Exercise Problems∗
§13 Basis for a Topology
1.
Let X be X be a topological space; let A be a subset of X . Suppose that for each x ∈ A there is an open set U set U containing containing x such that U ⊂ A. A . Show that A that A is open.
sol) For each a ∈ A let U a be an open set with a ∈ U a ⊂ A. A . Then A =
U a , so A is open.
a∈A
3.
Show Show that the collection collection T c of all subsets U of X , X , where X where X − U either U either is countable or X , is a topology on the set X set X ..
sol) (i) X (i) X − X = ∅ is countable and X and X − ∅ = X = X ,, so X so X,, ∅ ∈ T c . (ii) Let U Let U α ∈ T c for α for α ∈ J . J . Then
− X −
U α =
α∈J
Thus
− U α ) = (X −
α∈J
− U α = X if X − = X ∀ α ∈ J othe otherw rwis isee
X coun counta tabl blee
U α ∈ T c . (iii) Let U i ∈ T c for 1 ≤ i ≤ n. n . Then
α∈J
X − −
i
This shows
U i =
U i ∈ T c .
i
(X − − U i ) =
X coun counta tabl blee
if X − = X for some i − U i = X othe otherw rwis isee
i
4.
Let { T α } be a family of topologies on X on X .. (a) Is
U α a topology on X on X ??
sol) No. T 1 ∪ T 2 is not a topology since { a, b} ∩ {b, c} = { b} ∈ / T 1 ∪ T 2 (See (c) below).
(b) Show (b) Show that there is a unique smallest topology on X X containing all T α, and a unique largest topology contained in all T α . ∗
These are based on student student solutions. solutions.
1
sol) The unique unique largest topology contained contained in all T α is topology on X on X ,, let T s be the topology generated by as · · · U β,n U = U β,1 where β, 1 β,n β
β
T α . Noting Noting T α is a subbasis for a T α . If U U ∈ T s then U then U can can be written U β,1 β∈ β, 1 , . . . , Uβ,n
T α .
Thus U Thus U ∈ T for for any topology T that contains T α , i.e. T s ⊂ T . T . Suppose T s is a smallest topology containing all T α . In general general we can’t say say either T s ⊂ T s or T s ⊂ T s . Howeve However, r, since T s contains T α , so T s ⊂ T s . Then Then T s ⊂ T s since T s is a smallest topology that contains all T α . This complete the proof. The unique largest topology contained in all T α is T l = T α . The proof of uniqueness is similar to the above.
(c) If X = {a,b,c}, let T 1 = {∅, X, {a}, {a, b}} and T 2 = {∅, X, {a}, {b, c}}. Find Find the the smallest topology containing T 1 and T 2 , and the largest topology contained in T 1 and T 2 . sol) By (b), T s = {∅, X, {a}, {b}, {a, b}, {b, c}} and T l = {∅ , X, {a}}. 5.
Show Show that that if A is a basis for a topology on X , then the topology generated by A equals the intersection of all topologies on X that contain A .
sol) Let {T α } be a family of all topologies on X X that contains the basis A and T A be the topology generated by the basis A. Note T α is a topology on X and T α ⊂ T A since A ⊂ T A . On the other hand, any U ∈ T A is a union of elements in A , so U so U ∈ T α for all α all α and thus T A ⊂ T α . This completes the proof.
§16 The Subspace Topology
1.
Show Show that that if Y if Y is a subspace of X , and A is a subset of Y , Y , then the topology A inherits as a subspace of Y is Y is the same as the topology it inherits as a subspace of X . X .
sol) Let T be be the topology of X and T Y denote the subspace topology of Y . Y . We also denote denote by T A/Y (resp. T A/X ) the subspace topology of A A as a subspace of Y Y (resp. X ). ). Then W ∈ T A/Y ⇐⇒ W = V ∩ A for some V some V ∈ T Y ⇐⇒ W = (U ∩ Y ) ∩ A for some U ∈ T ∩ Y )
⇐⇒ W = U ∩ A for some U some U ∈ T ⇐⇒ W ∈ T A/X . 4.
A map f map f : X → Y is said to be an open map if for every every open set U set U of X , X , the set f set f ((U ) U ) is × Y → X is open in Y in Y .. Show that π that π 1 : X × X is an open map.
sol) Let W × Y . Let W be be an open set in X in X × Y . Then W Then W = and Y and Y respectively. So, π1 (W ) W ) = π 1 ( This shows π1 is an open map.
(U α × V α ) where U where U α and V and V α are open in X in X
(U α × V α ) ) =
2
π1 (U α × V α )) =
U α
§17 Closed Sets and Limit Points
6.
Let A, B and C denote subsets of a space X . Prove the followins: (a) If A ⊂ B, then A ⊂ B.
sol) Recall that A is the intersection of all closed sets that contain A. Since B is closed and it contains A, we have A ⊂ B. (b) A ∪ B = A ∪ B. sol) A ∪ B ⊂ A ∪ B since A ∪ B is closed set containing A ∪ B. On the other hand, A ∪ B is a closed set that contains both A and B , so A ∪ B ⊃ A ∪ B.
Aα ⊃
(c)
Aα ; give an example where equality fails.
sol)
Aα is a closed set that contains all Aα , so it contains all Aα , i.e., Aα ⊃ Aα (Note that Aα is, in general, not closed !) Let An = { n1 } for n ∈ N. Then 0 ∈ An but 0∈ / An .
8.
Let A, B and A α denote subsets of a space X . Determine whether the following equations hold ; if an equality fails, determine whether one of the inclusions ⊃ or ⊂ holds. (a) A ∩ B ⊂ A ∩ B.
sol) A ∩ B is a closed set that contains A ∩ B, so A ∩ B ⊂ A ∩ B. Let A = (0, 1) and B = (1, 2). Then A ∩ B = ∅ , while A ∩ B = { 1}. (b)
Aα ⊂
Aα
sol) Similarly as above,
Aα is a closed set that contains
(c) A − B ⊃ (A − B).
Aα . So,
Aα ⊂
Aα .
sol) Let x ∈ (A − B) = A ∩ (X − B) ⊂ A ∩ (X − B). Then for any neighborhood U of x, U ∩ A = ∅ (since x ∈ A) and U ∩ (X − B) = ∅ (since x ∈ U ∩ (X − B)). Thus x ∈ A ∩ (X − B) = A − B. Let A = [0, 1] and B = (0, 1). Then A − B = {0, 1}, while A − B = ∅ . 9.
Let A ⊂ X and B ⊂ Y . Show that A × B = A × B in X × Y .
sol) A × B ⊂ A × B since the latter is a closed set (see prob 3) that contains A × B. Conversely, let (x, y) ∈ A × B. Then for any basis element U × V containing (x, y) we have (U × V ) ∩ (A × B) = (U ∩ A) × (V ∩ B) = ∅. Thus x ∈ A × B.
