other points in in A  A.. It follows that A that  A is  is neither connected nor path-connected. [0,, 1] 1] and Further, A Further,  A  is not locally connected at 0 at  0 × 0  or  0 × 1. I’ll prove this result for   0 × 0  only  only.. If   T  is T  is a neighborhood of   00 × 0 element   (a, b) × (c, d)  where  where   a <  0  < b <  1   and  and   c <  0  < d <  1  1.. Fo Forr any b any b,, the vertical in   R2 , it must contain the basis element  /n × [0  N  as  n >  1 /b (i.e., lines 1 lines  1/n [0,, d)  are contained in this basis element where  n >  1/b  1 /b.. Designate Designate N   as the smallest  n  where  where n  1/b  (i.e., T ). As a resu  N  and the right-most vertical line in  in   T ). result lt,,   T   ∩ A  contains all vertical lines where  where   n   ≥  N   and is open in   A. We will will  T 1  be open in   R2 where  T 0  = (a, β ) × (c, d)  and  T 1  = (β, b) × (c, d)  where show that T  that  T  ∩ A  is separable in A in  A.. Let T  Let  T 0   and and T  where T  /n × [0 /N . We see that T  in  T    and contains 0 contains  0 × 0  as well as all vertical lines 1 lines  1/n [0,, d)  where 1/(N  +  + 1) < 1)  < β <  1  1/N  that  T 0  is contained in T   T 1  are disjoint and open in /N   × n > N ; N ; further,  T 1  is contained in T   and nd T  in  T    and contains the vertica verticall line line 1  1/N  × [0 [0,, d). Clearly T  Clearly  T 0  a further, T  R2 , and T  and T 0 ∩ A  and  and T   T 1 ∩ A  are open in A in  A.. Since ( Since  (T  T 0 ∩ A) ∪ (T 1 ∩ A) =  T  ∩ A, we see T  see  T  ∩ A  is separable. Because this is true for every neighborhood of  0 that  A  is not locally connected. We infer that  A  is not locally path-connected,  0 × 0, it follows that A either. The components and path components of  A of  A  are both  { 0 × 0}, {0 × 1}  and each 1 each  1/n /n × [0  where n  n  ∈ N. The quasi[0,, 1] 1] where /n × [0,  n  ∈ N. The vertical lines are obviously quasicomponents. components consists of   { { 0 × 0, 0 × 1}  and each 1 each  1/n [0, 1] where 1]  where n quasicomponents. You cannot place  place   0 × 0   and  and   0 × 1  is separate sets of a separation of   of   A  because doing so would require the separation of  some vertical line in A in  A,, which is not possible.

Section 26 Problem 26.1. Solution: Part (a)   Suppose  Suppose   X  X    is compact in   T   . If   A  is an open covering of   X  X    in   T  , then every element of   A  is also   X  is open in  T   , so X  so  X  is  is also an open covering in that topology. Since  X   is compact under  T  , there is some finite collection of  A  the covers X  covers  X .. Therefore X  Therefore  X  is  is compact under  T    as well.  X  is compact in  T   then By contrast, it does not follow that if  X  is   then it is compact in  T   . Let X  Let  X    = [0, [0 , 1] 1].. This set is compact X   is: under the standard topology of   R. Under the   R  topology, an open covering of   X   A  =

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 1 , ) [ n+1 n

∪ {0} ∪ {1}. N

n∈

But there is no finite subcollection of   A A  that covers [0 covers  [0,, 1]. 1]. Thus [0 Thus  [0,, 1] 1] is  is not compact under the   R   topology.  Part (b)   Suppose   T    is finer than   T   . The ident identit ityy map  map   id   :   X   →   X  X    from the topologies   T    to   T   is continuous: 1 U ) =  U   U  is  is clearly open under the (finer) topology  T  . Further, i Further,  i d   is Given an open set U  set  U    under  T   , the reverse image i image  i − D   (U )

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clearly bijective. Therefore i Therefore  i D  is a bijective, continuous function from a compact space to a Hausdorff space. By Theorem  i D  is a homeomorphism.  V   is open in  T  , then V  26.6, i 26.6, homeomorphism. Therefore Therefore if  V  is then  V  is  is open in  T   . We conclude that if the topologies are comparable, they must be equal.

Problem 26.2.  Y    be a subspace of   R  in the finite complement topology and  A  be a collection of open sets in   R Solution: Part Part (a)  Let  Y  Y .. If   R  ∈ A, then trivially there is a finite subcollection consisting only of   R   that covers  Y .. If   R  ∈ / A, each that covers  covers   Y  covers   Y  0  is a subset A  A\0A as some in  A . It follows that  Y  \A that Y  be infinite complement isefinite. Designate Designate A of  ∈   R\AA must former because must beits finite. finit e. If there ar are  N  elements  N  elements in  Y  in Y  , we mayset denumerate each element of  Y  of  Y  \A  a  ass  y n 0 , so the former N  Y    = n=0 An , which is composed of a finite Thus us   Y  for some  some   n   ∈   S N N  , and there is some   An   ∈ A  such that   yn   ∈   An . Th subcollection of   A A . By Lemma 26.1, every subspace of   R  in the finite complement topology is compact. Part (b)  The subspace  subspace   [0, compact. Define Define   Qn   =  { q   ∈   Q  :  q   ∈  [1 − 1/n, 1)}  for any [0, 1] in 1]  in this topology of   R   is not compact. n  ∈ N. The complement of the set  R \Qn  for any n any  n  is  Q n , which is countable; therefore   R\Qn  is open. Define  A  as:

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A  =  { R\Qn }∞ n=1 . This is a covering of   [0,  by open sets of   R R . Let  A  be a finite subcollection of   A A  and  and Y   Y    = An ∈A  An . Since  A  is finite, [0 , 1] 1] by if  P    P   =  {n  ∈ N :  An  ∈ A }, then there is a maximum element  M   M    of  P .  P . Since there is no  Q n   in  A  where where n  n > M , there is  no rational number greater than or equal to 1 to  1 − 1/M   in Y  in  Y .. Therefore [0 Therefore  [0,, 1]  is not contained in in Y   Y , , so  A is not a covering 1] is of  [0  [ 0, 1] 1].. Because Because this is true for every every finite subcollection subcollection of   A, we conclude from Lemma 26.1 that  [ [00, 1] 1] is  is not compact in this topology.

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Problem 26.3. Solution:  We will prove this propositio proposition n by mathe mathematical matical induction. induction. Suppose  Suppose   X 0   and  and   X 1  are compact subspaces of   of   X .  Y 1  by open sets in X  Y 1 , each is also Designate  Y 1  =  X 0  ∪ X 1 . Let  A  be a covering of  Y  Designate Y  in  X .. Since X  Since  X 0   and X  and  X 1  are subsets of   Y   X 1 , respectively, and covered by  A . Because Because each is compac compact, t, there are finite subcollection subcollectionss  A 0  and  A 1  that cover X  cover  X 0  and  a nd X   X 0  ⊆ A0   and X  Consequently,  Y 1  =  X 0  ∪ X 1  ⊆ A ∈A A0 ∪ A ∈A A1 . Moreover, because  A 0   and  A 1 and  X 1  ⊆ A1 . Consequently, Y  hence X  hence X    that covers Y  are finite, finite, their union is finite. Thus  A 0  ∪ A1  is finite subcollection of  A  of open sets in  in   X  covers  Y 1 . From Lemma 26.1, Y  26.1,  Y 1  is compact. Now suppose the inductive hypothesis holds for some n some  n  ∈ N. Suppose we have a collection  {X 0 , . . . , Xn  +1 }  of compact n subspaces of   Y . Y  . By the inductive hypothesis, Y  hypothesis,  Y n+1  = k=0 X 0   and and Y   Y n+1 ∪ X n+1  are each compact. Thus any finite union of compact subspaces of  X    X   is compact.

