MAT480 Hw8 Solution

 

MAT480 Topology Homework 8

Due Apr. 10th in class

1. (a) What are the components components and path component componentss of   Rω (in product topology)? (b) Consider   Rω in the uniform uniform topology. topology. Show Show that  x   and   y  lie in the same component of   Rω if and only if the sequence  x−y   = (x1  − y1 , x2  − y2 , · · ·   , ) is bounded. (c) Give   Rω the box topology. topology. Show that  x   and   y  lie in the same component of   Rω if and only if the sequence  x − y  is “eventually zero”. ω topology) is connected. connected. So there is only one Proof.   (a) As shown in class that   R (in the product topology) component. Given any two points, a, y  ∈ Rω , write  x  = (xn ), y  = (yn ).  Then define  f n  : [0, 1] → R, t  → xn t +  yn (1 − t).  Obviously   f n  is continuous. continuous. So the function  f   : [0, 1]  →   Rω is continuous. continuous. Hence  Rω is path-connected. (b) and (c) Dropped. 2. Let   X   denote the rational points of the interval [0, 1] ×  0 of   R2 .   Let   T  denote the union of all line segments joining the point   p  = 0 × 1 to points of   X . (a) Show that   T  is path connected , but is locally connected only at the point  p . T , there exist Proof.  It’s easy to see that  T  is path connected since for any two given points  x 1 , x2   of   T  line segments   L1   and   L2  connecting   x1   and   p  as well as   x2   and   p   respectively respectively.. It’s locally connected only at   p   because for any point   x   =    p, there exists an open neighborhood   U   of   x   on   R2 , such that U   ∩ T   is a union of disconnect disconnected ed line segments segments.. If   x   =   p, from the simple geometry, it’s easy to see that that for a small neighborhood   V    of   p,   V    ∩ T  is path connected, hence is connected.

3. Let  p  :  X   →  Y  be a quotient map. Show that if   X  is   is locally connected, then   Y   is locally connected. Proof.   Let   p   :   X   →   Y   be the quotient map and let   C  be a component of an open set   U   of   Y  . We want to show that  C  is open in  Y  . It suffices to show that  p −1 (C ) is open in  X . We claim that  p −1 (C ) is a union of components of   p−1 (U )).. Equivalen Equivalently tly it suffices suffices to sho show w that each each component of   p−1 (U ) either lies in   p−1 (C ) or doesn’t intersect   p−1 (C )).. If not, not, i.e., i.e., there exists exists a compone component nt   D   of   p−1 (D) such that   D ∩ p−1 (C 1 )   =  φ  and   D ∩ p−1 (C 2 )   =  φ,  where  C 1   and   C 2  are two components of   U . Then  p(D) ∩ C 1   =  φ,   and   , p(D) ∩ C 2   =  φ.

This is impossible since   D   is connected so   p(D) is connected thus   p(D) is connected in only one component of   U . Hence we showed that  p −1 (C ) is a union of components of   p p (U ). Since  X  is locally −1 connected then each component of  p (U ) is open in  X , Hence   p (C ) is open, i.e.,   C  is   is open as  p  is a 1

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quotient map.   T 

  T 

  T 

  T 

4. (a) L Let et   and be two topologies on the set   X ; suppose that compactness ss of   X  ⊃ . What does compactne under one of these topologies imply about the compactness under the other? (b) Show that if   X  X  is compact Hausdorff under both   T  and   T , then either   T   and   T  are equal or they are not comparable.  compactt too. But the conve converse rse is not true true in genera general. l. Here Here is Proof.   (a) If   T  is compact then   T  is compac  an example: Let   X   = [0, 1] and   T  be the standard topology and   T  the discrete topology, then   T  ⊃  T  and   T   is compact compact.. But   T  is not compact since each single point set is open, then the open covering {x}, x  ∈  X  doesn’t have finite subcovering. (b) If not suppose we have T  ⊂  ⊂  T  and   T     =  T  . Then the inclusion   i  : (X, T  )  →  ( X, T ) is continuous. Obviously   i  is a bijection so from the condition that  X  is Hausdorff compact under both   T   and  T  , it follows that  i  is homeomorphic. A contradiction. 5. (a) Show that in the finite complement complement topology topology on   R, every subspace is compact. (b) If   R R  has the topology consisting of all sets  A  such that  R − A  is either countable or all of  R  R , is [0, 1] a compact subspace? Proof.   For any   A  ⊂  R , let   {U α }  be any open covering of   A. Take   U α  open, then   R − U α  is finite, so is   A − U α . Write   A − U α   =  { a1 , · · ·   , an }.  Then there exist   U α , · · ·   , U αn   such that   ai   ∈  U αi ∀i.  Thus  A  =  U α (∪n i=1 U αi ).  Hence  A  is compact. 1

