SOLUTIONS SOLUTIONS TO EXERCISES EXERCISES
Here are solutions to some of the problems in Munkres. There may be other, and perhaps better, ones. They are sometimes more detailed than absolutely necessary, but not excessively so.
∈
Problem 13.1. For each x A, choose an open set U x with x A = x∈A U x , which is open because it is a union of open sets.
∈ U x ⊆ A.
The Then
Problem 13.3. We have X X = , which is countable, so X T c , and X = X , so Next suppose suppose that that U α all U α are T c . Next T c for α in some index set J . If all empty, so is their union, so α∈J U α T c . Otherwise, Otherwise, there is some some β J such that X U β is countable countable.. Since X U X U , X β α∈J α α∈J U α is countable, and again α∈J U α Finally,, suppose suppose U, V eithe herr of U and V is T c . Finally T c . If eit empty, so is U V . V . Otherwise, Otherwise, X U and X V are countable, and hence so is X (U V ) V ) = (X U ) U ) (X V ), V ), and U V T c . This completes the proof that T c is a topology. The collection T ∞ is not a topology, in general. For example, take X = R. Then ( , 0) and (0, (0 , ) are in T ∞ , but ( , 0) (0, (0, ) = R 0 is not. (In fact, T ∞ is a topology iff X is finite, in which case it is the indiscrete topology, but a single example is all that’s needed.)
∅∈ −
− ∩
−∞
−
∩
∅
∈
∈∈ −
∈
− ∪ −
∞
⊆ − ∈ − ∩ ∈
−
−
−∞ ∪ ∞
−∅
∈
−{ }
Problem 13.8. (a) Certainly B is a collection of open sets in the standard topology. Suppose that U is open in the standard topology and x U . U . By definition of the standard topology, there are real numbers a < b with x (a, b) U . U . Since a < x, there is a rational number q with a < q < x. x . (This (This is a standard standard fact fact about the real numbers, known as the density of the rationals; for completeness, I give a proof below.) below.) Similarly Similarly,, there there is a rational number number r with x < r < b. Then (q, r) B b . Then and x (q, r) U , U , and the result follows from Lemma 13.2. (b) For any x R, there exist rational numbers a and b with a x < b (for instance, a = x and b = x + 1), so that x [a, b) C. Also, the intersection of any two elements of C is either empty or an element of C, and so C is a basis for some topology T on R. Let Let x be an irrational irrational number. number. The interv interval al [x, [ x, x + 1) is open in the lower limit topology; we show that it does not belong to T . Supppose, on the contrary, that it does. Since x [x, x + 1), there exists an element [a, [ a, b) (a and b rational) of C with x [a, b) [x, x + 1). 1). Now x [a, b) implies a x, while [a, b) [x, x + 1) implies implies x a, so x = a, a contradiction.
∈ ∈
∈
⊆ ∈
⊆
∈
∈ ≤
⊆
∈
⊆
∈
≤
∈
∈
≤
−
Density of the rationals. Let a and b be real numbers with a < b. b . Since 0 < b a, 1 there is a positive integer n with n < b a. Let Let m be the smallest integer such m m−1 m 1 that n > a. Then Then n a, so n a + n < a + (b (b a) = b. Thus Thus the rational rational m number m satisfies . a < < b. b n n
≤
≤
−
−
While we’re at it, let’s prove that between any two reals there is an irrational. Density of the irrationals. Let a and b be real numbers with a < b. By what what was was just proved, there is a rational number q with a 2
−√
1
−√
√
Problem 16.1. Let T X and T Y be the topologies that A inherits as a subspace of X and Y , respectively. Suppose first that W T X . Then there is an open set U of X with W = U A. Set V = U Y . Then V is open in Y and W = V A, so A T Y . Suppose conversely that W T Y . Then there is a set V open in Y with W = V A. Since V is open in Y , there is an open set U of X with V = U Y . Then W = U A, so W T X .
∈
∩
∩
∩
∩
∈
∩
∈
∩
∈
Problem 16.9. Let T 1 be the dictionary order topology on R R, and T 2 the product topology on Rd R. As pointed out on page 85 of Munkres, the sets ( a b, a d) for b < d form a basis for T 1 . (This is because any set (a b, c d) with a < c is the union of all the sets ( a b, a y) for y > b, (x y1 , x y2 ) for a < x < c and y1 < y 2 , and (c y, c d) for y < d.) Now the singleton sets a form a basis for Rd (Example 3 on page 79), and the sets ( b, d) for b < d form a basis for R by definition of the usual topology on R, so the sets a (b, d) for b < d form a basis for T 2 by Theorem 15.1. But (a b, a d) = a (b, d), so the same collection of sets is a basis for both T 1 and T 2 , so these topologies are equal. I observed in class (after correcting myself) that the dictionary order topology is strictly finer than the usual topology on R2 , but it is easier to see this using the basis of rectangles (a, b) (c, d) (a < b and c < d) for the usual topology, rather than the circle basis. We have (a, b) (c, d) = x∈(a,b) x (c, d), which is open in T 1 . On the other hand, a (b, d) is not open in the usual topology, because it is non-empty and contains no rectangle.