13. Show that X is Hausdorff iff the diagonal = { (x, x)|x ∈ X } is closed in X × X . 3
sol) Suppose X is Hausdorff and let (x, y) ∈ X × X − . Since x = y, there are disjoint neighborhoods U and V of x and y respectively. Then (x, y) ∈ U × V ⊂ X × X − . This implies X × X − is open, or is closed. Conversely, if X × X − is open then for any x = y there is a basis element U × V such that (x, y) ∈ U × V ⊂ X × X − . This implies U and V are disjoint neighborhoods of x and y respectively. 19. If A ⊂ X , we define the boundary of A by the equation Bd A = A ∩ X − A. (a) Show that IntA and BdA are disjoint, and A = IntA ∪ BdA. sol) Suppose there is x ∈ IntA ∩ BdA. Then IntA is a neighborhood of x and x ∈ X − A, so IntA ∩ (X − A) = ∅ which is impossible since IntA ⊂ A. On the other hand, note that A contains both IntA and BdA. Let x ∈ A − IntA. Suppose there is a neighborhood U of x such that U ⊂ A. Then x ∈ U ⊂ IntA, which contradicts to the assumption on x. Any neighborhood of x thus intersects X − A and hence x ∈ BdA. This shows A = IntA ∪ BdA.
(b) Show that BdA = ∅ ⇐⇒ A is both open and closed. sol) If BdA = ∅ then IntA ⊂ A ⊂ A = IntA where the last equality follows from (a). Conversely, if A is both open and closed, then IntA = A = A, so again by (a) BdA = ∅.
(c) Show that U is open ⇐⇒ BdU = U − U . sol) U is open ⇐⇒ U = IntU ⇐⇒ BdU = U − U where the last equivalence follows by (a).
(d) If U is open, is it true that U = Int(U )? Justify your answer. sol) Not true; let U = (−1, 0) ∪ (0, 1). Then Int(U ) = (−1, 1).
20. Find the boundary and the interior of each of the following subsets of R2 : (d) D = {(x, y)|x is rational}. sol) IntD = ∅ (since D contains no open sets in R2 ) and hence BdD = D = R2 .
§18 Continuous Functions
2.
Suppose that f : X → Y is continuous. If x is a limit point of the subset A of X , is it necessarily true that f (x) is a limit point of f (A)?
sol) Not true; constant functions. 3.
Let X and X denote a single set in two topologies T and T , respectively. let i : X → X be the identity function. 4
(a) Show that i is continuous ⇐⇒ T is finer than T . sol) i is continuous ⇐⇒ i −1 (U ) = U ∈ T for all U ∈ T ⇐⇒ T ⊂ T . 6.
Find a function f : R → R is continuous ar precisely one point.
sol) Define a function f : R → R by f (x) =
x −x
if x is rational, if x is irrational
Then, f is continuous at 0 since f (−/2, /2) ⊂ ( −, ) for any > 0. On the other hand, if x = 0 then the interval (f (x) − |f (x)|/2, f (x) + |f (x)|/2) does not contain f (x − δ, x + δ ) for any δ > 0. This shows f is not continuous at x = 0.
§20 and §22
3.
Let X be a metric space with metric d. Show that d : X × X → R is continuous.
sol) Triangle inequality shows if (x, y) ∈ Bd (x0 , /2) × Bd (y0 , /2) then
|d(x, y) − d(x0 , y0 )| < . This shows d is continuous at any (x0 , y0 ) ∈ X × X . 2.
(a) Let p : X → Y be a continuous map. Show that if there is a continuous map f : X → Y such that p ◦ f equals to the identity map of Y , then p is a quotient map.
−1 −1 −1 sol) Let idY be the identity map of Y . Then U = id−1 Y (U ) = f ( p (U ) ). Thus p (U ) is open in X =⇒ U is open Y . This implies p is a quotient map since p is continuous and surjective — subjectivity of p follows from p ◦ f = id Y .
4.
(a) Define an equivalence relation on the plane X = R2 as follows,: if x0 + y02 = x 1 + y12 .
(x0 , y0 ) ∼ (x1 , y1 )
Let X ∗ be the corresponding quotient space. It is homeomorphic to a familiar space. What is it? sol) Let p : X → X ∗ be the quotient map induced from the equivalence relation ∼ and define a function g : X → R by g(x, y) = x + y 2 . Then g is constant on the preimage p −1 ([(x0 , y0 )]) for each point [(x0 , y0 )] in X ∗ . Thus there is a continuous function f : X ∗ → R with g = f ◦ p. Since g is onto and g(x0 , y0 ) = g(x1 , y1 ) ⇐⇒ (x0 , y0 ) ∼ (x1 , y1 ), f is onto and one-to-one. In order to prove f is a homeomorphism, we will show that f is an open map (i.e. its inverse function is continuous). Let G : X → X defined by G(x, y) = (g(x, y), y). Then G is a homeomorphism with the inverse function G−1 (x, y) = (x − y 2 , y). Since
5
g = π 1 ◦ G, where π : X × X → X is the projection map onto the first factor, and both π 1 and G are open map, so is g . On the other hand, subjectivity of p and g = f ◦ p give g( p−1 (U )) = f ( p( p−1 (U ))) = f (U ). Consequently, U is open =⇒ p −1 (U ) is open =⇒ g ( p−1 (U )) = f (U ) is open.
§23 Connected Spaces
2.
Let { An } be a sequence of connected subspaces of X , such that An ∩ An+1 = ∅ for all n. Show that ∪ An is connected.
sol) We will use induction together with the following claim : Claim : Let A and B be connected subsets of X with A ∩ B = ∅. Then A ∪ B is connected. proof : Suppose A ∪ B is a disjoint union of open sets C and D of A ∪ B. If C = ∅, since C is both open and closed in A ∪ B, C ∩ A and C ∩ B are respectively nonempty both open and closed in A and in B (check using subspace topology!). Thus C ∩ A = A and C ∩ B = B , or equivalently, A ⊂ C and B ⊂ C . This implies D = ∅ , so there is no separation of A ∪ B, namely A ∪ B is connected. The claim shows A 1 ∪ A2 is connected. Suppose Bk = ∪ Ai is connected. Then Bk ∪ Ak+1 i≤k
is also connected by the claim. The induction principle thus completes the proof. 6.
Let A ⊂ X . Show that if C is a connected subspace of X that intersects both A and X − A, then C intersects Bd A.
sol) First, recall Bd A = A ∩ (X − A) and A = Int A Bd A. Suppose C ∩ Bd A = ∅. Then C ∩ A = C ∩ A
and
C − (C ∩ A) = C ∩ (X − A) = C ∩ (X − A).
C ∩ A is thus both open and closed in C . Since C is connected, either C ∩ A = ∅ or C ∩ A = C where the latter is equivalent to C ∩ (X − A) = ∅. This contradicts to the assumption C intersects both A and X − A. 11. Let p : X → Y be a quotient map. Show that if each p−1 ({y}) is connected, and if Y is connected, then X is connected. sol) Let A be both open and closed in X . Then for each y ∈ Y , p −1 ({y}) ∩ A is both open and closed in p−1 ({y}). Since p−1 ({y}) is connected, p−1 ({y }) ⊂ A or p−1 ({y}) ⊂ (X − A). This implies p(A) ∩ p(X − A) = ∅ , so we have
p−1 p(A) ∩ p−1 p(X − A) = ∅.