(b) [0, 1] is NOT compact. compact. Since rational rational numbers numbers are coun countable table,, we may write the rational number number in [0, 1] as   a1 , · · ·   , an , · · · . Defin Definee   U n   = ([0, 1]  −  Q )  ∪ {an }   then   U n   is open and [0 , 1] =   ∪nn=1 U n . But there is no finite subcovering since   an   ∈   U n , an   ∈ =   n.  So if one   U n   is dropped, then    U m   if   m    an   ∈  U 1  ∪ U 2  ∪ · · · ∪ U n−1  ∪ U n+1  ∪ · · ·  .

6. Show that that a finite union union of compa compact ct subspaces subspaces of   X  is compact. n Proof.   Let  A 1 , · · ·   , An  be compact and  A  =  ∪ i=1 Ai . Let  { U α }  be an open covering of   A  then  ∪ U α   is Ai ∀i.  From the compactness of   A A i , one obtains that there exist  U αi , · · ·  , U αiki also an open covering of   A such that i Ai  ⊂ ∪kj =1 U αij , ∀i  = 1, · · ·   , n. 1

i Hence  A  =  ∪ni=1  ∪kj =1  U αij ,  i.e.,  A  is compact.

7. Let   A   and   B  be disjoint compact subspaces of the Hausdorff space   X . Show Show that there exist disjoin disjointt open sets  U   and   V    containing  A  and   B   respectively. Proof.   Let   A, B  be two compact subsets of Hausdorff space   X A ∩ B  =  φ.  Fix   a  ∈  A,  for any   b  ∈  B , there exist open subsets   V  a,b   and   U b  such that   a  ∈  V  a,b   and   b  ∈   U b   and   V  a,b  ∩ U b   =   φ.  Obviously we have B  =  ∪ b∈B U b . From the compactness of   B , it follows that there exist   b1 , · · ·   , bn  such that i=1 U bi . B  ⊂ ∪n

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Let  U a  =  ∪ni=1 U bi .  Then the set  V  a  =  ∩ni=1 V  a,b  is a neighborhood of  a  a  and  V  a ∩ U a  =  φ.  It’s easy to see that this process works for any   a. So for any   a  ∈  A,  we have   V  a   and   U a  such that   V  a ∩ U a   =  φ.  Note that   U a  changes as   a  varies.  varies. We see that A  ⊂ ∪a∈A V  a .

Now the compactness of   A A  implies that there exist  a 1 , · · ·   , am  such that  A  ⊂ ∪im=1 V  ai .  Let  V    = ∪ im  V  ai and   U   = ∩ m  V , B  ⊂  U. ∩ V    =  φ  and   A ⊂  V, i   U ai .  Then we have   U   ∩ 8. Sho Show w that if  f   f   :  X   →  Y   is continuous, where  X  is compact and   Y  is Hausdorff, then   f  is closed map. X . Since  X  is compact, then  C  is compact, thus  f (C ) is compact. Proof.   Let  C  be a closed subset of   X  By the condition that   Y  is Hausdorff, it follows that   f (C ) is closed. 9. Sho Show w that if   Y  Y   is compact , then the projection   π1  :  X  ×  × Y   →  X  is a closed map. Proof.   Let  C   ⊂  X  ×  × Y  be any closed subset. We want to show that  π 1 (C ) is closed. Equivalently we have to show that  X  − π1 (C ) is open. Since   C  is   is closed, so   X  ×   is open. We may write  × Y   − C  is X  ×  × Y   − C   = ∪ α U α  × V  α

where   U α   ⊂  X  is open and   V  α   ⊂   Y   is open. No Now w take any any point  x0   ∈   X  −  − π1 (C ) we want to find a neighborhood  W   of  x  x 0  such that  W   ⊂  X − π1 (C ) We first note that {x0 }× Y   =  π 1−1  ({x0 })  ⊂  X × Y  −  − C. By tube lemma , there exists   W  a neighborhood of   x0  such that

{x0 } × Y   ⊂  W   × Y   ⊂  X  ×  × Y   − C. Hence W   =  π 1 (W   × Y  )  ⊂  X  −  − π1 (C ).