×
×
×
×
×
×
× { }×
×
× ×
× × {}
×
×
{ }× { }×
×
×
{ }×
For the problem on the handout for the second assignment, see Theorem N16.1 in the supplementary notes. Problem 17.21. We shall write A for X A. (a) First we show that if U is open then U − = U −◦− . Since U is open and U U − , U U −◦ and hence U − U −◦− . Conversely, U −◦ U − and U − is closed, so U −◦− U − . Now let A1 be any subset of X , and define A2n = A− 2n−1 and A2n+1 = A2n for − n Z+ . Also let B1 = A1 , and similarly define B2n = B2n−1 and B2n+1 = B2n for Z n + . Clearly any set obtained from A1 by repeated closure and complementation is either An or Bn for some n Z+ . Now A3 is open, since it is the complement of − −◦− −◦− −− A2 = A− . But A− = A◦− = A8 . Hence 1 , so A3 = A3 3 = A4 and A3 4 = A4 A8 = A4 , and so An+4 = An for n 4. Similarly Bn+4 = Bn for n 4, and so every set obtainable from A1 is one of the fourteen sets A1 , . . . , A7 , B1 , . . . B7 . (b) Let us first try simply to find A1 so that A1 , . . . , A7 are distinct. We have A3 open, as observed above, and A7 = A−− = A−◦ 3 3 , so we need an open set that is different from the interior of its closure. A simple example of such a set is an open interval or ray with a point removed; we shall work with rays since they have the convenient property that the complement of a ray is a ray. If A3 is obtained by removing a point from an open ray with endpoint a, then A4 , . . . , A7 are the four rays with endpoint a. (For instance, if A3 = (0, ) 1 then A4 = [0, ), A5 = ( , 0), A6 = ( , 0], and A7 = (0, ).) Now A2 , being the complement of A3 , will be a closed ray with endpoint a together with an isolated point (( , 0] 1 in the example), and if A1 is any non-closed set with closure of this form then A1 , . . . , A7 will be distinct. Next we try to get B1 , . . . , B7 also distinct. If A1 is a ray (either open or closed) with a point removed, together with an isolated point, then B1 has the same form,
−
⊆
⊆
⊆
⊆
∈ ∈
⊆
∈
≥
−∞
−∞
≥
∞ −{ }
∞
2
∞ −∞ ∪ { }
and each of A1 and B1 is a non-closed set whose closure is a closed ray together with an isolated point. Here A1 , . . . , A7 are distinct, as are B1 , . . . , B7 . Unfortunately, A4 , . . . , A7 are the same four rays as B4 , . . . , B7 (in a different order). Here is a specific example.
−∞
− {−1} ∪ {1} A2 = ( −∞, 0] ∪ {1} A3 = (0, ∞) − {1} A4 = [0, ∞) A5 = ( −∞, 0) A6 = ( −∞, 0] A7 = (0, ∞) A1 = (
, 0)
{−1} ∪ [0, ∞) − {1} B2 = {−1} ∪ [0, ∞) B3 = (−∞, 0) − {−1} B4 = (−∞, 0] B5 = (0, ∞) B6 = [0, ∞) B7 = (−∞, 0) B1 =
The trick here is to use that fact that there is a dense set whose complement is also dense (for instance, the rationals) to get the endpoints of the rays in A2 and B2 to be different. This leads to the following example achieving the maximum of 14.
−∞ − − {− } ∪− ∩ ∪{ }
A1 = (
, 1)
2
B1 =
{−2} ∪ [−1, 1] − Q [ 1, 1] Q 2 ∪ (1, ∞) − {2} A2 = (−∞, 1] ∪ {2} B2 = {−2} ∪ [−1, ∞) A3 = (1, ∞) − {2} B3 = ( −∞, −1) − {−2} A4 = [1, ∞) B4 = ( −∞, −1] A5 = (−∞, 1) B5 = ( −1, ∞) A6 = (−∞, 1] B6 = [ −1, ∞) A7 = (1, ∞) B7 = ( −∞, −1) ¯ we have A ⊆ B. ¯ Since B ¯ is closed, and A¯ is the Problem 17.6. (a) Since B ⊆ B, ¯ ¯ smallest closed set containing A, A ⊆ B. ¯ and A¯ ∪ B ¯ is closed, A ∪ B ⊆ A¯ ∪ B. ¯ Conversely, since (b) Since A ∪ B ⊆ A¯ ∪ B ¯ ⊆ A ∪ B, so A¯ ∪ B ¯ ⊆ A ∪ B. A ⊆ A ∪ B, A¯ ⊆ A ∪ B by part (a). Similarly, B (c) For each α, A¯α ⊆ Aα by part (a), so A¯α ⊆ Aα . For the counterexample, any collection (necessarily infinite) of closed sets whose union is not closed will do. For a specific example, let X = R and Aq = {q} for q ∈ Q. Then q∈ A¯q = Q,
while
¯ q∈Q Aq = Q = R.