6
This shows A = p−1 p(A) and X − A = p −1 p(X − A) since
X = p−1 (Y ) = p−1 p(A) ∪ p(X − A) = p−1 p(A) ∪ p−1 p(X − A)
A ⊂ p−1 p(A)
X − A ⊂ p−1 p(X − A)
and
Consequently, p(A) is both open and closed in Y since p is a quotient map. By Connectedness of Y we have either A = ∅ or A = X . 12. Let Y ⊂ X ; let X and Y be connected. Show that if A and B form a separation of X − Y , then Y ∪ A and Y ∪ B are connected. sol) Suppose Y ∪ A is not connected. Then there is a separation of Y ∪ A, i.e. Y ∪ A is a disjoint union of nonempty open sets C and D of Y ∪ A. Since Y is connected, applying Lemma 23.2, we can assume Y ⊂ C . This, together with the fact Y A = C D, gives D ⊂ A. On the other hand, Lemma 23.1 shows C ∩ D = ∅
and
C ∩ D = ∅
(0.1)
Similarly, together with the fact D ⊂ A, again by Lemma 23.1 we have B ∩ D ⊂ B ∩ A = ∅
and
B ∩ D ⊂ B ∩ A = ∅.
(0.2)
Now, note that X is a disjoint union of (B C ) and D : X = Y (X \ Y ) = Y A B = C D B = (B C ) D.
(0.3)
By (0.1) and (0.2), (B ∪ C ) ∩ D = (B ∪ C ) ∩ D = ∅, so C ∪ B = C ∪ B by (0.3). Thus C ∪ B is closed in X . Repeating the same argument gives D = D is also closed in X . This contradicts to the fact X is connected. Similar solution without using Lemma 23.1 and Lemma 23.2 :
Since A and B are nonempty open sets in X − Y , there exist nonempty open sets U A and U B of X such that A = U A ∩ (X − Y ) and B = U B ∩ (X − Y ), so X − Y can be written as X − Y = A B = [U A ∩ (X − Y )] [U B ∩ (X − Y )] Suppose Y ∪ A is not connected. Then Y ∪ A has a separation, so (similarly as above) there exist nonempty open sets U C and U D of X such that Y ∪ A can be written as : Y ∪ A = [U C ∩ (Y ∪ A)] [U D ∩ (Y ∪ A)] where both U C ∩ (Y ∪ A) and U D ∩ (Y ∪ A) are not empty. It then follows that Y = Y ∩ (Y ∪ A) = (U C ∩ Y ) (U D ∩ Y ). Since Y is connected, we can assume U C ∩ Y = Y , i.e. Y ⊂ U C , and U D ∩ Y = ∅ . In this case, A ∩ U D = ∅ and A = (U C ∩ A) (U D ∩ A). 7
This, together with the facts Y ⊂ U C , A ⊂ U A and B ⊂ U B , implies X = Y A B ⊂ (U B ∪ U C ) ∪ A ⊂ (U B ∪ U C ) ∪ (U D ∩ A)
⊂ (U B ∪ U C ) ∪ (U A ∩ U D ) ⊂ X Note that (i) ∅ = A ∩ U D ⊂ U A ∩ U D and (ii) (U B ∪ U C ) ∩ (U A ∩ U D ) = ∅ since (U A ∩ U D ) ∩ U B ⊂ U A ∩ U B = ∅ and
(U A ∩ U D ) ∩ U C ⊂ U D ∩ U C = ∅.
We have contradiction to the fact X is connected.
§24 Connected Subspaces of the Real Line
1.
(a) Show that no two of the spaces (0, 1), (0, 1], and [0, 1] are homeomorphic.
sol) Suppose there exists a homeomorphism f : (0, 1] → (0, 1). Then the restriction map g = f |(0,1) : (0, 1) → (0, 1) − {f (1)} is also a homeomorphism. (0,1) is connected, but its image g(0, 1) = (0, 1) − f (1) is not connected. Contradiction! (c) Show that Rn and R are not homeomorphic if n > 1. sol) Suppose there exists a homeomorphism f : Rn → R. Then the restriction map f |
Rn \{f −1 (0)}
: Rn \ {f −1 (0)} → R \ {0}
is also a homeomorphism. Rn \ {f −1 (0)} is path connected, while its image R \ {0} is not. Contradiction! 2.
Let f : S 1 → R be a continuous map. Show there exists a point x of S 1 such that f (x) = f (−x).
sol) Suppose not. Consider a continuous function g : S 1 → R
defined by
g(x) = f (x) − f (−x).
The image g(S 1 ) is connected since S 1 is connected and g is continuous. Since g(x) =0 1 1 1 for all x ∈ S , we have either g(x) > 0 for all x ∈ S or g(x) < 0 for all x ∈ S . This is impossible since g(−x) = − g(x). 3.
Let f : X → X be continuous. Show that if X = [0, 1], then f (x) = x for some x ∈ X . What happens if X equals to [0, 1) or (0, 1]? 8
sol) Define g : X → X by g(x) = f (x) − x. Then g is continuous and g(0) = f (0) − 0 ≥ 0
and
g(1) = f (1) − 1 ≤ 0.
Since g([0, 1]) is connected, g(x) = 0 for some x ∈ [0, 1], i.e., f has a fixed point. If X = [0, 1) (resp. X = (0, 1]) then the function f (x) = 12 x + has no fixed point. 8.
1 2
(resp. f (x) = 12 x)
(b) If A ⊂ X and A is path connected, is A necessarily path connected?
sol) No. (cf. p157 Example 6) Let S = {( x, sin( x1 ) ) | 1 < x ≤ 1}. Since S is the image of (0, 1] under the continuous map g : (0, 1] → S defined by g(x) = ( x, sin( x1 ) ), S is path connected. In particular, S is connected, so is its closure S in R2 . Recall V = {0} × [−1, 1] ⊂ R2 .
S = V ∪ S where
Suppose there is a path f : [a, c] → S with f (a) ∈ V and f (c) ∈ S . Since V is closed in S , its preimage f −1 (V ) is closed in [a, c] ; f −1 (V ) is bounded and closed in R. The preimage f −1 (V ) thus has a maximum b and b < c since f (c) ∈ S . Write h(t) = (x(t), y(t)) for the composition map h : [0, 1] −→ [b, c] −→ S
where the first map is t → (1 − t)b + tc /2 and the second map is the restriction of f to [b, c]. WLOG, we may assume h(0) = (0, 0). Since x(t) > 0 for t > 0, for each n > 0 we can choose u satisfying 0 < u < x( n1 )
and
sin( u1 ) = (−1)n .
Applying the Intermediate Value Theorem to the function t → x(t), we can then find tn with 0 < tn < n1 such that x(tn ) = u. We have contradiction since h is continuous and tn → 0, but h(tn ) dose not converge to (0, 0). (d) If {Aα } is a collection of path-connected subspaces of X and if ∩Aα = ∅, is ∪Aα necessarily path connected? sol) Yes. Let x, y ∈ ∪Aα . Then x ∈ Aα1 and y ∈ Aα2 for some Aα1 and Aα2 . Fix a point z ∈ A α1 ∩ Aα2 and choose paths from x to z in Aα1 and from z to y in Aα2 respectively. One can then paste two paths to make a path from x to y in Aα1 ∪ Aα2 ⊂ ∪Aα (see The Pasting Lemma p108). 9.