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Problem 26.4. Solution:   Suppose Y  Suppose  Y  is  is a compact subspace of metric space  X   X .. Let  A  be the following collection: A =  {Bd (y, 1)}y∈Y  . covers   Y  Y .. Define Define Obviously   A  covers  covers   Y  Y  with  with sets open in  in   X . The There re is a finit finitee subcollect subcollection ion   A =  { Bd (yαn , 1)}N  n=1  the covers    set  D  =  {d(yαn , yαm ) : 1 ≤  n, m ≤  N }  is finite, so C  =  = A ∈A  A . Because there are a finite number of balls in  A , the set D  M .. Since it has a maximum element M  element  M .. Accordingly, the maximum distance between the centers of any two balls in  A  is is M  Y  must be in some ball in  A  , it follows that for any y  Y ,, the two points are contained in the balls every element of   Y  must any  y 0 , y1  ∈  Y  centered by some y some  y αn   and y and  y αm , respect respectively ively.. It follo follows ws from the Triangle Inequality Inequality that:

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d(y0 , y1 )  ≤  d(  d (y0 , yαn ) + d(yαn , yαm ) + d(yαm , y1 )  < M  +   + 2.  Y  is Hence Y  Hence  is bounded by this metric.

Problem 26.5.

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Solution:   Let  Let   A   and  and   B  be disjoint compact subspaces of a Hausdorff space   X . Give Given n   a   ∈   A, by Lemma 26.4 for each b   ∈   B   there are disjoint open sets   T a,b   U    X  X     a   and   in   that are neighborhoods of  of     and   b, respect respective ively ly.. Let   B  be the a,b a,b a,b N  Sincee   B  is a covering of   B   in  in   X , there is a finite subcollection   {U aa,b collection of   U a,b ,bk }k=1  that covers a,b   for all   b   ∈   B . Sinc N  B . Des Design ignate ate   V a   = k=1 V a,b Obviously sly   V a   is open since the intersecti intersection on is finite. If   b  ∈  B , then  then   b  ∈  B a,bk  for some a,bk . Obviou bk , so T  so  T a,b  b . Theref Therefore ore b  b ∈ /  V a . Consequently Consequently,,   V a  is a neighborhood of   a a  that does not intersect B intersect  B . If  a,bk  does not contain  b. V  V    = a∈A V a , then V  then  V  is  is open, contains A contains  A,, and does not intersect B intersect  B . A similar argument shows there is an open set in X  in  X  that contains B contains  B  and does not intersect A intersect  A..

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Problem 26.6. Solution:   Suppose  Suppose   f  f    :   X   →   Y   Y   is a continuous map where   X   X   is compact and   Y  Y    is Hausdo Hausdorff. rff. Given Given a clo closed sed set  set  U  U    of  X , by Theorem 26.2 subspace   U   U   is compact. The restriction restriction   f |U  is U  is continuous continuous by Theo Theorem rem 18.2(d). 18.2(d). By Theorem Theorem 26.5, f |U ( U (U ) U )  is compact, and by Theorem 26.3 f  26.3  f |U ( U (U ) U )  is a closed subspace of   Y . Y  . Since f  Since  f ((U  U )) =  f |U  U ((U  U )), it follows that f  that  f    is a closed map. Problem 26.7. Y   and Solution:   I found this one very tricky. tricky. Let Let   A  be closed in  in   X   × × Y   and   B   =  π 1 (A). We will will show that that   B  is equal to its Y )\A, which must be open. If   x x 0  ∈ /  B  B,, then  { x0 } × Y  Y  does closure, so it is closed. Designate Z  Designate  Z  =  = (X   × × Y )  does not intersect A intersect  A,, x Y  Y  is    Z . Z   Y  is  W    W  x 0   in X  hence  { 0 } ×  is a slice of  . Because Y  Because  is compact, the tube lemma implies that there is a neighborhood   of   x in  X  Y )) = W  Y  is × Y  =  W  is  is open. such that the tube W  tube  W   × × Y   is contained in Z  in  Z ,, and therefore does not intersect A intersect  A.. It follows that π that  π 1 (W   ×  W ,,  Y .. But x  A for  W ,, then x  W    cannot intersect B  x 1  is in both B But  x 1  × y1  ∈  W   for some y some  y 1  ∈  Y  both  B   and and W  then  x 1  × y1  ∈  A Further, W  Further, intersect  B  because if  x  A A  W  is /  B  that does not intersect B which contradicts that that W    A, and A and   don’t intersect. Accordingly ,  W   is a neighborhood of  x of  x 0  ∈ intersect  B . By Theorem 17.5(a), 17.5(a), A  A W   , so so A  is closed. ThusAccordingly, Thus π  π 1  is a closed map.  =  = A Whew!

Problem 26.9.  A,, we have Solution:  The first part of this proof is nearly identical to the proof of the tube lemma in the text. Given  x 0  ∈  A {x0 } × B  contained in N  in  N .. Since N  Since  N  is  is open, we may cover  { x0 } × B  by the collection  A  =  {U α × V α }  where  where U   U α  and  V α  are basis elements of   X   X   and Y  and  Y  that  that are contained in N  in  N  (which  (which must exist since N  since  N  is  is the union of basis elements). Observe that since  { x0 } × B  is homeomorphic with B with  B,, this set is compact. Therefore there is a finite subcollection  A  =  {U n × V n }N  n=1 that covers  { x0 } × B . We may assume each set in   A intersects  { x0 } × B  and thus x thus  x 0   ∈  U n  for each  each   n; otherwise, that N  set contributes contributes nothing nothing to the covering covering and we may discard it from the subcolle subcollection ction.. Define  Define   U x   = n=1 U n , which is show w that that U   U x  × B  is covered by the collection   A . If  open in X  in  X .. Clearl Clearlyy  U x  × B  contains the slice   {x0 } × B . We will sho then  x  ∈  U n  for all n all  n and  and  y  ∈  V k  for some k some  k.. Therefore x Therefore  x × y  is contained in U  in  U k × V k . It follows that  A  x × y  ∈  U x × B , then x 0

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x × B ; since this covering is contained in N   A there  U  {  N ,, itin X   there is a ”tube” U  ”tube”  U x × B  containing covers U  covers that for each x each  x 0  ∈  A x0 } × B  contained in N   X .. the slice in  N    where U  where  U x  isinopen infollows By an analogous argument, for each  y 0  ∈  B  there is a tube A tube  A × V y  containing the slice A slice  A × {y0 }  that is contained in N   N   where V  where  V y  is open in Y  in  Y .. U   = x∈A U x   and   V   V   = y∈B V b , each of which is open in   X  X    and  Y ,, respectively  V    is open in Define   U   Define and   Y  respectively.. Thus  Thus   U   × V  U    and  V .. Next X   × Y .. Given  so   x × y  is contained in  in   U  and   V  Next we will show show that and   y  ∈  U y , so × Y  Given   x × y   ∈  A × B , we have  have   x  ∈  U x   and U  × V    is contained in   N  N .. Let x V . It follow some   y0   ∈  B . and   y   ∈  V y  for some  some   x0   ∈  A   and   × V  Let  x × y   ∈  U  × V . followss that  that   x  ∈  U x  for some  Thus x Thus  x  ∈  U n  for all n all  n  and  y  ∈  V m  for all m all  m,, 0

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Problem 26.10. Solution: Part (a)  Note that this is Dini’s Theorem. Since the sequence sequence ( f n )  is increasing and converges to   f  f ,, it must  (f  be that  that   f n (x)  ≤  f   f ((x)  for all   n  ∈   N   and and   x  ∈  X   X .. If the opposite opposite were true, true, there there woul would d be som somee   f n (x)  > f  f ((x)  for some n  ∈  N  and  x  ∈  X   X .. If      =  f n (x) − f ( f (x)  > 0  >  0,, then for m for  m > n  we have f  have  f m (x)  ∈ /  B d (f  f ((x), / /2) contradicting ng that that ( f n (x)) 2),, contradicti  (f  converges to f  to  f ((x).  g n  :  X  → R  as  g n (x) =  f (  f (x) − f n (x)  for all n Define g Define all  n  ∈ N. Since f  Since  f  and  and each f  each  f n  are continuous, each g each  g n  is continuous.  X  is  f n (x)  ≤  f   f ((x)  in all cases, g Because X  Because  is compact, it follows that g that  g n (X )  is compact. Further, Further, since since f  cases,  g n (x)  ≥  0  0.. Finally, by f n ), the sequence ( the convergence and monotonicity of  ( of  (f  sequence  (ggn )  converges to the zero function and is decreasing.