Q
∈ × × × ∩ × ∩ × ∈ ∈ × ∈ × × × ∈ × ∩ × ∩ × ∩
Problem 17.9. First suppose (x, y) A B. If U is any neighborhood of x, U Y is a neighborhood of (x, y), so (U Y ) (A B) = (U A) B is non-empty. ¯ Similarly, y B, ¯ so (x, y) A¯ B. ¯ Hence U A = , so x A. ¯ Any neigborhood of (x, y) contains one of Conversely, suppose (x, y) A¯ B. ¯ U the form U V , and it suffices to show that U V meets A B. Since x A, meets A, and similarly V meets B. Hence (U V ) (A B) = (U A) (V B) is non-empty.
∩ ∅ ×
∈
∩
× ∩
Problem 17.13. Note that if A and B are subsets of X , A B is empty iff A B ∆ is empty. Suppose first that X is Hausdorff. We show that if (x, y) / ∆, then (x, y) is not a limit point of ∆. For x = y, so there are disjoint open sets U x and V y. Now U V is a neighborhood of (x, y) disjoint from ∆.
×
3
∈
∈
Now suppose ∆ is closed, and let x = y X . Then (x, y) is not in ∆, so it is not a limit point of ∆, and hence there is a neighbourhood of ( x, y) of the form U V disjoint from ∆. Now U and V are disjoint neighborhoods of x and y.
×
Problem 18.6. Define f by
0 if x Q; x if x / Q.
f (x) =
∈ ∈
Recall that f is continuous at x if, for every neighborhood V of f (x), there is a neighborhood U of x with f (U ) V . If V is a neighborhood of f (0) = 0, then f (V ) V , so f is continuous at 0. If x is irrational, then V = R 0 is a neighborhood of f (x) = x, but any neighborhood of x contains a rational y and f (y) = 0 / V , so f is not continuous at x. Finally, if x is rational and non-zero then V = ( x , x ) is a neighborhood of f (x) = 0, but any neighborhood of x contains an irrational y with y > x and f (y) = y / V , so f is not continuous at x. Thus f is continuous precisely at 0.
⊆
⊆
∈
−| | | |
−{ }
|| ||
∈
Problem 18.7(a). The condition on f means that, for all a R and > 0, there exists δ > 0 such that a x < a + δ implies f (x) f (a) < . Let V be open in R and a f −1 (V ). There exists > 0 such that (f (a) , f (a) + ) V . Let δ be as above. Then [a, a + δ) is a neighborhood of a in R and [a, a + δ) f −1 (V ). It R is continuous. follows that f −1 (V ) is open in R , and hence f : R
≤
∈
|
−
−
|
∈
⊆ ⊆
→
{ | ∈
≤ ∞
}
∈
Problem 18.8. (a) Let A = x f (x) g(x) , and suppose a / A, so that f (a) > g(a). If there is some y0 Y with f (a) > y0 > g(a), let V 1 = (y0 , ) and V 2 = ( , y0 ); if not, let V 1 = (g(a), ) and V 2 = ( , f (a)). In either case, V 1 and V 2 are neighborhoods of f (a) and g(a), respectively, and if y1 V 1 and y2 V 2 then y1 > y2 . Then U = f −1 (V 1 ) g −1 (V 2 ) is a neighborhood of a, and if x U then f (x) > g (x), so U is disjoint from A. Hence A is closed. (b) Let A be as in (a), and B = x g(x) f (x) . Then X is the union of the closed sets A and B. Since h A = f A and h B = g B are both continuous, h is continuous by the pasting lemma.
−∞
−∞
∩ { | | |
≤ |
}
∞
∈
∈ ∈
|
Problem 19.7. In the box topology, R∞ is closed (so it is equal to its closure). Suppose x = (xn ) / R∞ . Set
∈
U n =
∞
R R
− { 0}
if xn = 0; if xn = 0.
∈ ∈
Then U = n=1 U n is a box neighborhood of x, and if y U then yn = 0 whenever xn = 0, so y / R∞ . In the product topology, the closure of R∞ is Rω . Let x Rω . Any neighborhood ∞ of x contains one of the form U = n=1 U n , where there is some N with U n = R for n N . Set xn if n < N ; yn = 0 if n N .
∈
≥
Then y = (yn )
≥
∈ U ∩ R∞.
Problem 20.2. By problem 16.9 (assignment 2), the dictionary order topology is the same as the product topology Rd R, where Rd denotes R with the discrete
×
4
topology. Since Rd and R are both metrizable, so is Rd giving the dictionary order topology is
× R.