Show that if A is a countable subset of R2 , then R2 − A is path connected.
sol) Let L p denote a line passing through a point p in R2 . Note that (i) there are uncountably many such lines L p and (ii) there are at most countably many lines L p with L p ∩ A = ∅ . 2 So, for any distinct points x, y in R − A, we can choose lines Lx and Ly such that their intersection Lx ∩ Ly is one point. We can then join x and y by line segments in R2 − A using L x and L y . 9
10. Show that if U is an open connected subspace of R2 , then U is path connected. sol) Fix x0 ∈ U and define P = { x ∈ U | x can be joined to x0 by a path in U } . It suffices to show P is open in U (or in R2 since U is open) and closed in U . Let x ∈ P . We can then choose an open ball B(x, ) ⊂ U for small > 0. Since any point in B(x, ) can be joined to x by a line in B (x, ) ⊂ U , by pasting two paths, any point in B(x, ) can be joined to x0 by a path in U . This implies P is open. In order to prove P is closed in U , one can show U − P is open as done in class. Let’s do differently. First recall that the closure of P in U is P ∩ U , namely the closure of P w.r.t the subspace topology of U is P ∩ U . Let x ∈ P ∩ U . Since U is open in R2 and x ∈ U , we can choose an open ball B(x, ) ⊂ U for small > 0. Since x ∈ P , there exists y ∈ B(x, ) ∩ P . We can then join x to x0 by pasting a path in U from x0 to y with a line segment in B (x, ) ⊂ U from y to x. Thus, x ∈ P . Since x is arbitrary, P = P ∩ U is closed in U . 11. If A is a connected subspace of X , does it follows from that Int A and Bd A are connected? Does the converse hold? sol) (i) X = R and A = (0, 1): Bd A = {0, 1}.
(ii) X = R2 and A = B (0, 0), 1 ∪ B (2, 0), 1 : Int A = B((0, 0), 1) B((2, 0), 1). (iii) X = R2 and A = B((0, 0), 1) ∪ X where X = { ((x, 0)|1 < x < 2 } : Int A = B((0, 0), 1)
and
Bd = S 1 ∪ Y
where S 1 is the unit circle and Y = { ((x, 0)|1 ≤ x ≤ 2 }
§25 Components and Local Connectedness
8.
Let p : X → Y be a quotient map. Show that if X is locally connected, then Y is locally connected.
sol) Let U be a neighborhood of y ∈ Y . We need to show there is a neighborhood C of y with C ⊂ U . Let C be a component of U containing y. It remains to show C is open. For each x ∈ p−1 (C ) ⊂ p−1 (U ), since X is locally connected and p−1 (U ) is open, there is a connected neighborhood V x ⊂ p−1 (U ) of x. It then follows that = p(V x ∩ p−1 (C )) ⊂ p(V x ) ∩ p( p−1 (C )) ⊂ p(V x ) ∩ C. ∅ This shows p(V x ) ⊂ C because C is a component of U and p(V x ) is connected subset of U — any point in p(V x ) is equivalent to any point in p(V x ) ∩ C (and hence any point in C ) under the equivalence relation on U defining (connected) components as in p159 Definition. Consequently, V x ⊂ p−1 ( p(V x )) ⊂ p−1 (C ). Therefore, p−1 (C ) is open, so is C since p is a quotient map. 10
9.
Let G be a topological group; let C be the component of G containing the identity element e. Show that C is normal subgroup of G.
sol) For each x ∈ G, xCx−1 = Rx−1 ◦Lx (C ) is connected where Lx and Rx are homeomorphisms defined by Lx (g) = xg and by Rx−1 (g) = gx−1 . On the other hand, since e = xex−1 ∈ xCx −1 and C is a component containing e, xCx −1 ⊂ C . A similar argument shows that for any a, b ∈ C , ab −1 C ⊂ C which implies ab −1 ∈ C .
§26 Compact Spaces
5.
Let A and B be disjoint compact subspaces of the Hausdorff space X . Show that there exists disjoint open sets U and V containing A and B , respectively.
sol) Since A is compact and A ∩ B = ∅, for each b ∈ B there are disjoint neighborhoods U b and V b of A and b respectively. By Compactness of B there are finitely many V b1 , · · · , V bn whose union contains B. We set U A = U b1 ∩ · · · ∩ U bn
and
V B = V b1 ∪ · · · ∪ V bn .
Then A ⊂ U A , B ⊂ V B and U A ∩ U B = ∅ since x ∈ V B ⇒ x ∈ V bi for some b i ⇒ x ∈ / U bi ⇒ x∈ / U A . 6.
Show that if f : X → Y is continuous, where X is compact and Y is Hausdorff, then f is a closed map.
sol) Let A be a closed set in X . Then A is compact and hence f (A) is compact. Since Y is Hausdorff, f (A) is closed. 7.
Show that if Y is compact, then the projection π1 : X × Y → X is a closed map.
sol) We will show that X − π1 (A) is open if A is closed in X × Y . Let x ∈ X − π1 (A). Noting (x, y) ∈ / A for any y ∈ Y and X × Y − A is open, for each y ∈ Y choose neighborhoods U x,y of x and V y of y satisfying (x, y) ∈ U x,y × V y ⊂ X × Y − A. Since {V y | y ∈ Y } is an open covering of Y and Y is compact, there is a finite subcover {V yi | i = 1, · · · , n}. Now, set U = U x,y1 ∩ · · · ∩ U x,yn . U is then a neighborhood of x such that U × Y ⊂ X × Y − A. This shows U ∩ π1 (A) = ∅ and hence X − π1 (A) is open.
11
8.
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 .
sol) Suppose f is continuous. Let (x, y) ∈ X × Y − Gf and choose neighborhoods V y of y and V f (x) of f (x) with V y ∩ V f (x) = ∅. Then U = f −1 V f (x) is a neighborhood of x such that
U × V ⊂ X × Y − Gf . This shows X × Y − Gf is open. Now, suppose Gf is closed. Fix x0 ∈ X and let V be a neighborhood of f (x0 ). Then Gf ∩ X × (Y − V ) is closed and hence by Problem 7
U = X − π1 Gf ∩ X × (Y − V )
is open and x0 ∈ U since (x0 , f (x0 )) ∈ / X × (Y − V ). Let x ∈ U . Then (x, f (x)) ∈ / X × (Y − V ) and hence f (x) ∈ V . Thus f (U ) ⊂ V . This shows f is continuous at x0 . Since x 0 is arbitrary, f is continuous. 12. 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 is compact, then X is compact. sol) Let { U α } be an open covering of X . Then for each y ∈ Y since p −1 ({y }) is compact it can be covered by a finite subcollection { U αi | i = 1, · · · , ny } of the covering { U α }. Set U y = U α1 ∪ · · · ∪ U αny
W y = Y − p(X − U y ).
and
Then, we have (i) W y is open since p is closed, (ii) y ∈ W y since p−1 {y} ⊂ U y implies y ∈ / p(X − U y ),
(iii) p−1 (W y ) = p −1 Y − p(X − U y ) = X − p−1 p(X − U y ) ⊂ X − (X − U y ) = U y . (i) and (ii) show that { W y | y ∈ Y } is an open covering of Y . Since Y is compact, we can choose a finite subcover { W yi | i = 1, · · · , n}. Then by (iii) n
X = p
−1
(Y ) = p
−1
n
W yi =
i=1
n
−1
p
i=1
(W yi ) ⊂
U yi .
i=1
Since each U yi is a finite union of open sets in { U α }, this implies that the open covering {U α } has a finite subcover. 13. Let G be a topological group. (a) Let A and B be subspaces of G. If A is closed and B is compact, show A · B is closed.