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We will now show that g that  g n (X )  is bounded for each n each  n  ∈ N. Obviously g Obviously  g n (X )  is bounded below by 0 by  0.. Because g Because  g n (X )  is compact, there is a collection  C  =  { (ay , by ) :  y  ∈  g n (X )  and  a y  < y < by }  of open sets in   R. By Lemma 26.1, there is a some  b yk  that is the largest right-hand endpoint of the covers  g n (X ). There is some b finite subcollection  C  =  {(ayj , byj }N  j =1  that covers g  intervals of   C  . Thus Thus every every   y  ∈  g n (X )  is less than or equal to  b yk . Sinc Sincee  g n (X )  is a bounded above, it has a supremum M n . Since ( Since  (ggn )  is decreasing, it follows that ( that  (M  M n )  is decreasing. Define the sequence ( sequence  (G Gn )  where  where G  G n  = [0, each  n  ∈ N. Clearly each G each  G n  is closed and contains g contains g n (X ). Further, [0, M n ]  for each n this is a nested sequence where G where  G n+1  ⊆  Gn . Since each G each  G n  is closed and containe contained d in the compact subspace subspace G  G 1 , Theorem 26.9 implies that n∈N Gn   is non-empty non-empty.. If   c   ∈ n∈N Gn , then  then   c   ≥   0. Sinc Sincee   (gn (x)) )) converges  converges to   0  for all   x   ∈   X , it P    we have   gn (x)   ∈   Bd (0 δ/2) follows that given any positive   δ < c, for each   x  there is some   P   ∈   N  where for   n   ≥   P  (0,, δ/ 2)..

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N n  and therefore is not an element of  n∈N Gn . Thus the only element of  /n  G Gn   is Consequently, positive positive c is 0  0.. /n ∈G We conclude that   c(g∈ )   conver con verges ges uniforml uniformlyy to the zero function function. . Remembe Rememberr that that if positiv positivee   α ∈ n , then since /  Gm   for Gn )  is a nested sequence, α  β > α  then  β ∈ /  Gn . Further, since ( Gn  = [0 sequence,  α ∈ since  (G  then β that  α > M n ; hence if  β [0,, M n ]  it follows that α m > n. Therefore given   g N () (X )  ⊆  GN ()  ⊆  Bd (0 M n )  is decreasing, it given   >  0,  0 , there is some N  some  N (()  ∈ N  where  where g (0,, ). Since ( Since  (M  follows if   n > N  N ((), then g then  g n (X )  ⊆  B d (0, (0, ). Thus if   x  ∈  X ,  X , then  then   gn (x)  ∈  B d (0 (0,, )   for for   n  ≥  N   N ((), from which it follows that f  that  f ((x) − f n (x)  ∈  B d (0 Therefore  d((f ( f (x), f n (x)) < definition,  (f  f n )  converges uniformly to f  to  f .. (0,, ). Therefore d ))  < . By definition, ( Part (b) Where sequence is not increasing.  See sequence in exercise 21.6. Where  X  is not compact.  The sequence g sequence  gn  = arctan(x  for n  n  ∈ N, which converges converges point-wise only to f  to  f ((x) = 1. arctan(x + n)  for

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Problem 26.11. Solution:   Since   A  is simply-ordered by proper inclusion, the elements of   A   fo form rm a nested nested sequence sequence.. By Theorem Theorem 26.9, Y  Y    = A∈A A   is non-empty Y   is close C    and  non-empty. Since  Since   Y   closed d (being the intersection intersection of closed sets), sets), it is compact. Assume  Assume   C  and   D Y  . It follows that C   D  are closed and, by Theorem 26.2, compact. Applying exercise 26.5, there form a separation of   Y . that  C    and and D

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U   and  V   in  X   of   C   C   and U    and  V    cannot form a separation of a given are disjoint open sets  sets   U   and   V   in   X   and   D, respectively. respectively. We know  know  U  and   V  Aα  ∈ A  because  A α  is connected. Since ( U  ∪ V ) ∩ Aα  is open in A U   ∪ V )) ∩ Aα ) = A V ))  because A Since  (U   ∪ V ) in  A α , it follows that A that  A α \(( ((U  ∪ V  =  A α \(U   ∪ ∪ V  V )). Further, if   A Aβ  ∈ A V )) =  A α \(U   ∪ V )  is closed in X  ∪ V  ∪ V  since  X \(U   ∪ ∪ V ) in  X ,, so is A is  A α  ∩ X \(U   ∪ is closed in A in  A α . In addition, since X  V )). Hence the collection V ), it follows that a  A α , then for each a ∪ V  therefore  a  ∈  A β \(U   ∪  ∪ V ) that  a  ∈  A β  and therefore a each  a  ∈  A α \(U  ∪ contains A contains {A\(U   ∪ ∪ V  V )) :  A  ∈ A}  is a nested sequence of closed sets in X  in  X .. By Theo Theorem rem 26.9, Z  26.9, Z    = A∈A A\(U   ∪ ∪ V  V ))  is non-empty. But Z  But  Z  is  is a subset of   Y  that Y   that does not intersect U  intersect  U    or or V   V ,, contr contradictin adictingg that C  that C    and D and  D  form a separation of   Y . Y  . We conclude that Y  that  Y    is connected.

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Section 27 Problem 27.1. Solution:   Suppose  Suppose   X  is X  is an orde ordered red set where where every every closed closed set is compact. compact. Let  Let   A   be a non-empty subset of   X  X  that  that is  X .. If  A  A  contains the  X    and m A . Otherwise, bounded above by m by  m  ∈  X  largest element of  X  of  X ,, then m then  m  ∈  X  and  m  is the supremum of   A.  A  :  a  ≥  a 0 }. As a result, any upper bound of   A A  is an upper bound of  let let a  a 0  be an arbitrary element of  A  A  a  and nd A  A  =  { a  ∈  A : A , and vice-versa. vice-versa. Define C  Define C a  = (−∞, a)  for each a each  a  ∈  A  and  D a  =  X \C a  = [a, ∞), which is closed. Observe that D that  D a   is  m . Designate: not empty because it contains  m. D  = Da .

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closed sed and, by hypothe hypothesis sis,, compac compact. t. First First we will show show that that   D   is non-empty non-empty.. Let   D = Since each  each   Da   is closed,   D  is clo  N   D . It follows that if  a  am  < an  for any m, any m, n in D , then D then Dm ∩ Dn  = [am , ∞) ∩ [an , ∞) = {Daj }j =1  be a finite subcollection of  D.  [an , ∞). Accordingly Accordingly,, the intersectio intersection n of the elements elements in   D is   [m, ∞)  where  where   m  is the largest of the finite set of   {aj }.  Consequently,  D has the finite intersection property. Since every D every  Da  is contained in the compact space [ space  [a a0 , ∞), by Theorem 26.9 the intersection a∈A  Da  is non-empty. Next we will show that D that  D equals  equals the set of all upper bounds of  A  A  (and (and A  A,, as well). well). If   d d  ∈  D,  D , then d then  d  ≥  a  a for  for all a all  a  ∈  A ; /  D.  D . Conversely /  D α  for some  case   d ∈ Conversely,, if   u  is an upper bound of   A, then if the contrary were true,  true,   d ∈ some   α  ∈  A  , in which case  u  ≥  a  for any a  A,, so   D . We conc  u  is an upper bound of   any  a  ∈  A so   u  ∈  D a  for all  all   a; hence,  hence,   u  ∈  D. conclude lude that an element element u of   A  if and

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 D.  D . onlyLet if it  i Dis :contained  D  →  D  beinthe Let i identity function, which is continuous. Since D Since  D  is compact, by the Extreme Value Theorem  i D X    has M . Accordingl  M  is consequently,   A. We conclu conclude de that  that  X  has an absolute minimum of   M . Accordinglyy  M   is the supremum of   of   A and, consequently,  the least upper bound property.