A specific metric
|
− | if x1 = y1 ; { | − y2|} if x1 = y1. Let us verify the triangle inequality. For x, y, z ∈ R2 , we have |x2 − z2| ≤ |x2 − y2| + |y2 − z2| ≤ do(x, y) + do(y, z). If x1 = z1 , |x2 − z2 | = do (x, z), and we are done. Otherwise, y1 is different from at least one of x1 and z1 , so do (x, y) + do (y, z) ≥ 1, and hence do (x, z) ≤ do (x, y) + do (y, z), as required. For ≤ 1, Bd (x, ) = {x1 }× (x2 − , x2 +), so do induces the right topology. Problem 20.3. (a) Let (x1 , x2 ) ∈ X × X , and let V be any neighborhood of s = d(x1 , x2 ). There exists > 0 such that |s − t| < implies t ∈ V . Let U be the neighborhood B(x1 , 2 ) × B(x2 , 2 ) of (x1 , x2 ) in X × X . For (y1 , y2 ) ∈ U , set do (x, y) = do (x1
× x2, y1 × y2) =
x2 y2 max 1, x2
o
t = d(y1 , y2 ). Then
s = d(x1 , x2 )
≤ d(x1, y1) + d(y1, y2) + d(y2, x2) < t +
and
≤ d(y1, x1) + d(x1, x2) + d(x2, y2) < s + . Hence |s − t| < , so t ∈ V . That is, d(U ) ⊆ V , and it follows that d is continuous. (b). Let U be open in X and x ∈ U . Take > 0 so that Bd (x, ) ⊆ U . Since d(x, x) = 0, d−1 (−∞, ) is a neighborhood of (x, x) in X × X . Hence there are neigborhoods V 1 and V 2 of x in X with d(V 1 × V 2 ) ⊆ (−∞, ). For any y ∈ V 2 , (x, y) ∈ V 1 × V 2 , so d(x, y) < . Thus V 2 ⊆ Bd (x, ) and it follows that U is open in t = d(y1 , y2 )
X , as required.
Problem 21.4. In R , a countable base of neighborhoods of x is
{[x, x + 1/n) | n ∈ Z+}.
In I o2 , consider first x y with 0 < y < 1. A countable base of neighborhoods at x y consists of the intervals (x c, x d) with c and d rational, 0 c < y < d 1. Next consider x 0 with 0 < x. Here a countable base of neighborhoods consists of the intervals (a 1, x d) with a and d rational, 0 a < x, 0 < d 1. A countable base of neighborhoods at 0 0 consists of [0 0, 0 d) with d rational, 0 < d 1. The remaining cases are similar.
×
×
× × ×
× ×
×
×
≤ ×
≤ ≤
≤ ≤
∈
≥
Problem 21.8. Let > 0. Take M Z+ such that d(f n (y), f (y)) < /2 for n M and all y X . Since f is continuous (Theorem 21.6), there is a neighborhood U of x such that d(f (y), f (x)) < /2 for y U . Since xn x, there exists N M such that xn U for n N . Now, for n N ,
∈ ∈
≥
d(f n (xn ), f (x)) Hence f n (xn )
→ f (x).
∈ ≥
→
≥
≤ d(f n(xn), f (xn)) + d(f (xn), f (x)) < /2 + /2 = .
Problem 22.2. (a) Clearly p is surjective. (For y Y , y = p(f (y)).) Suppose −1 −1 V Y and U = p (V ) is open in X . Then f (U ) is open in Y . But f −1 (U ) = f −1 ( p−1 (V )) = ( p f )−1 (V ) = id−1 Y (V ) = V . Hence p is a quotient map.
⊆
∈
◦
5
→
◦
(b) The inclusion j : A X is continuous and r j is the identity of A. (The equality r j = id A is equivalent to the condition that r(a) = a for all a A.) The result follows from part (a).
◦
∈
×{ } ⊆ ∩ ∞ ∞
Problem 22.3. Clearly q is continuous and surjective. Let B = R 0 A. The restriction of q to B is a homeomorphism. Suppose V R and U = q −1 (V ) is open in A. Then U B is open in B, so q(U B) is open in R. But q(U B) = V , so q is a quotient map. The set U = x y A y > 0 is open in A, but q(U ) = [0, ) is not open in R. Also F = x y A xy = 1 is closed in A, but q(F ) = (0, ) is not closed in R. Hence q is neither open nor closed.
∩
⊆
∩
{ × ∈ | { × ∈ |
} }
R by g(x y) = x + y 2 . Then g is a continuous Problem 22.4(a). Define g : R2 surjection, and x0 y0 x1 y1 iff g(x0 y0 ) = g(x1 y1 ). By Corollary 22.3, g induces a continuous bijection f : X ∗ R. Also define h: R R2 by h(x) = x 0. ∗ −1 2 If p : R X is the quotient map, then f = p h, which is continuous. Therefore ∗ X is homeomorphic to R.
× ∼ ×
→
×
→
→
×
×
→
◦
×
Note: problems 22.3 and 22.4 can both be done by using problem 22.2(a). Problem 22.6. Note that if U is open in R, then U K is open in RK . (For U is a union of open intervals, U = α∈J (aα , bα ), and U K = α∈J (aα , bα ) K .) In particular, R K is open in RK , so K is closed in RK . (a) Recall that a quotient space is T 1 iff the elements of the partition are closed sets. Here the elements of the partition are the singletons x for x / K , which are closed in R and therefore in RK , and K itself, which is closed in RK . Hence Y is T 1 . Now p(0) and p(K ) are distinct points of Y . Let V 1 and V 2 be neighborhoods of p(0) and p(K ), respectively. Then U 1 = p−1 (V 1 ) and U 2 = p−1 (V 2 ) are open sets in RK containing 0 and K , respectively. There exist a < 0 < b such that 1 (a, b) K U 1 . Take n < b. Then n1 U 2 , so there exist Z+ such that n 1 c < n < d with (c, d) U 2 . (Every basic neighborhood of a point of K is an 1 interval.) We may assume that n+1 c. Then (c, n1 ) U 1 U 2 , so V 1 V 2 is non-empty. Hence Y is not Hausdorff. (b) By problem 17.13, a space X is Hausdorff iff the diagonal is connected in X X . Let ∆1 be the diagonal in R2 , which is closed in the usual topology, and hence in RK RK . Note that since K is closed in RK , K K is closed in RK RK . The diagonal ∆2 in Y Y is not closed by part (a), but ( p p)−1 (∆2 ) = ∆1 K K , which is closed, so p p is not a quotient map.