12
sol) We will use the fact every topological group is regular (see p146 #7 (c)). Let c ∈ / A · B. −1 Then cB ∩ A = ∅ and hence by regularity of G for each b ∈ B there exist disjoint neighborhoods U b and V b of cb−1 and A respectively. Note that cB −1 = Lc ◦ µ(B) is compact where Lc is the left translation by c and µ : G → G defined by µ(g) = g −1 . One can thus find finitely many U b1 , · · · , U bn whose union contains cB −1 . We set U = U b1 ∪ · · · ∪ U bn
and
V = V b1 ∩ · · · ∩ V bn .
Then cB −1 ⊂ U, A ⊂ V and U ∩ V = ∅ (cf. # 5 above). So, W = U B is a neighborhood of c such that W ∩ AB ⊂ U B ∩ V B = ∅ . This shows G − AB is open. (b) Let H be a subgroup of G; let p : G → G/H be the quotient map. If H is compact, show that p is a closed map. sol) Recall that p−1 ( p(C )) = C H . If C is closed, then p−1 (G/H − p(C )) = G − CH is open by (a). Thus G/H − p(C ) is open, i.e. p(C ) is closed. (c) Let H be a compact subgroup of G. Show that if G/H is compact, then G is compact. sol) The proof directly follows from #12 and (b) since p−1 (gH ) = gH is compact for any gH ∈ G/H .
§27
5.
Compact Subspaces of the Real Line
Let X be a compact Hausdorff space; let {An } be a countable collection of closed sets of X . Show that if each A n has empty interior in X , then the union ∪ An has empty interior in X .
sol) We will show that U − ∪ An = ∅ for any open set U (this implies Int(∪An ) = ∅). We first show that given A1 and nonempty open set U there exists a nonempty open set V 1 satisfying V 1 ⊂ U and V 1 ∩ A1 = ∅ . Since IntA1 = ∅ , there exists y ∈ U − A1 . Noting A 1 is compact, choose a neighborhood of W 1 of y and an open set W 2 containing A 1 with W 1 ∩ W 2 = ∅ (by Lemma 26.4). Similarly, since X − (W 1 ∩ U ) is closed and hence compact, choose a neighborhood W 1 of y and an open set W 2 containing X − (W 1 ∩ U ) with W 1 ∩ W 2 = ∅ (again by Lemma 26.4). Let V 1 = W 1 . The nonempty open set V 1 then satisfies V 1 ⊂ W 1 ∩ U ⊂ U and V 1 ∩ A1 = ∅ . Now, applying the same arguments inductively, givenAn and V n−1 one can find a nonempty open set V n satisfying V n ⊂ V n−1 and V n ∩ An = ∅ . Consider the nested sequences V 1 ⊃ V 2 ⊃ · · · of nonempty closed sets of X . Since X is compact, there exists x ∈ ∩V n (by Thm 26.9) such that x ∈ / An for each n since V n ∩ An = ∅ for each n. Thus, x ∈ U − ∪An .
13
§31
2.
The Separation Axioms
Show that if X is normal, every pair of disjoint closed sets have neighborhoods whose closures are disjoint.
sol) Let A, B are disjoint closed sets. Then X − B is open and A ⊂ X − B and hence there exists an open set U A with A ⊂ U A and U A ⊂ X − B (see Lemma 31.1 (b)). In particular, B ⊂ (X − U A ) since U A ∩ B = ∅ . Again by Lemma 31.1 (b) applied to B ⊂ (X − U A ) gives U B open in X with B ⊂ U B and U B ⊂ (X − U A ). The latter is equivalent to U B ∩ U A = ∅. Thus U A and U B is the desired neighborhoods of A and B respectively. 5.
Let f, g : X → Y be continuous; assume that Y is Hausdorff. Show that { x|f (x) = g(x)} is closed in X .
sol) Consider the graph Gg = {(x, y) ∈ X × Y | y = g(x)} of g. Since g is continuous and Y is Hausdorff, the graph Gg is closed in X × Y (cf. p171 ex 8). Define a function F : X → X × Y
by
F (x) = (x, f (x)).
Then, F is continuous and hence F −1 (Gg ) = { x|f (x) = g(x)} is closed in X . 6.
Let p : X → Y be a closed continuous surjective map. Show that if X is normal, then so is Y .
sol) Let’s first prove the given Hint : Claim : If U is an open set containing p −1 (y), then there is a neighborhood W of y such that p −1 (W ) ⊂ U . Proof : Let W = Y − p(X − U ). Then W is open in Y since p is a closed map and y ∈ W ; suppose y ∈ P (X − U ). Then there exists x ∈ (X − U ) such that p(x) = y. This implies x ∈ p −1 ({y}) ⊂ U . Contradiction! On the other hand, we have p−1 (W ) = p−1 (Y − p(X − U )) = X − p−1 p(X − U ) ⊂ X − (X − U ) = U. This completes the proof of the claim. Let A and B be disjoint closed sets in Y . Then p−1 (A) and p−1 (B) are disjoint closed sets in X . Using normality of X , choose disjoint neighborhoods U A and U B of p −1 (A) and p−1 (B) respectively. The claim then shows that for each a ∈ A (resp. b ∈ B) there is a neighborhood W a of a (resp. W b of b) with p−1 (W a ) ⊂ U A (resp. p−1 (W b ) ⊂ U B ). The open sets and W A = W a W B = W b
a∈A
b∈B
are desired neighborhoods of A and B respectively;
p−1 (W A ∩ W B ) = p −1 (W A ) ∩ p−1 (W B ) ⊂ U A ∩ U B = ∅ . This together with subjectivity of p shows W A ∩ W B = ∅ . 14
§32
1.
Normal Spaces
Show that a closed subspace of a normal space is normal.
sol) Let X be normal and A be a closed subset of X . Let A 1 and A 2 be disjoint closed subsets of A. Since A is closed in X , A 1 and A 2 are also closed in X . Normality of X then shows there are disjoint neighborhoods U 1 and U 2 of A1 and A2 respectively in X . Thus the open sets U 1 ∩ A and U 2 ∩ A in A are disjoint neighborhoods of A1 and A 2 in A.
§33
1.