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Problem 27.2. Solution: Part Part (a)  Given  Given x  x  ∈  X ,  X , if  d(  d (x, A) = 0, then 0 then  0 = inf {d(x, a) :  a  ∈  A }. Therefore given any  any   >  0  0,, there is some a  ∈  A such  A  such that d that  d((x, a )  <   (otherwise,  (otherwise, d  d((x, A)  ≥   >  0 by  0  by the definition of the infimum). As a result, the ball  B d (x, ) intersects A intersects  A  at  a   for any  any   >  0. 17.5(b),  x  ∈  A.  A .  0. By Theorem 17.5(b), x Conversely,, suppose Conversely supp ose   x   ∈   A. For an anyy    >   0, the ball   Bd (x, )   intersects   A, so there is some point   a   ∈   A   where d(x, a )   < . Sinc Sincee the the me metr tric ic is bou bound nded ed below below by zero zero an and d any  any    >   0   cannot cannot be a lo lowe werr bound, bound, it follow followss that that d(x, A) = inf {d(x, a) :  a  ∈  A}  = 0.  A  is compact, we know that  d(  d (x , A )  is a continuous function of  x Since  A  is compact, by the Extreme of  x  . Since A Part (b)  If   If  A  A  such that d  d (a, {x}) for all a  X .. Since d Value Theorem there is some c some  c  ∈  A such that d((c, {x})  ≤  d( all  a  ∈  X  Since d((a, {x}) = inf {d(a, x)}  = d  =  d((a, x),  d (c, x)  is the greatest lower bound of   { it follows that d that  d((c, {x}) =  d( { d(a, x) :  a  ∈  A }. Therefore d Therefore  d((x, A) = d =  d((c, x).  A  in  X  corresponds  X  corresponds to all points in  X   X  that Part (c)  Note that the  the  -neigh -neighborhood borhood of  A  that are within a distance  distance    of some point in A in  A.. It obviously includes all of  A of  A.. Let    >   0   be give Let given. n. If   a0   ∈   A   and and   x   ∈   Bd (a0 , ), then  then   d(x, a)   < ; hence  hence   d(x, A)   < . Theref Therefor oree   x   ∈   U  U ((A, ), so U ((A, ). On the the other other hand, hand, if   x   ∈   U ( U (A, ), then  then   inf {d(x, a) :   a   ∈   A}   < . Theref Therefor oree there is some a∈A Bd (a, )   ⊆   U  a   ∈   A  such that  U ((A, )  ⊆ a∈A Bd (a, ), and the that   d(x, a)   < , from which it follows that   x  ∈  B d (a, ). Consequent Consequently ly,,   U  two sets are equal. U  contains the union of the sets in the collection   {Bd (a, a )}a∈A . Part (d)   Suppose   A   is compa compact. ct. The ope open n set  set   U  contains Obviously then, it also contains union of the collection   {Bd (a, a /2)}a∈A , which covers  covers   A  with sets open in  in   X . Sinc Sincee   A N   A.    a }  (which must be greater   that covers covers A . Let Let   = inf  { is compact, there is a finite subcollection   {Bd (aj , aj /2)}N  j j =1 j =1 2) for  for some  some   aj . then   a  is contained in  in   Bd (aj , aj /2) than zero since the collection collection is finite). It follows that if   x  ∈  B d (a, ), then  We then have: d(x, aj )  ≤  d(  d (x, a) + d(a, aj )  < aj /2 + aj /2 =  aj .

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But as indicated above, B above,  B d (aj , aj )  is contained in U  in  U .. Therefore U  Therefore  U    contains U  contains  U ((A, ). /x  : x  x >  0} Part (e)  We need a closed closed set that has no finite subcovering. subcovering. In the metric space R>0 × R>0 , the set  { x × 1/x : is closed and not compact (since it is unbounded in the Euclidean and square metrics), and no open set contains an -neighborhood of it.

Problem 27.3. Solution Solu tion:: Pa Part rt (a)   Let   A   = [0, [0, 1]\K   ∪ {(1/ (1/(n  + 1), 1), 1/(n  −  1)) :   n   ∈   N} ∪ {1}, which is an open covering of   of   [0 [0,, 1] N    sets. It follows as a subspace of   RK . Le Lett   A be a finite subcollection of   A  that contains  contains   N  follows that there there are at most most   N  /n  in the union of the sets in   A ; let  distinct values of   1/n in let   1/M   be the greatest such such value. It follows follows that if   j > M , then  / A ∈A  A . Thus there is no finite subcollection of   A A  that covers [0 covers  [0,, 1] 1],, so so [0  [0,, 1] 1] as  as a subspace of   RK   is compact. 1/j ∈ Part (b)  Let  A  = (−∞, 0) a 0)  and nd (0  (0,, ∞)  be subspaces of   RK . The set ( set  (a, a, b)  is open in ( in  ( −∞, 0) 0) where  where a  a  ≤  0  0;; there is, however, no open set in ( in  ( ∞, 0) excluding  K  because every element of  K  is  K  is greater than 0 than  0.. Thus ( Thus  ( −∞, 0) 0)  excluding any elements of  K  because has the same topology as a subspace of  R K  as it does as a subspace of  R  R ’s standard topology. By Corollary 24.2, ( 24.2,  ( −∞, 0) is connected. a, b)  is open in the subspace  B   = (0, The set ( set  (a, (0, ∞)   of   RK   where a where  a  ≥  0  0.. In additio addition, n,   (c, d)\K   ∩ ∩ (0 (0,, ∞)  is open in   RK   c  ≥  0. c, d)\K  ∩ /n for when c when  0 . No set ( set  (c,  ∩ (0, (0, ∞)  can, however, be included in a separation of  (0 of  (0,, ∞)  because it excludes all  1  1/n  for n  ∈   N. Consequent /n  and therefore contains neighborhoods of   Consequently ly,, the other set in the separation separation must contain contain all   1/n and of   1/n that intersect  intersect   (c, d)\K   ∩ ∩ (0, (0, ∞). There Therefore fore any separation separation of   (0, (0, ∞)  as a subspace of   RK  must be of sets also open in (0, (0, ∞)  as a subspace of   R. Since Since   (0, (0, ∞)  as a subspace of   R  is connected, it follows that this interval also connected as subspace of   RK . The union of the closures A closures  A = point  0  in  = (−∞, 0) = (−∞, 0] and 0]  and  B  = (0, (0, ∞) = [0, [0, ∞)  equals  R K , and both have the point 0 common. Because A Because  A  and  B  are connected by Theorem 23.4,   RK  is connected by Theorem 23.4. Part (c)   If   RK  is path-connected, there is a path from   0   to to   1. Let  Let   f  f    : [a, b]  →   RK  be such a path from the interval −1  is a closed subset of   [a, Because  [a, a, b]  is compact, U  compact,  U    is Since  f    is continuous by definition, f  definition,  f  ([0, ([0, 1]) = U  =  U  is [ a, b]. Because [ [a, b]  of  R . Since f  U ) = [0, compact by Theorem 26.2, so f  so  f ((U ) [0, 1] is 1]  is compact in   RK , a contradiction. Therefore   RK   is not path-connected.