− −
−
−
{}
− ⊆
×
×
∈
⊆
∈
∈ ⊆ ∩
≤
∩
× ×
× ×
× ∪ ×
n
Problem 23.2. Let Bn = i=1 Ai . We prove by induction on n that Bn is connected, the case of B1 = A1 being trivial. Suppose Bn is connected. Then Bn+1 = Bn An+1 and Bn An+1 An An+1 = , so Bn+1 is connected by Theorem 23.3. Now An = Bn and Bn = A1 A1 A2 = , so An is connected by the same theorem.
∪
∩
⊇
∩
∅ ⊇ ∩ ∅
Problem 23.8. Let ρ¯ be the uniform metric on Rω , let A Rω be the set of bounded Rω , Bρ¯(x, 1) is sequences, and B the set of unbounded sequences. For any x contained in A if x A, and is contained in B if x B. Hence A, B is a separation of Rω in the uniform topology, which is therefore not connected.
⊆ ∈
∈
6
∈
{
}
{ }
Problem 24.8. (a) Yes. Let X α α∈J be a family of path-connected spaces, and x = (xα ) and y = (yα ) points of X = X α . For each α J , let f α : [0, 1] X α be a path from xα to yα , and define f : I X by f (s) = (f α (s)). Then f is continuous because each πα f = f α is, and is a path from x to y in X . (b) No. If S R2 is as in Example 7 on p. 156, S is path connected, but the ¯ is not. topologist’s sine curve S
∈
→
◦ ⊆
→
{ }
Problem 24.11. Neither holds. In R, [0, 1] is connected but its boundary 0, 1 is not. In R2 , the union of two closed (Euclidean) discs of radius 1 with centers ( 1, 0) is connected, but its interior is the union of the two open discs, which is not connected. To disprove the converse, consider the disconnected set Q in R. Its interior is empty, and its boundary is R, both of which are connected. (If you’d prefer an example with non-empty interior, take Q ( , 0).)
±
∪ −∞
Problem 25.10. (a) The reflexive and symmetric properties are obvious. Suppose x y and y z, and let A, B be a separation of X with x A. Then x y implies y A, and now y z implies z A. Hence x z. (b) Let C be a component. For any separation A, B , either C A or C B, so x y for all x, y C . That is, C is contained in a single quasicomponent. Suppose X is locally connected. For x X let C be the component and Q the quasicomponent containing x. We already know that C Q. By Theorem 25.4, C is open, as is X C (being the union of the other components of X ), so C, X C is a separation of X . For y Q, x C and x y imply y C , so Q C . (c) Suppose x1 y1 and x2 y2 are in A with x1 < x2 . There exists some x / K with x1 < x < x 2 . Then A ( , x) R and A (x, ) R form a separation of A showing that x1 y1 x2 y2 . Hence each quasicomponent (and therefore each component) of A is contained in n1 [0, 1] for some n Z+ , or in 0 0, 0 1 . Since n1 [0, 1] is connected (indeed path connected), it is both a component and a quasicomponent of A. Since 0 0, 0 1 is disconnected, the remaining components of A are the singletons 0 0 and 0 1 . Suppose that U, V is a separation of A with 0 0 U and 0 1 V . There is some n Z+ with n1 0 U and n1 1 V . But now the connected set n1 [0, 1] meets both U and V , a contradiction. Hence the remaining quasicomponent of A is 0 0, 0 1 . The set B 0 1 is path connected, and hence connected. Hence B = B 0 1 is connected, and is therefore its own unique component and quasicomponent. Let C 1 = K [0, 1], C 2 = K [ 1, 0], C 3 = [0, 1] K and C 4 = [ 1, 0] K . 1 Let m = n Z+ , and suppose that U, V is a separation of C with m 0 U 1 1 −1 1 −1 and n 0 V . There is some k Z+ with m U and n V . But now k k −1 the connected subset [0, 1] k of C meets both U and V , a contradiction. Hence 1 1 0 0, and it follows that C 1 is contained in a single quasicomponent of m n C . Similarly, each of C 2 , C 3 and C 4 is contained in a quasicomponent. Suppose ¯1 U . But C ¯1 contains points of C 4 , U, V is a separation with C 1 U . Then C and it follows that C 1 and C 4 belong to the same quasicomponent, and then similar arguments show that C is its own unique quasicomponent. This implies that there is no separation of C , so C is also its own unique component.