The Urysohn Lemma
Examine the proof of the Urysohn lemma, and show that for given r,
f −1 (r) =
p>r
U p −
q
U q
where p, q rational. sol) We will use the following facts : (1) x ∈ U r ⇒ f (x) ≤ r or equivalently f (x) > r ⇒ x ∈ / U r , (2) x ∈ / U r ⇒ f (x) ≥ r or equivalently f (x) < r ⇒ x ∈ U r . Suppose x ∈ f −1 (r), i.e. f (x) = r. Then f (x) < p for any rational p > r, so x ∈
p>r
U p
by (2). Similarly, for any rational q < r, f (x) > q , so x ∈ / U q ⊂ U q by (1). This implies x∈ / U q . Therefore,
q
f −1 (r) ⊂
p>r
Conversely, suppose x ∈
p>r
U p −
q
U p −
q
U q .
U q . (2) then implies f (x) ≥ q for any rational q < r,
so f (x) ≥ r. Since for any rational p > r, x ∈ U p ⊂ U p , by (1) f (x) ≤ p. This shows f (x) ≤ r. Therefore, f (x) = r, i.e. x ∈ f −1 (r). 2.
(a) Show that a connected normal space having more that one point is uncountable.
sol) Let X be a connected normal space and x and y are distinct points of X . Then by Uryshon Lemma, there is a continuous function f : X → [0, 1] with f (x) = 0 and f (y) = 1. Since the image f (X ) is connected, f (X ) = [0, 1]; suppose not. Then f (X ) is a disjoint union of nonempty open sets f (X ) ∩ [0, r) and f (X ) ∩ (r, 1] for some r ∈ / f (X ) where 0 < r < 1. −1 Consequently, the preimage f (s) = ∅ for any irrational s ∈ [0, 1]. Thus X is uncountable.
3.
Give a direct proof of the Urysohn lemma for a metric space (X, d). 15
sol) Let A and B are disjoint closed sets of X and let f (x) =
d(x, A) . d(x, A) + d(x, B)
Recall that (i) d(x, A) = inf { d(x, a) | a ∈ A }, (ii) it is a continuous function on X into R and (iii) for closed A, d(x, A) = 0 iff x ∈ A (see p175-177). It thus follows that f is a continuous function on X (since A ∩ B = ∅) such that f (x) = 0 iff x ∈ A, f (x) = 1 iff x ∈ B and 0 ≤ f (x) ≤ 1 for all x ∈ X . 4. Recall A is a Gδ set in X if A is the intersection of a countable collection of open set of X . Show that : Let X be normal. Then there exists a continuous function f : X → [0, 1] such that f (x) = 0 for x ∈ A, and f (x) > 0 for x ∈ / A, if and only if A is a closed Gδ set in X . sol) (by Jonathan) Suppose there exists a continuous function f : X → [0, 1] such that f (x) = 0 for x ∈ A, and f (x) > 0 for x ∈ / A. Then A is closed since A = f −1 (0) and A is a Gδ set since 1 A = f −1 [0, ) . n n Conversely, suppose A is a closed Gδ set. Then A = ∩ U n for some open sets U n . Since X is normal, for closed A and open U 1 ⊃ A, there is an open set V 1 with A ⊂ V 1 and V 1 ⊂ U 1 (see Lemma 31.1). Inductively, for each n ∈ N one can define an open set V 1 n+1 satisfying A ⊂ V 1 and V 1 ⊂ U 1 ∩ U 2 ∩ · · · U n+1 ∩ V 1 .
n+1
Then we have A ⊂ V 1 and V n
A ⊂
V
1 n+1
⊂
n+1
1 n+1
⊂ V 1 for all n ∈ N and n
(U 1 ∩ · · · U n+1 ) ∩ V 1 ⊂ n
n
V 1 = A; n
(U 1 ∩ · · · ∩ U n+1 ) ⊂
U n = A.
Now, similarly as in the proof of Urysohn lemma, writing rational numbers in the set (0, 1) − { n1 |n ∈ N} as a sequence { p1 , p2 , · · · } and using the open sets V 1 and induction on n finite subsets of { p1 , p2 , · · · }, for any rational p ∈ (0, 1) one can define an open set U p ⊃ A satisfying U q ⊂ U p if q < p and U p = V 1 if p = n1 . The function f (x) = inf { p | x ∈ U p } n is then continuous and f (x) = 0 for x ∈ A. Moreover, f (x) > 0 for x ∈ / A; in this case, 1 x∈ / V 1 for some n, so f (x) ≥ n (see p210 (2)). n
(Another) Let A = ∩ U n . Then by Urysohn lemma, for each n there is a continuous function f n : X → [0, 1] with f n ≡ 0 on A and f n ≡ 1 on X − U n . Define a function f n (x) . 2n Since 0 ≤ f n (x) ≤ 1 for all n, F is a well-defined function with F ≡ 0 on A. Let x ∈ / A. f n (x) 1 Then x ∈ X − U n for some n (since A = ∩ U n ) and hence F (x) ≥ 2n = 2n > 0. Thus it remains to show F is continuous. Let F n = ni=1 f i2(x) i . Note that F n is continuous for all n and ∞ 1 1 0 ≤ F (x) − F n (x) ≤ = (0.4) 2i 2n+2 F : X → [0, 1]
by
F (x) =
i=n+1
16
for any x ∈ X . This shows F n uniformly converges to F , so F is also continuous. It proof is : fix x ∈ X and > 0. By (0.4) there is N > 0 ( 2N 1+1 < ) such that F n (x) ∈ (F (x) − , F (x) + ) 2 2 for all n > N . Continuity of F n then shows there is a neighborhood U n of x with F n (U n ) ⊂ (F (x) − 2 , F (x) + 2 ) where n > N . It then follows that F (U n ) ⊂ (F (x) − , F (x) + ). This shows F is continuous. 5.
Prove: Let X be normal space. There is a continuous function f : X → [0, 1] such that f (x) = 0 for x ∈ A, and f (x) = 1 for x ∈ B, and 0 < f (x) < 1 otherwise, if and only if A and B are disjoint Gδ sets in X .
sol) (by Mohamed and Jonathan) Let A and B are disjoint closed Gδ sets. By the above Problem 4 there are continuous function f A (resp. f B ) into [0, 1] such that f A (x) = 0 for x ∈ A (resp. f B (x) = 0 for x ∈ B) and f A (x) > 0 for x ∈ / A (resp. f B (x) > 0 for x ∈ / B). The desired function is then f A (x) f (x) = f A (x) + f B (x) The proof of the converse is the same as that in Problem 4.
§35
4.
The Tietze Extension Theorem
(a) Show that if Z is Hausdorff and Y is a retract of Z , then Y is closed in Z .
sol) Let r : Z → Y be the retraction. Then r(z) = z iff z ∈ Y . Problem 5 in p199 thus implies Y = { z | j ◦ r(z) = id Z (z)} is closed where j : Y → Z is the inclusion map. (b) Let A be a two-point set in R2 . Show that A is not a retract of R2 . sol) There is no retraction r : R 2 → A; otherwise, r(R2 ) is connected, A is not connected and r(R2 ) = A. (c) Can you conjecture whether or not S 1 is a retract of R2 ? sol) S 1 is a not retract of R2 since π1 (R2 ) = 0 while π1 (S 1 ) = Z where π1 is the fundamental group. 7.