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Problem 27.4. Solution:   My solution solution is a bit long and relies of two two pr proofs oofs by contradi contradicti ction, on, but here here it goes. Choose Choose an arbit arbitra rary ry x0   ∈  X   X .. Defin  d (x0 , y). Sinc X  by Definee   f x   :  X   →   R  where  where   f x (y) =  d( Sincee there is more than one point point in   X   by hypothesis,  hypothesis,   f x (X ) /   f x (X )   but  contai con tains ns zero and some point greate greaterr than than zer zero. o. Assume Assume that is an  an   r   ∈   R  such that   r ∈ but   r  is not an upper 0

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X   a greater distance from   x0   than  bound of   f x (X )  (which means there is some point in   X   than   r ). Fo Forr each each   y   ∈   X , define  r ), and let   C  be the collection of   C y   =   Bd (y, f x (y )  −  r) Given n   C y   ∈ C , we will show of   C y   for all   y   where   f x (y )   > r. Give Suppose the opposite opposite were were true true and there there were a   z   in both sets. Consequently Consequently,, that   C y   does not intersect   Bd (x0 , r). Suppose f x (y ) = d =  d((x0 , y )  ≤  d  d((x0 , z ) + d(z, y )  < r + d(x0 , y ) − r  = d  =  d((x0 , y ), which is impossible. Therefore each C  each  C y   and B and  B d (x, r )   are disjoint. In addition, every point in z in  z ∈  X  is  X  is either in some C  some  C y   or or B  B d (x0 , r). Obviously d Obviously  d((x0 , z )    r.. If   d( d (x0 , z  )  < r, =  r then z then  z  ∈  Bd (x0 ,); otherwise, z otherwise,  z  ∈  C z . Therefore the union of sets in   C   and   Bd (x0 , r)   are a separation of   X , a contra contradic dictio tion. n. We infe inferr that there there can be r∈ /  f x X  unless X  unless it is an upper bound of  f   m  ∈  f x (X ), then the closed interval [0 of  f x (X ). As a result, if  m interval  [0,, m]  is contained  f x (X ). Let A  X  :  f x (y )  ≤  m}. in in f  Let  A =  =  {y  ∈  X  : 0

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, mthe  X  is  X . :  L  →  [0  [0, ]  as interval the restriction suppose in which case there isx a ◦surjection  : , m→ .  g   :   N   → g  [0   L. Sinc f x countable, [0, ]   is X  surjective, implying that   [0 [0,, m] is surjective, it follows that   f  of   of   f Now tosuppose X  Since e   is x   to  X    is is counta countable ble.. But  But   [0 [0,, m]   is uncountable, so  so   L   m must ust be uncountable. uncountable. Because Because   L   is a subset of   of   X , it follows that   X  uncountable. 0

Define  f x0 Define f 

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Section 28 Problem 28.1.  j ) :  j  ∈ N}  where  s ( j)  j )  is an omega-tuple whose elements are: Solution:   Designate S  Designate  S  =  =  {  {ss( j)  j ))k  = (s( j))

   1   for   j = k,  =  k, 0   for   j   =  k.

S  is S  is In other words,  words,   S   is a set of omega tuples that shift the value   1  to the right right among among a sea of zer zeros. os.... ... Obviou Obviously sly   S   is an x  ∈  S , then B then  B ρ¯(x, 1/2) that  S  has  has no limit points. Given  x  x ∈  ∈  [0  intersects S   S  infinite subset of   [0,  [0,, 1]ω , if   x ∈ 2) intersects [0 , 1]ω . We will show that S  only at itself. On the other hand, if  x  ∈ /  S , there is some neighborhood of   xx  that does not intersect S  intersect  S .. Therefore S  Therefore  S  has  has no limit points.

Problem 28.2. Solution:   Let S  Let  S  = /n : n  n  ∈ N}  and  x  ∈  [0,  x  = 1, then the set [1] set  [1],, which which is open in the [0 the  [0,, 1]  subspace of   R R ,  =  { 1 − 1/n :  [0 , 1]. 1]. If  x = 1] subspace is a neighborhood of  x that  x  that does not intersect intersect  S . If  x  x   an  m  ∈ N such that 1 that  1 − 1/m ≤  x <  1 − 1/(m + 1) = 1, then there is an m 1)..  x  that does not intersect S  Therefore [1 Therefore  [1 − 1/m, 1 − 1/(m + 1)) is 1))  is a neighborhood of  x intersect  S .. Accordingly, Accordingly,  [0  [0,, 1] 1] as  as a subspace of  R  is not limit point compact. Problem 28.7. Solution: Part (a)   Suppose f  Suppose  f  is  is a contraction of the compact metric space  space   X . We can easily easily establish establish that there there is at  X    where f  f (b) =  b  b.. We have 0  d((f  f ((a), f  f ((b)) =  d  d((a, b)  ≤  αd  αd((a, b). most one fixed point of  f  of  f .. Let  Let   a, b  ∈  X  where  f ((a) =  a  and  a nd   f ( have  0  ≤  d  α  ∈  (0  b , so any fixed point of   f  f  is Because α Because  (0,, 1) 1),, we infer that   d(a, b) = 0. There Therefore fore   a  =  b,  is unique. unique. No Now w we need to see if  there is a fixed point! f    is continuous:  X   and  d((x, y)  < , then d f ((x), f  f ((y))  ≤  αd  αd((x, y )  <   since The function  function   f  continuous: Given any  any   >  0,  0 , if   x, y   ∈  X   and d then  d((f  1 n n−1 α  ∈  (0 sequence  (f  f n )  where  where f   f  =  f    f   and and f   f  =  f   ◦ ◦ f  for for n  n >  1 Since  X  is  is compact, f  compact,  f n (X )  is compact  (0,, 1) 1).. Define the sequence (  1.. Since X  and, by Theorem 26.3, closed. We will show that   (f n (X )) is nested d sequence of non-empty non-empty sets. As a general general matter for any function function   g , if   C 0   of  ))  is a neste a given  given   C , then  then   g(C 0 )   ⊆   f ( f (C ). If   If   b0   ∈   g (C 0 ), there is some  some   c0   ∈   C 0   where  where   g (c0 ) =   b0 . Sinc Sincee   c0   ∈   C , it follows that b0  ∈  g(  g (C ). On the hand, if  b  b 1  ∈ /  g  g((C ), there is no c  A  where g  g((c1 ) =  b 1 , so there is also no such element in  C 0 . Hence no  c 1  ∈  A where b1  ∈ /  g(  g (C 0 ). Further, er,   f 2 (X ) = We’ll We ’ll now now apply apply thi thiss princ principle iple in the inductio induction. n. Obvious Obviously ly   f 1 (X )  is a non-empty subset of   X . Furth  f  is f   ◦ f 1 (X ). Since   under f   is a non-empty subset of   f 1 (X )  under Since   f 1 (X )  is a non-empty subset of   X , it follows that the image of   the Now w suppose suppose the proposit proposition ion holds for some  some   n   ∈   N. For   n  + 1, we have   f n+1 (X ) =   f   ◦  f n (X ). By the of   of   f 1 (X ). No n n+1 n−1 n X  X   f  f  X    f  X    f    f  ( ) =   ◦ ( )  is a non-empty subset of  ( ). Th Thus us  ( )  is a non-empty subset of   of  inductive hypothesis,  hypothesis, f n (X ) =  f   ◦ ◦ f n−1 (X ). Thus the proposition holds for all  n  ∈ N.

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)  is a nested sequence of non-empty, closed sets in compact space   X . Define Define   An   =   f n (X )   and In summary,  summary,   (f n (X ))) is A0  = X   =  X .. By Theorem 26.9, A 26.9,  A  = n∈N  is nonempty and contains some element a element  a..  f ((A)  ⊆  A  and  A  contains only a  a.. Given z  f ((A), there is some t We will show that  f  only  a,, establishing that f  that  f ((a) =  a Given  z  ∈  f  some  t  ∈  A such that f  that  f ((t) =  z  z.. For each n each  n  ∈ N, it follows that t that  t  ∈  A n , so so z  z  ∈  f n+1 (X ) =  f  ◦  ◦ f n (X ) =  f   f ((An ) =  A n+1 . As a result, z  ∈  A,  A , and f  and  f ((A)  is a subset of  A.  A . All that is left is to show that  A consists  A  consists of a single element. Since each A each  A n  is compact, d(An , An )  ⊆ R  is compact and, by Theorem 27.3, bounded by some  M   ∈ R. We will show by induction that the diameter of   An  bounded by  by   αn M . If   x, y   ∈  X ,  X , then  then   d(f ( f (x), f ( f (y ))  ≤  αd  αd((x, y)  < α1 M . M . Assuming Assuming the proposition proposition holds for some n−1   n−1 n+1 n+1    < α(αn d(x, y)) =  α n+1 d(x , y  ).  αd((f  (x ), f  (y ))  X  we have d n  ∈ N, then for some x )) < have  d((f  (x), f  (y )) ≤  αd some  x , y ∈  X  we Hence the proposition holds for all n all  n  ∈ N.