∼
∈
∼
{ ∼
∼
× }
}
× ∈
−{ × }
∈ × ∈ × ∼ × { }
{
∈
∈
×
∈
{
∈
∈
−
{ ×
}
∈ ∼ × ∩ −∞ × × ∼ ×
− ×− { ∈
}
⊆
∼ ⊆
⊆
{
∈
− }
⊆ ∩ ∞× × ∈ { × × } { × } { × } × ∈ ∈ × {× ×}
× ∈
× ∈ −{ × }
×
}
⊆
×
×
∼
∈
×−
× ∈
−
× ∈
× × ∈
⊆
Problem 26.1. (a) If X is compact under T , then it is compact under T , since every cover of X by sets of T is a cover by sets of T , and so has a finite subcover. The 7
converse does not hold: if T is the indiscrete topology, X is compact under T , while if T is the discrete topology, X is not compact under T if X is infinite. (b) Suppose that the topologies are comparable; without loss of generality, T (X, T ) is a continuous bijection from a T . Then the identity map id : (X, T ) compact space to a Hausdorff space, so it is a homeomorphism by Theorem 26.6. This shows that T = T .
⊇
→
×
Problem 26.7. Let A be a closed subset of X Y and B = π1 (A). We show that X B is open. Let x X B. Then x Y is disjoint from A, so it is contained in the open set X Y A. By the tube lemma, there is a neighborhood U of x such that U Y X Y A. Now U X B, and we are done.
−
∈ − × − × ⊆ × −
{ }× ⊆ −
Problem 26.8. Suppose first that f is continuous. Since Gf is the inverse image under the map f idY : X Y Y Y of the diagonal, and the diagonal is closed in Y Y since Y is Hausdorff, Gf is closed. Now suppose that Gf is closed. The hint involves the unnecessary taking of complements, so I’ll ignore it. For B Y and x X ,
×
×
× → ×
⊆
∈ x ∈ f −1 (B) ⇔ x × f (x) ∈ Gf ∩ X × B ⇔ x ∈ π1 (Gf ∩ X × B). That is, f −1 (B) = π1 (Gf ∩ X × B). If now B is closed in Y , then Gf ∩ X × B is closed in X × Y , so f −1 (B) is closed in X by the previous problem. Hence f is continuous.
Problem 27.1. Let A be a non-empty subset of X that is bounded above, and suppose that A does not have a least upper bound. Let B be the set of upper bounds for A. Then B = b∈B (b, ), so B is open. Also X B = a∈A ( , a). Now pick a0 A and b0 B; we have a0 < b 0 . The collection ( , a) a A B is a cover of [a0 , b0 ] by sets open in X . Hence there exist a1 , . . . , an A such n that [a0 , b0 ] B , ai ). Let a be the maximum of a0 , a1 , . . . , an . Then i=1 ( n a [a0 , b0 ] but a / B , ai ), a contradiction. i=1 (
∈
⊆
∈
∞ ∈ ∪ −∞ ∈ ∪ −∞
−
−∞ { −∞ | ∈ } ∪ { } ∈
∈
Problem 27.4. Let (X, d) be a connected metric space and x0 = x1 X . Let r = d(x0 , x1 ) > 0. Define f : X R by f (x) = d(x0 , x). Then f is continuous, f (x0 ) = 0 and f (x1 ) = r, so by the Intermediate Value Theorem, the image of f contains the interval [0, r] and is therefore uncountable. Hence X is uncountable.
→
Problem 28.1. Let ρ¯ be the uniform metric on Rω , and let A Rω be the set consisting of all sequences of 0’s and 1’s. Suppose first that x A. For any y A different from x, ρ¯(x, y) = 1, so Bρ¯(x, 1) A = x and x is not a limit point of A. If x / A, pick m Z+ such that xm = 0 and xm = 1. Let be the minimum of xm and 1 xm . Then Bρ¯(x, ) is disjoint from A, so again x is not a limit point of A.
| |
∈
| − |
∈
∩
{}
∈
⊆
∈
Problem 29.3. The answer to the first part is no. Let X be the rationals with the discrete topology, Y the rationals with the usual topology, and f the identity. Certainly f is a continuous surjection. For x X , x is both a neighborhood of x and a compact set, so X is locally compact. I said you could assume without proof that Q is not locally compact, but I’ll give a proof for completeness. If Q is locally compact at 0, then since Q is Hausdorff there exist a < 0 < b such that the closure of (a, b) Q relative to Q is compact. But the relative closure is [a, b] Q, which is not closed in R and therefore not compact.
∈
∩
{}
∩
8
∈
If f is continuous and open then f (X ) is locally compact. Suppose y = f (x) f (X ). There is a neigborhood U of x and a compact set C X with U C . Now f (U ) is open in f (X ), f (U ) f (C ), and f (C ) is compact by Theorem 26.5.
⊆
⊆
⊆
Problem 30.2. Let J be the set of all pairs (m, n) of positive integers such that there is some C C with Bm C Bn , and for (m, n) J pick C m,n C with Bm C m,n Bn . Then C m,n (m, n) J is a countable subcollection of C; we must show that it is a basis. Let x X and let U be any neighborhood of x. Since Bn U . Since C is a basis, there B = Bn is a basis, there is some n with x is some C C with x C Bn . Finally, since B is a basis, there is some m with x Bm C . Then (m, n) J , and x Bm C m,n Bn U .