Therefore  X  − π1 (C ) is open. 10. Let  f   :  X   →  Y  ; let  Y   be compact Hausdorff. Then   f  is continuous if and only if the  graph   of   f , Gf   =  { x × f (x)|x  ∈  X },

is closed in   X  ×  × Y. Proof.  Let’s first assume that  f   :  X   →  Y  is continuous. To show that the graph  G f  is closed in  X  × Y, is suffices to show that   X  × =  f (x0 ).  × Y   − Gf  is open. Take any point (x0 , y0 )  ∈  X  ×  × Y   − Gf , then  y 0   Since   Y  is Hausdorff, there exist   U, V  ,   y0   ∈  U ,   f (x0 )  ∈  V  , and   U   ∩ V    =  φ.  Then   f  (V  )  ⊂  X   is open and we claim that   f  (V  )  ×  U   ∩  G f    =   φ,   thus   f −1 (V  )  ×  U   ⊂   X   ×  Y   −  g f .  If not, say there exists (x, y )  ∈  f −1 (V  ) × U  ∩ Gf , then  y  =  f (x) and  f (x)  ∈  V    and  y  ∈  U . This is impossible since  U  ∩ V    =  φ. Now let’s assume that  g f  is closed in  X  × Y.  We want to show that   f   :  X   →  Y    is continuous. For any open subset V    of  Y   Y  , then Gf ∩X ×(Y  −V  ) is closed in X ×Y,  so by the last problem,  π 1 (Gf ∩X ×(Y  −V  )) )) is closed in   X . It’s easy to see that 1

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π1 (Gf   ∩ X  ×  × (Y   − V  )) )) =  X  −  − f −1 (V  ),

since  G f  ∩ X  × (Y   − V  ) =  {x × f (x)|f (x)  ∈  Y   − V  }.  Hence  X  − f −1 (V  ) is closed, i.e.,  f −1 (V  ) is open. 3

 

11. Let   p  :  X   →  Y   be a closed continuous surjective map such that   p−1 ({y }) is compact, for each   y   ∈  Y  . Show that if   Y  Y   is compact, then   X   is compact. −1 Proof.  We first claim that if   U  is an open set containing   p ({y }),  then there is a neighborhood   W  of   y  such that  p −1 (W )  ⊂  U.  To show the claim, we look at the closed subset  X  − U,  then  p (X  −  − U ) is closed, i.e.,  Y   − p(X − U ) is open. Then there is  W , a neighborhood of  y  y  such that  W   ⊂  Y   − p(X − U ), hence   p−1 (W )  ⊂  U. Now we show that   X  is   is compact. Let  X   =  ∪α U α  be an open covering. For any  y  ∈  Y  , there are some U αi ’s of   U  U α  such that ∪i U αi   ⊃  p −1 ({y }). Since  p −1 ({y }) is compact, so there exist a finite subcovering  ∪ in=1 U αi   ⊃  p −1 ({y}).  Let  U y  =  ∪ in=1 U αi . This process works for any point   y , so to each   y, we obtain an open subsetU y  containing   p−1 ({y }). Then by the claim, there exists  W y   ⊂  Y   open, such that U y   ⊃  p −1 (W y ).

Obviously we have  Y   =  ∪y∈Y  W y , so there exist  y 1 · · ·  , ym  such that  Y   = ∪ jm=1  =  W yj   as  Y   is compact. So m −1 X   =  f −1 (Y  ) =  f −1 (∪m (W yj )  ⊂ ∪jm=1 U yj . j =1 W yj ) =  ∪ j =1 f  Since each   U yj   is a union of finitely many   U α   thus thus we obtaine obtained d a finite finite subcove subcoverin ring. g. Hence Hence   X   is compact.

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