n n , it must be that 0  P  be  A that P   P <  diam An  = α  α M   N   N .. Let P  theisdiameter of  A. . Since P  Since  is a subset of each  A that  0  ≤empty,  =that A   for all  n  ∈only all n  PLet >  0,  P   P  is An . Therefore P   P    A for If  P  0 ,  be there some  n  such some n that ≥  diam Therefore   = 0. 0 . Since A Since  A  is not it follows that  contains a. Since f   A,, we conclude that f   a. Thus a f . Since  f ((A)  ⊆  A that  f ((a) =  a. Thus  a  is the unique fixed point of   f . Part (b)  Most of the argument here is identical to the argument in part (a). The only differences are where the metric of an image are concerned. We’ll address only those parts. First, there is still at most one fixed point of  f .  f . Let a, Let a, b  ∈  X  be   be two such fixed points. It follows that 0 that  0  ≤  d  d((f  f ((a), f (b)) = d(a, b)  < d(a, b), which is only possible if  a  a  = b  =  b.. Second,   f  f    is conti continuo nuous. us. Given Given    >   0, if   if   x   =    y   and   d(x, y )   < , then   d(f  f ((x), f  f ((y))   < d(x, y) =   ; otherwise, otherwise, d(f (x), f  f ((y)) = 0 < 0  < .  f ((A)  (which, combined with the Finally, we need to show that  A  has a single element a element  a.. First we will show that A that  A  ⊆  f  f ((A)  is a subset of   A, establishes that   f ( f (A) =  A  A). result in part (a) that  that   f  ). Let Let   x  be an element of   of   A, in which case for n  x . Clearly each   n  ∈   N  there is some element   xn   ∈  A n  such that  each Clearly the then, n,   (xn )  is sequence in   X . Theor that   f  (xn ) =  x. Theorem em 28.2 xnk )  that converges to some point c xn )  has a consequence subsequence ( point  c.. subsequence  (x holds that X  that  X  must  must be sequentially compact, compact, so ( so  (x  m  each point x An )  is a nested sequence of sets, every for  n  ≥  m each in  A n . By the definition point  x n  is contained in A Observe that since ( since  (A of convergence, convergence, any neighborhood neighborhood of  c intersects  c  intersects an infinite number of points in ( in  (x xn )  (and therefore ( therefore  (x xnk )). Consequently,

Since  A n each  A n . Since A every neighborhood neighborhood of   cc  intersects each A each  A n  at some point other than itself; hence  c  is a limit point of each A is closed,  closed,   c  ∈  A n , so  so   c  ∈  A.  A . By the definition of continuit continuityy, since  since  (xnk )  converges to c to  c  and  a nd   (f (xnk ))  converges to  to   x, we )) converges conclude that f  that  f ((c) =  x  x.. Thus x Thus  x  is contained in f  in  f ((A), so so A  A =  f ((A).  = f  Now we will show that diam A diam  A   = 0. Beca Becaus usee   A  is compact,   A  ×  A  is compac compact. t. By the Extreme Extreme Valu Valuee The Theor orem, em, d(A × A)  has an absolute maximum, so there is some pair of points x,  A such points  x, y  ∈  A  such that d that  d((x, y ) = diam =  diam A, which constitutes an upper bound on the distance between any two points in   A. Assu Assume me   x   and   y  are distinct distinct.. There There are likew likewise ise some  u, t  ∈  A where  A  where f   f ((u) =  x  and  f (  f (t) =  y.  y . As a result, d  d((f (u), f  f ((t))  < d(u, t)  ≤  d  d((x, y ), which impossible. distinct u, distinct result,  d((x, y) =  d )) <  A  has a single point  a.  a . Since f   A, we conclude that f   a,, so  a  is  f   f ’s Thus A Thus Since  f ((A) =  A, that  f ((a) =  a so a ’s unique fixed point.  and   x   ∈   [0, [0, 1], 1], the func functio tion n is inc increa reasin singg ove overr its entire entire domain domain.. Since Since   f (0) ( 0) = 0   and Part (c)  Since  Since   f  (x) = 1 − x   and f  f (1) that  f  is  is a map from [0 from  [0,, 1] to (1) = 1/ 1/2, we conclude that f  1]  to  [0,  [0 , 1]. 1]. Using the standard metric, given x, given  x, y  ∈  [0,  [0 , 1]: 1]: 2

2

   x  y     1   1 d(f (x), f  f ((y)) = x −   − y + = (x − y ) − (x − y ) = | x − y | 1 − (x + y ) 2 2 2 2       d (x, y ) 1 −  1 (x + y ) . =  d( 2

2

2

 x    y , then x    0  is contained in U  ω  C  is because C  because  is a compact metric subspace, and thus  B  is compact. But for the same reasons that  [0  [0,, 1] is not compact, B compact,  B is not compact (see exercise 28.1), a contradiction. Therefore [0 Therefore  [0,, 1]ω is not locally compact.

Problem 29.5. f    :   X 1   →   X 2   is a homeomorphism. Solution:   Suppose  Suppose   X 1 , X 2  are locally compact Hausdorff spaces and   f  homeomorphism. By Theorem Theorem  Y 2  that are the one-point compactifications of  X  29.1, there exist compact Hausdorff spaces Y  spaces  Y 1  and  a nd Y  of  X 1   and X  and  X 2 , respectively.  g  : : Y   Y 1  →  Y 2  as follows: Define g Define   f ( f (y )   for   y  ∈  X 1 , g (y ) = y2   for   y  = y  =  y1 ,



where  y 1  is the single point in  Y 1 \X 1   and where y and y  y 2  is the single point in  Y 2 \X 2 . Obviously g Obviously  g  is injective and surjective. We will −1 show that g that  g   and g and  g are continuous.  U  is  is open in Y  in  Y 1 , then g then  g (U  U ))  is open in Y  in  Y 2 . If   U  is U  is contained To show g show  g −1 is continuous, we will show that if some set  U  U ) =   f  f ((U ) U ), which is open in   X 2 . We kno because   Y 2 \X 2   =   {y2 }, which must be know   X 2   is open in   Y 2   because  in   X 1 , then   g (U ) U  is therefore open in  Y 2 . On the other U   contains  closed since  closed since  Y 2   is Hausdorff. Hausdorff. By Lemma 16.2,  16.2,  U  is other hand, hand, if   U   contains   y1 , designate C   =  Y 1 \U , U , which must be closed and contained in   X 1 . Since   f ((C )  is compact Since   Y 1  is compact, C  compact,  C  is  is compact, so g so  g (C ) =  f   g (Y 1 )\g(C ) =  G(  G (Y 1 \C ) =  G  G((U  U ))  is open (we used exercise 3.2(h) and and thus closed in  in   Y 2 . It follow followss that  that   Y 2 \g(C ) =  g(

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continuous. An analogous argument argument shows that  that   g   is continuous. continuous. We conclude conclude the fact that  that   g   is injective) injective).. Thus Thus   g−1 is continuous.  g  is a homeomorphism. that g that