⊆ ⊆ { } ∈ ∈ ⊆
∈
⊆ ⊆ | ∈ } ∈
{
∈ ⊆ ∈
∈
∈ ⊆
∈
⊆ ⊆
∈
⊆
Problem 30.5(a). Let d be a metric for the space X , and let D be a countable dense subset of X . Then Bd (x, 1/n) x D, n Z+ is a countable collextion of open subsets of X ; we show that it is a basis. Let x X and let U be a neigborhood of x. There is some n Z+ such that Bd (x, 2/n) U . Since D is dense, there is some y D with d(x, y) < 1/n. Now x Bd (y, 1/n) Bd (x, 2/n) U , and we are done.
{
| ∈
∈ } ∈ ⊆ ∈ ⊆
∈
∈
⊆
Problem 30.16. (a) We show that RJ is separable, where J = (0, 1]. Elements n R. Define φ : RJ by of RJ will be written as functions f : J n∈Z+ Q φ(q1 qn )(s) = qns . (Thus φ(q1 qn ) is a rational-valued step function.) We show that the countable set Im φ is dense. It is enough to show that any nonempty basis element U = s∈J U s meets Im φ. Here U s1 , . . . , Us are non-empty open subsets of R for some s1 , . . . , sk J , and U s = R for s / s1 , . . . , sk . Take n so large that nsi = nsj for i = j. For 1 m n, there is at most one i, 1 i k, with nsi = m. If there is such an i, let qm be a rational number in U s , and if not set qm = 0. Then φ(q1 qn ) U . (b) Let D be a dense subset of RJ . Define φ : J P(D) by φ(α) = πα−1 ((0, )) D. If α = β J , πα−1 ((0, )) πβ−1 (( , 0)) is a non-empty open set, so it meets −1 −1 D. Any point of πα ((0, )) πβ (( , 0)) D is in φ(α) φ(β ), so φ is injective and hence D is uncountable.
×···×
∈
∈{
≤ ≤ ×···× ∈ → −∞ −∞ ∩ −
i
∈
→
k
≤ ≤
→ ×···×
∞ ∩ ∞ ∩
}
∞ ∩
Problem 33.4. If f exists, then A = f −1 (0) is closed. For n Z+ , set U n = ∞ f −1 ([0, n1 )), an open set. Then n=1 U n = A, so A is a Gδ . Suppose then that A ∞ ¯n+1 U n . is a closed Gδ . We show first that A = n=1 U n where U n is open and U ∞ Let A = n=1 V n , V n open, and set U 1 = V 1 . Suppose that open sets U 1 , . . . , Un ¯i+1 have been constructed with U U i for 1 i < n and A U n V n . Since U n V n+1 is an open set containing A, there is an open U n+1 containing A with ¯n+1 U U n V n+1 . Inductively, we construct U n for all n Z+ with the desired properties. ¯n−1 U n for n > 0 and Let U 0 = X , and consider the closed sets Bn = U ¯n U n for n 0. We have Bn Bn+1 = C n , Bn Bm = for m > n + 1, C n = U ∞ and n=1 Bn = X A. Since C n−1 and C n are disjoint closed subsets of the normal 1 space Bn , there is a continuous function f n : Bn [ n+1 , n1 ] such that f n (x) = n1 1 for x C n−1 and f n (x) = n+1 for x C n . For x Bn Bn+1 , f n (x) = f n+1 (x), so there is a well-defined function f : X A (0, 1] with f Bn = f n . To see that f is continuous, we can either observe that Bn is locally finite (see problem 18.9), or use the following cheap trick. By the pasting lemma, f is continuous
∩
⊆
∈
⊆
∈
⊆
≤
⊆
∩
−
−
≥
∩
∩
→ ∈ ∈ ∩ − → { } 9
∈ −
⊆
∅
|
(a) (b) (c) (d) (e) (f ) (g) (h) (i) (j) (k) connected path connected locally connected locally path connected compact limit point compact locally compact T 2 Hausdorff regular completely regular normal first-countable second-countable Lindel¨ of separable locally metrizable metrizable
×
• • • ◦ ◦ • ◦ ◦ ◦ • − − − ◦ − − −
×
×
− − • ◦ ◦ • ◦ ◦ ◦ − ◦ • − −
− − • • ◦ ◦ •
− − • • ◦ ◦ ◦ • ◦ −
×
×
×
×
×
×
×
×
×
×
×
×
Table 1.
×
×
×
×
×
◦ ◦ • − − • − − • ◦ ◦ ◦ − − • • − • − − − − − − − −
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
◦ ◦ ◦ • • • • −
◦ ◦ • • − − • −
◦ ◦ ◦ ◦ ◦ • ◦ ◦ ◦ •
◦ ◦ ◦ ◦ ◦ − − ◦ •
◦ ◦ • ◦ ◦ • • − − ? ◦ − − • − ◦ • ◦ − − • − − −
×
×
×
×
×
Answers to Problem 4S.1
×
×
×
×
×
×
×
×
¯n−1 U n+1 , and therefore on the open set U n−1 U ¯n+1 . But on Bn Bn+1 = U ∞ ¯ X A = n=1 U n−1 U n+1 . Now extend f to f : X [0, 1] by setting f (x) = 0 for x A. Since the restriction of f to the open set X A is continuous, f is continuous at each point of X A. Also f is continuous at each point of A since f (U n ) [0, n1 ).