(a) Show the logarithmic spiral C = { (0, 0) } ∪ { (et cos t, et sin t | t ∈ R } is a retract of R2 . Can you find a specific retraction r : R2 → C ?
17
sol) Let C ∗ = C − {(0, 0)} and (R2 )∗ = R2 − {(0, 0)} and define continuous functions f : R → C ∗
by
G : (R2 )∗ → R
by
f (t) = (et cos t, et sin t), 1 G(x, y) = ln (x2 + y 2 ). 2
Then f is bijective and the restriction of G to C ∗ is the inverse function of f . The composition f ◦ G : (R2 )∗ → C ∗ is thus a retraction. Note that for any seq (xn , yn ) in (R2 )∗ converging to (0,0) the sequence f ◦ G(xn , yn ) also converges to (0,0). So, the extension r : R2 → C of f ◦ G defined by r(0, 0) = (0, 0) is continuous (cf. Theorem 21.3) and thus r is a retraction, i.e. C is a retract of R2 . (b) Show that the knotted x-axis of Figure 35.2 is a retract of R3 . sol) Noting K is homeomorphic to R, fix a homeomorphism f : K → R. Then, since K is closed in R3 , by Tietze extension theorem, there is a continuous extension F : R 3 → R of f . Then the composition f −1 ◦ F : R3 → K is a retraction.
§36
2.
Imbeddings of Manifolds
Let X be a compact Hausdorff space. Suppose that for each x ∈ X , there is a neighborhood U of x and a positive integer k such that U can be imbedded in Rk . Show that X can be imbedded in RN for some positive integer N .
sol) Since X is compact, there is a finite open covering { U i | i = 1, · · · , m } of X such that for each i there is an imbedding ϕi : U i → R ki for some ki > 0. Note that X is normal since X is compact Hausdorff. So, there is a partition of unity { φi } subordinated to { U i } that gives rise to an imbedding : Φ : X → R × · · · × R ×Rk1 × · · · × Rkm
m times
defined by Φ(x) = φ1 (x), · · · , φm (x), φ1 (x) · ϕ 1 (x), · · · , φm (x) · ϕ m (x) where for x in X − supp(φi ) we define φi (x) · ϕi (x) = 0 so that it is a continuous function on X . 5.
Let X be the union of the set R −{0} and the two-point set { p, q }. Topologize X by taking as basis the collection of all open intervals in R that do not contain 0, along with all sets of the form (−a, 0) ∪ { p} ∪ (0, a) and all sets of the form (−a, 0) ∪ {q } ∪ (0, a), for a > 0. (b) Show that each of the space X − { p} and X − {q } is homeomorphic to R.
sol) Noting the collection of all open intervals in R that do not contain 0, along with all sets of the form (−a, a) = (−a, 0) ∪ {0} ∪ (0, a) for a > 0 is a basis for the standard topology on R , we define f : X − {q } → R by f (x) = x for x ∈ R − {0} and f ( p) = 0. This function f is bijective and both f and its inverse f −1 send basis elements to basis elements. Thus f is a homeomorphism. 18
§43
Complete Metric Spaces
2. Let (X, dX ) and (Y, dY ) be metric spaces; let Y be complete. Let A ⊂ X . Show that if f : A → Y is uniformly continuous, then f can be uniquely extended to a continuous function g : A → Y , and g is uniformly continuous. sol) Let a ∈ A. Then there is a sequence { an } in A converging to a (see Lemma 21.2). Since f is uniformly continuous, {f (an )} is a Cauchy sequence in Y ; given > 0 there exists δ > 0 (depending only on ) such that if d X (an , am ) < δ then d Y (f (an ), f (am )) < . Since an → a, in particular {an } is a Cauchy sequence in X , one can choose N > 0 satisfying : if n,m > N then dX (an , am ) < δ . Using Y is complete, set g (a) = lim f (an ). Then (i) g : A → Y is a well-defined function; suppose we have two sequences {an } and { bn } both converging to a ∈ A. Let {cn } be a sequence defined by c2n−1 = an and c2n = b n . Then a similar argument as above and the inequality dX (an , bm ) ≤ dX (an , a) + dX (a, bm )
∀ n, m ∈ N
imply that { f (cn )} is a Cauchy sequence in a complete metric space Y , so by uniqueness of the limit we have lim f (an ) = lim f (c2n−1 ) = lim f (c2n ) = lim f (bn ). (ii) g is an extension of f ; for a ∈ A let an = a for all n. Then g(a) = lim f (an ) = f (a). (iii) g is uniformly continuous; given > 0 choose δ > 0 corresponding to the uniform continuity of f and let a, b ∈ A with dX (a, b) < δ/2. One can then find sequences {an } and { bn } respectively converging to a and b such that there is N > 0 satisfying : if n , m > N then dX (an , bm ) < δ , dY (g(a), f (am )) < and dY (f (bm ), g(b)) < . Consequently, dY (g(a), g(b)) < dY (g(a), f (am )) + dY (f (an ), f (bm )) + dY (f (bm ), g(b)) < 3. (iv) g is the unique continuous extension of f to A; suppose h is another continuous extension of f to A. Then for any a ∈ A and any sequence {an } converging to a, h(a) = lim h(an ) = lim f (an ) = g(a) (cf. Theorem 21.3). 4.
Show that the metric space (X, d) is complete if and only if for every nested sequence A1 ⊃ A2 ⊃ · · · of nonempty closed sets of X with diamAn → 0, the intersection of the sets A n is nonempty.
sol) The necessary condition for completeness of (X, d) is given by Lemma 48.3. Let’s prove sufficient condition. Let {xn } be a Cauchy sequence in (X, d). There exists N 1 > 0 such that if n,m > N 1 then d(xn , xm ) < 1. Choose n 1 > N 1 and let A 1 = B(xn1 , 1). Similarly, there exists N 2 > N 1 such that if n,m > N 2 then d(xn , xm ) < 12 . Choose n2 > N 2 and let A2 = A 1 ∩ B(xn2 , 12 ). Note that A2 ⊂ A 1 and if n > N 2 then xn ∈ A 2 since N 2 > N 1 . Repeating the same argument with induction then shows for all k ∈ N there exists N k and closed set A k such that (i) A k ⊂ A k−1 and diamAk → 0 and (ii) x n ∈ Ak if n > N k . Now, the assumption and (i) imply that there is a point x in the intersection of the sets Ak . Then for any k > 0, if n > N k then d(x, xn ) < k2 since both x, xn ∈ A k by (ii). This shows xn → x. 19
5.
Show that if f is a contraction of a complete metric space (X, d), then there is a unique point x ∈ X such that f (x) = x.
sol) Fix x0 ∈ X and define a sequence { xn } recursively: x1 = f (x0 ) and xn+1 = f (xn ). Then for some 0 < α < 1 d(f (xn+1 ), f (xn )) < αd(xn+1 , xn ) < αd(f (xn ), f (xn−1 )) < · · · < αn+1 d(f (x0 ), x0 ). Together with triangular inequality, this shows { xn } is a Cauchy sequence. Thus xn → x for some x ∈ X . Since every contraction is continuous, we have f (x) = lim f (xn ) = lim xn−1 = x.