Problem 29.7. The set set   R   is Hausdor Hausdorff. ff. Given  Given   a, b   ∈   R Solution:   Let  Let   R   =  S Ω  ∪ {∞}  where   ∞  is some element not contained in   S ω . The  b    b  is the immediate successor of  a,  a , then ( a, ∞)  are disjoint neighborhoods of   a a where  a < b, if  b where a =  ∞, then if  b then  ( −∞, b)  and  (  (a,  b,, respectively.  a < z < b, b, so z, ∞)  are disjoint neighborhoods of   a a  and and b and respectively. Otherwise, Otherwise, there is a  z  where  where a so   (−∞, z )  and  (  (z, , respectively. On the other hand, if   and , there is some countable subset   of   that contains . Since b a least element   b  =  ∞  a and < b A 0  (being countable) has an upper  A 0 bound c  R  c 0  by Theorem  a 10.3, R has element   a0  (because it is  b = well-ordered) bound it follows that [ that  [a a0 , c0 ]  is closed and contains b contains  b;; by Theorem 27.1, [ 27.1,  [a a0 , c0 ]  closed. The set R set  R \[a0 , c0 ]  is a disjoint neighborhood of  ∞  ∞ . Hence R Hence  R  is Hausdorff.  R . This collection must have an open set C  It also follows follows that R that  R is  is compact. Let  C  be an open covering of  R. set  C 0  containing    compact. Since Since   C  is a covering of   C  , there is a finite subcollection of   C   that where   C  is compact. ∞  and be of the form   R\C  where    C  . This subcollection and R of  R.. and  R \C  are a finite open covering of  R covers C  covers  C  is compact.  R . Any neighborhood of  ∞ where C   ∞  is of the form R form  R \C  where Finally,   S Ω  is a subspace whose closure equals  R.  R \C  must intersect S  S Ω , so  S Ω  is not compact (see example 28.2),  C  is a proper subset of   S  intersect  S Ω . so R Since S  Since  R  S  equals  R,, and R and  R \S Ω  equals To sum up, R up,  is a compact Hausdorff space of which S  which Ω  is a proper subset whose closure equals R the single point  {∞} . By definition, R definition,  R  is the one-point compactification of   S  S Ω . It follows from Theorem 29.1 that   S Ω   is locally compac compact. t. We can also show show that   S Ω  in the order topology is the one-point compactification of   S Ω . Obviou Obviously sly they differ by the single single point Ω point  Ω,, and the closure of   S  S Ω   equals  equals   S Ω . Fo Forr the same reasons as above, any two distinct distinct points in S  in  S Ω  have disjoint neighborhoods (where ( (where  (d, d, Ω]  for d  d <  Ω  is a neighborhood neighborhood Ω] for  Ω is S Ω  must include a neighborhood of  Ω of  Ω).  Ω ). Finally, S  Finally,  S Ω  is compact. Any open covering  C   of   S  of  Ω  that includes some element  c,, so there is a finite a0 , c0 ]  containing S Ω ). As shown above, there is a compact space [ c  of   S  S Ω  (since  containing c space  [a  (since Ω  Ω  is a limit point of   S  subcollection of   C  the covers  covers   [a0 , c0 ]; therefore this finite subcollection along with the neighborhood of   Ω  is a finite open S Ω . S Ω . We conclude that S  that  S Ω  is a one-point compactification of   S  covering of   S   R and  S  It follows from Theorem 29.1 that since R since  and S Ω both meet the requirements (1) through (3), there is a homeomorphism between them.

Problem 29.8. /n  : n  n  ∈ N}  (just define Solution:  As subspaces of   R R  in the standard topology, obviously   N  is homeomorphic with T  with  T    =  { 1/n : f  f    :  N  →  T   T    where  where   f  f ((n) = 1/n). /n). We will show that  that   S   =  { 0} ∪ T   T   as a subspace of   R  is the one-point compactification of  T  T .. Clear Clearly ly 0  T ; conversely, any point other than   0  that is not in   T   T   is not a limit point. point. Thu Thuss   T   T   =  S   S ..  0  is a limit point of  T ; Moreover,   S  Moreover, S    is Hausdo Hausdorff. rff. Given  Given   x, y   ∈   S   S   where  where   0  < x < y, y , it follows that   {x}   and   {y }  are disjoint neighborhoods of  those points. If  x  x  = 0, then y then  y  = 1/n0  for some n some  n 0  ∈ N. Thus U  Thus  U n   =  {0} ∪ {1/n : /n : n  n > n0 }  is a neighborhood of   00  that is  S  is S  must contain a neighborhood of  0 disjoint from  { y}. Finally, Finally,  S   is compact. Any open covering covering  C   of   S  must of  0  that contains U  contains  U n 0

1

in  U n1 .

Thus Thus there are a finite number number of ele elemen ments ts in  in   S  that   that are not contained Accordingly Accordingly,, there is a for S . finite subcollection subcollection of   C  C  that covers S  covers  S .. We conclude that T  that  T  is  is the one-point compactification of   S . T   are locally compact. Applying As subspaces of   R, it follows from example 29.1 that   N   and and   T   Applying exercise 29.5, we infer that the one-point one-point compactificat compactification ion of   N  is homeom homeomorph orphic ic with T  with T .. some  n 1   ∈  N . some n

Problem 29.10. Solution:   Because  Because   X  is X  is locally compact at point   x, there is a compact space   C  C  that  that contains a neighborhood   V  V    of   x. The set W  set  W    =  U  ∩ V  V  is  is open and non-empty because U  because  U    and and V   V  are  are each neighborhoods of   x. x . Further, the closure of  W    W   is contained conta ined in C  in  C .. Every limit point of  W  is  W  is also a limit point of  C  of  C ;; since  since   C   is closed, it contains all of its limit points. By Theorem 26.2, W  26.2,  W    is compact.  W  is  D   by Now we need to show that  W   is contained in U  in  U .. The set D set  D  = U   =  U  ∩ C  is   is contained in C  in  C .. Let  C  be a covering of  D W  is U ,, then c /  U  ∩ V  V    because sets open in X  in  X .. The set X  set  X \W   is open and contains points in C  in  C  that  that are not in D in  D   (if   cc  ∈  C \U  then  c ∈ W ))  is a covering of  C  by  C  by sets open in X  V   ⊆  C ).  C ). The collec in  X ;; a finite subcollection covers C  covers  C ,, so that collection tion A∈C A ∪ (X \W  W    covers D same finite subcollection subcollection without without  X \W  covers  D.. It follows that  D  is compact, and by Theorem 26.3  D  is closed.  D . Thus any limit point of  W  is  W  is a limit point of  D,  D , and since D Observe that W  that  W  is  is a subset of  D. since  D  is closed, W  closed,  W    must be  D.   D  W  is  U .  W     x  W  is a subset of  . Therefore Therefore W  is contained in U  in . We conc conclude lude that there there is a neighborhood neighborhood W   of   where  where W   is contained in in U   U ..

 

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Section Secti on 29* (Sup (Supplem plementa entary ry Exerci Exercises: ses: Nets) Problem S29.1*. Solution: Part Part (a)  Given  a, b  ∈  A  where  a  ≤  b,  b , it follows that a  b and  b,, so  a    b  b and  b..  Given a,  where a that  a  ≤  b  and  b  ≤  b so a  and  b    b  A and  B in  C   A B Part (b)  Given  Given A  and B  in the collection, it follows that both are subsets of  C  of    = ∪  (which must be in the collection S ), so A  C   and  B    C .  C . as a subset of   S ), so  A    C   and B  A, B   ∈ A, it follows that C   C  is  A    C  Part (c)  Given  Given A, that  C    =  A ∩ B  is in  A   and and C   is contained in both A both  A  and   B . Therefore Therefore A  B    C . and B and  C    and B  C .. Part (d)  Given closed sets A sets  A  and  B  of space X  space  X ,, the set C  set  C  =  =  A ∪ B  is closed, so A so  A    C  and  B    C  Problem S29.2*. Solution:   Suppose J  K  is  K , then   J . Accordingly Suppose  J  is  is a directed set and subset   K   is cofinal in   J . If   α, β   ∈  K , then   α, β   ∈  J . Accordingly,, there is a γ   ∈  J   J  such  γ    and   γ . Since  K  such that   such that   α    γ  and   β     γ . Since there there is also a  a   δ   ∈  K  such that   γ      δ , the transitivity of the partial ordering  δ    and  β    δ . Hence K  relation ensures that α that  α    δ  and β  Hence  K  is  is a directed set. Problem S29.3*. Solution:   Let ( Let  (x xα )α∈N  be a net that converges converges to c to  c  ∈  X .  X . By definition, given a neighborhood U  neighborhood  U    of   c there c  there is some α some  αU   ∈ N where, for any β  any  β  ∈  N , if  α  α U     β , then x then  x β   ∈  U   U .. From the result in exercise S29.1*(a), the simple-ordering of   N  by  ≤  i iss a directed set. Therefore if   β  β  ≥  α U , then x then  x β   ∈  U   U ,, which meets the definition of convergence of a sequence to c to  c.. Problem S29.4*. xα )α∈J   and  Y .. The set  U  must Solution:  Let nets ( nets  (x and   (yβ )β∈J   be given. given. Let U  Let U    be a neighborhood  neighborhood   x × y   in X  in  X  × Y  set   U   must contain  B y  are basis elements of   X   X   and Y   J  where if  the basis element B element  B x  × By   where B where  B x   and and B and  Y ,, respectively respectively. There is some ζ  some  ζ   ∈  J  where ζ     α then  α  then x  x α  ∈  B 0 , and there is some ξ   J  where if  ξ   ξ    β , then y  J  where ζ   ζ    µ  µ and  ξ    µ  µ.. some  ξ  ∈  J  where then  y β   ∈  B1 . There is some µ some  µ  ∈  J  where  and ξ   U .. Thus ( xα  × yα )α∈J   converges to x  x γ  × yγ   ∈  Bx  × By  ⊆  U   y γ   ∈ B 1 , so  µ    γ , then x to  x × y . Thus  (x so x and y Therefore if  µ then  x γ   ∈  B0   and

Problem S29.6*. X  the converges to a point   x   ∈  X   X .. It follows follows that given given any Solution:  Suppose there is a net   (xα )α∈J  of the set   A  ⊆   X  the neighborhood  U   U    of  x,  x , there is an ξ  an  ξ U   ∈  J  such  J   such that if   ξ    ξ      α  then x  then  x  ∈  U   U . . Since Since x  x  ∈  A  for all all α  α  ∈  J ,  J  , it follows that  U  that U  U   U U   α α intersects A intersects  A.. Accordingly, x Accordingly,  x  ∈  A.  A. Conversely, suppose   x   ∈   A. Let   K  K  be  be a collection of all neighborhoods of   x   and     be a partial ordering based on reverse rever se inclusi inclusion. on. By exercise exercise 29.1*(c), K  29.1*(c), K  is  is a directed set. Define ( Define  (yyβ )β ∈K  to be a net where y where  y β   ∈  β  ∩ A  (which must be  V    of   x, x , if   V  V     β   β  then  β  ⊆  V   V  and non-empty) for all β  all  β  (using  (using the axiom of choice, of course). Given a neighborhood  V   then β   and thus yβ   ∈  V   V .. We conclude that ( that  (yyβ )β∈K  converges to x to  x..

Problem S29.7*.  U  be  X ,, let ( xα )α∈J  be a net converging to x to  x  and  U   be a neighborhood Solution:   Let f  Let  f  be  be a continuous function. Given x Given  x  ∈  X  let  (x U )  is open and contains x  J  such that if   µ µ    α  α,, then f (x)  in   Y  Y .. Since f  contains  x.. Thus there is some  some   µ  ∈  J  such of   f ( Since  f    is continuous, f  continuous,  f −1 (U ) −1  U ,, implying that ( f ((xα ))α∈J   converges to f   U .. Accordingly, if  µ  µ    α,  α , then f  U ), so f  xα  ∈  f  (U ) to  f ((x). that  (f  then  f ((xα )  ∈  U  so  f ((xα )  ∈  U  Conversely, suppose that for every for net ( net  (x xα )α∈J  that converges to x to  x,, the net ( net  (f  f (xα ))α∈J   converges to f  to  f ((x). If we U ))  is not open. We will show that paradoxically assume f  assume  f  is  is not continuous, there is some open set  U   U    in Y  in  Y    such that f  that  f −1 (U  U ))  that converges U ). By exercise S29.6*, there is a net  net   (xα )α∈J   in  in   X \f −1 (U  U ). Let x Let  x  ∈  X \f −1 (U ) U ) =  X \f −1 (U ) X \f −1 (U ) f ((xα ))α∈J    converges to   f ( f (x). If   f (x)   ∈   U  U ,, then   U  U    is a neighborhood of   f ((x), from which it to   x. By h hypot ypothes hesis, is,   (f  If   f ( of   f   J  such that if  ξ   ξ    α  then f ((xα ))  ∈  U   U .. But this contradicts that every  x α  is contained in follows that there is some  ξ  ∈  J  such  then (  (f  U ))  is equal to its closure, so it must be closed. It U ). Thus X  U ). Consequently, f  /  U   U ,, so  x  ∈  X \f −1 (U ) X \f −1 (U ) Thus  X \f −1 (U  Consequently,  f ((x)  ∈ so x −1 U )  is open, a contradiction. We conclude that  f    f   is continuous. follows that f  that  f  (U )

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Problem S29.9*. xα )α∈J  (given by the function f   X )) with a subnet ( f ◦ g )β ∈K  (where  g : : K   K  →  J   J )) Solution:  Suppose there is a net ( net  (x function  f    :  K   → X  subnet  (f   (where g U   of   x0 , there is some   ξ   ∈   K  such K  such that if   ξ     β   β  then U .. Let that converges to  to   x0 . Giv Given en a neighborhood neighborhood   U    then   (f   ◦ g )β   ∈   U  Let  γ   ∈  J ,  J , by the definition of a directed set there is some δ   J  such that γ   δ    and g  J   :  x α  ∈  U }. If  γ  LU   =  { α  ∈  J   some  δ  ∈  J  such that  γ    δ  and  g((ξ )    δ . K  such  ξ ;; otherwise,   g (µ)     g(ξ ), Since   g (K )  is cofinal in  Since in   J , there is some  some   µ   ∈   K   such that   δ     g (µ). It cann cannot ot be tha thatt   µ    ξ   g))µ   ∈   U , U , so   g (µ)   ∈   LU . Th Thus us   LU    is cofinal in   J , from which it follows that   x0   is an a contradiction. contradiction. There Therefore fore   (f   ◦ g accumulation point. On the other hand, suppose ( suppose  (x xα )α∈J  has an accumulation point x point  x 0 . We must construct a subnet that converges to  x 0 . Define:  x α }. K   = { (α, U ) U ) :  α  ∈  J    J   and U  x 0  that contains contains x and  U  is  is a neighborhood of   x The set K  set  K  is  is directed. directed. If   (α, U )  and  (β, V  )  are elements of  K ,  K , the set U  set  U  ∩ V  V    is a neighborhood of   x x 0  contained in both ( α, U )  ( β, V ) U   U   and V  and  V .. Because L Because  L U ∩V    is co-final in J  in  J ,, there is some γ  some  γ   ∈  J  such  J  such that α that  α    γ   γ    and  and   β      γ   γ    and  and   xγ   ∈  U  ∩ V  V  (we  (we skipped U ))    ( γ, U   ∩ V ))  and  ( β, V  V ))    ( γ, U   ∩ V )). Thus K  over the step of finding an intermediate value). It follows that  (α,  ( α, U   (γ, ∩ V   (β,  (γ, ∩ V  Thus  K    is a directed set. J   where   g((α, U )) =   α. This func  g )β ∈K . Fi Now let   g   :   K   →   J   ((α, U )) functio tion n constr construct uctss the subnnet subnnet   (f   ◦  g) Firs rst, t, if  if   i      j , the g ((i, U )) =  i    j   =  g((  g (( j, V )) V )).. Sec  J    because  K  contains a tuple for each ((i, U )) Second ond,,   g (K )  is cofinal in  in   J . The image  image   g (K ) =  J  because   K  contains element of   J . Thu Thuss given given any any   α   ∈   J , then  then   α   ∈   g(K )  so trivially there is some   β   ∈   g(K )  where  where   α      β . No Now w all that’s that’s left to show is that this subnet converges to  to   x0 . Let  Let   W  W    be a neighborhood of   x x 0 . There There mus mustt be some ν  some  ν   ∈  J  such  J  such that xν   ∈  W   W    because L because  L W   is non-empty (otherwise, it couldn’t be cofinal in J  in  J ). ). There exists exists an element element ( ν, W  W ))  i in n  K    K   because  (ν,  W .. W  W  is  is a neighborhood of  x  x 0  containing  containing x  x ν . It follows that if  (ν, W ))    ( β, Z )  then  then Z   Z  ⊆  W   W ,, so so x  x β   ∈  W   W ;; thus ( thus  (f  f  ◦  ◦ g)β   ∈  W   ( ν, W   (β, to  x 0 . We conclude that the subnet  (f  ◦ g )β ∈K  converges to x  ( f   ◦