§48
2.
Baire Spaces
The Baire Category theorem implies that R cannot be written as a countable union of closed subsets having empty interiors. Show this fails if the set are not required to be closed.
sol) Write the set Q of the rational numbers as a sequence Q = {q 1 , q 2 , · · · } and set An = (R − Q) ∪ {q n }. Then ∪ An = R where each A n has empty interior since R − An = Q − {q n } is dense in R. 5.
Show that if Y is G δ set in X , and if X is compact Hausdorff or complete metric, then Y is a Baire space in the subspace topology.
sol) Let Y = ∩ W n where W n is open in X and B n be a closed set in Y with empty interior in Y . Given U 0 open in X with U 0 ∩ Y = ∅ , we need to show that U 0 ∩ Y − ∪Bn = ∅ . Note that B n = B n ∩ Y , where B n is the closure of B n in X , since B n is closed in Y . Also note that there is a point y 0 ∈ U 0 ∩ Y − B 1 ∩ Y = (U 0 − B 1 ) ∩ Y since B 1 has empty interior in Y . Using normality of X , since y 0 ∈ U 0 ∩ (X − B 1 ) ∩ W 1 , one can find U 1 open in X with y0 ∈ U 1 (so U 1 ∩ Y = ∅ )
and
U 1 ⊂ U 0 ∩ (X − B 1 ) ∩ W 1 .
In addition, in the metric space, by replacing U 1 by a small ball at y 0 if necessary, one can also assume diam U 1 < 1. Repeating the same argument inductively yields a sequence of open sets U n in X satisfying : = ∅ , U n ∩ Y
U n ⊂ U n−1 ∩ (X − B n ) ∩ W n , diam U n <
1 n
(for metric space)
Then there exists a point x ∈ ∩ U n (by Theorem 26.9 and Lemma 48.3) such that (i) x ∈ U 0 ∩ Y since x ∈ U n ⊂ U n−1 ∩ W n ⊂ U 0 ∩ W n for all n and (ii) x ∈ / ∪ Bn = ∪ (B n ∩ Y ) since x ∈ X − B n for all n. (i) and (ii) show that U 0 ∩ Y − ∪ Bn = ∅ as desired. 6.
Show that the irrationals are a Baire space. 20
sol) Let Q = { q 1 , q 2 , · · · } and W n = R − {q n }. Then R − Q = ∩ W n and hence the irrationals R − Q is a Baire space by Problem 5. 7.
Prove the following: If D is a countable dense subset of R, there is no function f : R → R that is continuous precisely at the points of D. Proof.
(a) Show that if f : R → R, the set C of points at which f is continuous is a G δ set in R . sol) Let U n be the union of all open sets U of R such that diam f (U ) < each n > 0 there exists δ n > 0 such that f (x − δ n , x + δ n ) ⊂ (f (x) −
1 n.
If x ∈ C then for
1 1 3n , f (x) + 3n ).
Thus x ∈ ∩ U n . Conversely, let x ∈ ∩ U n . Then given > 0 and for n1 < since x ∈ U n there exists an open set U containing x with diam f (U ) < n1 < . So, f (U ) ⊂ (f (x) − , f (x) + ). This shows x ∈ C . (b) Show that D is not a Gδ set in R. sol) Suppose D = ∩ W n where W n is open in R. Since D is dense, each W n is dense; suppose W n is not dense. Then Int(R − W n ) = ∅, so (a, b) ⊂ Int(R − W n ) ⊂ R − W n for some a < b. Thus D ∩ (a, b) ⊂ W n ∩ (a, b) = ∅ which contradicts to the assumption D is dense. For each d ∈ D, set V d = R − {d}. Then V d is also open and dense, so Baire Category Theorem shows ∩ n,d∈D (W n ∩ V d ) is dense. But, ∩ n,d∈D (W n ∩ V d ) = D ∩d∈D V d = ∅.
Other Homework
HW Let G be a topological group, H be a subgroup of G and p : G → G/H be the quotient map. Show that p −1 p(U ) = U H for any subset U of G. sol) Noting gH = uH ⇐⇒ g ∈ uH we have g ∈ p −1 p(U ) ⇐⇒ p(g) ∈ p(U ) ⇐⇒ gH = uH for some u ∈ U
⇐⇒ g ∈ uH for some u ∈ U ⇐⇒ g ∈
uH = U H.
u∈U
HW Let G be a topological group and H be a subgroup of G. Show that if H and G/H are connected, so is G. sol) Let U be both open and closed in G. Since H is a connected subgroup and for any g ∈ G the left translation L g : G → G is a homeomorphism, uH = L u (H ) ⊂ U ∀u ∈ U
vH = L v (H ) ⊂ G − U ∀ ∈ G − U.
and 21
(0.5)
On the other hand, since the quotient map p : G → G/H is an open map, p(U ) and p(G−U ) both open in G/H . Suppose there exists gH ∈ p(U ) ∩ p(G − U ). Then gH = uH = vH for some u ∈ U and v ∈ G − U . This is impossible by (0.5). Thus p(U ) ∩ p(G − U ) = ∅ . This together with the fact p(U ) ∪ p(G − U ) = p(G) = G/H implies p(U ) are both open and closed in G/H . Now, by the assumption G/H is connected, we have either p(U ) = ∅ or p(U ) = G/H . Consequently, either U = ∅ or U = G. HW Let (H, d) be a metric space; H is the set of all sequences { an } in R satisfying and the metric d on H is defined by d({an }, {bn }) =
a2n < ∞
(an − bn )2 .
Show that S = { A ∈ H | d(A, 0) = 1 } is bounded and closed, but not compact where 0 is the zero sequence (0, 0, · · · ). sol) S is bounded since S ⊂ Bd (0, 2) and S is closed since S = f −1 (1) where f is a continuous map defined by A → d(A, 0). Let Ai be a sequence {a j } with a j = δ ij , i.e a j = 0 if j = i and ai = 1. The sequence {Ai }i∈N in S has no convergent subsequence since d(Ai , A j ) = δ ij . Thus S is not compact (cf. Theorem 28.2). HW Show that (H, d) is a complete metric space. sol) Let || · || be the norm on H satisfying || A − B || = d(A, B) and let A i denote the i-th term of the sequence A ∈ H . Let { An |n ∈ N} be a Cauchy sequence in H . For each i
|Ain − Aim |2 ≤ ||An − Am ||2 shows the sequence { Ain |n ∈ N} is a Cauchy sequence in R, so Ain → A i for some Ai ∈ R. Let A = { Ai |i ∈ N }. Given > 0 choose N > 0 such that
||An − Am ||2 =
|Ain − Aim |2 <
i
for all n, m > N . Taking limit n → ∞ then gives ||A − A m ||2 ≤ . This implies (i) A − Am ∈ H and hence A = (A − Am ) + Am ∈ H and (ii) Am → A in H . HW Find an example of a metric space that is not second countable. sol) Let X be an uncountable set and define a metric d by d(x, y) = 1 if x = y and d(x, y) = 0 if x = y. Then the metric topology is discrete, so X is not second countable.
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