−
∪
−
−
−
→
−
∈
−
⊆
Problem 33.5. It follows from the previous problem that if such an f exists A and B are closed Gδ sets, and obviously they are disjoint. Suppose then that A and B are closed, disjoint Gδ sets. By the previous problem there are continuous functions f 1 , f 2 : X [0, 1] such that f 1 (x) = 0 iff x A and f 2 (x) = 1 iff x B. By Urysohn’s Lemma, there is a continuous g : X [0, 1] with g(x) = 0 if x A and g(x) = 1 if x B. Set min 12 , max f 1 (x), g(x) if g(x) 12 ; f (x) = max 12 , min f 2 (x), g(x) if g(x) 12 .
→
∈
{ {
→ { {
∈
}} }}
∈ ∈ ≤ ≥
You can check that f is well-defined (hence continuous), and has the desired properties. Problem 33.9. In the uniform topology, RJ is normal, so there is a continuous function f : RJ [0, 1] that vanishes outside B(0, 1) and has f (0) = 1. Now give J R the box topology. Since this is finer than the usual topology, f is still continuous, and it vanishes outside ( 1, 1)J . Let U be a neighborhood of a point a, and take a box neighborhood (aα α , aα + α ) contained in U . The function h: RJ RJ defined by h(x) = ((xα aα )/α ) is easily seen to be a homeomorphism, and the continuous function g = f h vanishes outside U and has g(a) = 1.
→
− − − ◦
→
Problems from the supplementary exercises to Chapter 4 will be numbered as 4S.x. 10
Problem 4S.1. The answers are summarised in Table 1. We have the following implications:
⇒ connected ⇒ locally connected ⇒ limit point compact, locally compact and Lindel¨of ⇒ normal regular and Lindel¨of ⇒ normal normal ⇒ completely regular ⇒ regular ⇒ Hausdorff second-countable ⇒ first-countable, Lindel¨of and separable metrizable ⇒ locally metrizable and normal Lindel¨ of and metrizable ⇒ second countable separable and metrizable ⇒ second countable locally metrizable ⇒ first countable In the table, • and ◦ both mean that the space has the property, but ◦ indicates that this follows from other entries in the same column by the above implications. Similarly, and − both mean the space does not have the property, with − indipath connected locally path connected compact compact and Hausdorff
×
cating that this follows from other entries. We shall make use of the following easy facts. (1) If X is a linearly ordered set in which every non-greatest element has an immediate successor, then X is totally disconnected in the order topology. In particular, this is true of a well-ordered set. (2) A totally disconnected space with at least two points is disconnected. (3) A totally disconnected space is locally connected iff it is discrete. (4) If X Y has any of the properties except limit point compactness, so do X and Y . Below we justify the 90 and × entries. (a) S Ω There is a non-least element of S Ω with no immediate predecessor, so this space is neither connected nor locally connected by (1) – (3) above. That it is limit point compact but not compact is shown in Example 2 of 28. For any α S Ω , [α0 , α] (where α0 is the least element) is a neighborhood of α, and is compact. Hence S Ω is locally compact. It is normal by Theorem 32.4. It is not Lindel¨ of because the open cover [α0 , α) α S Ω has no countable subcover. If A S Ω is countable, then A has an upper bound α and A¯ [α0 , α] = S Ω , so S Ω is not separable. It is locally metrizable because each α has the neighborhood [α0 , α] which is regular and secondcountable (because it has only countably many intervals), and therefore metrizable by Theorem 34.1. Finally, as observed in Example 3 of 28, S Ω is not metrizable because it is limit point compact but not compact. ¯Ω (b) S This space is neither connected nor locally connected by (1) – (3) above. It is compact by Theorem 27.1, and Hausdorff because every order topology is Hausdorff. It is not first countable because it does not satisfy the sequence ¯Ω , but any sequence in S Ω has an upper bound in S Ω , and lemma: Ω S ¯Ω then therefore does not converge to Ω. If A is any countable subset of S
×
•
§
∈
⊆
{
⊆
| ∈ }
§
∈
11
¯Ω , so S ¯Ω A S Ω has an upper bound α S Ω . Hence A¯ [α0 , α] Ω = S is not separable. ¯Ω (c) S Ω S This space is not normal by Example 2 of 32. All other negative results follow from (4) above and the first two columns of the table. To show that it is limit point compact, it is enough to show that any countably infinite subset A has a limit point. Now π1 (A) is countable, so it has an upper ¯Ω , and bound α S Ω . Now A is contained in the compact space [α0 , α] S ¯Ω , so has a limit point. Any point has a neighborhood of the form [α0 , α] S ¯Ω is locally compact. Finally, S Ω S ¯Ω is completely regular because so S Ω S it is a product of completely regular spaces. (d) The ordered square I o2 Example 6 of 24 shows that this space is connected but not path connected. It has a basis consisting of intervals, which are connected since I o2 is a linear continuum. We show that the path components of I o2 are the sets x I , x I ; since these are not open sets, I o2 is not locally path connected. Each such set is homeomorphic to I , so it is path connected. Suppose f is a path from x1 y1 to x2 y2 with x1 < x 2 . By the Intermediate Value Theorem 24.3, the image of f contains (x1 , x2 ) I . For x1 < x < x 2 , the non–empty open subset f −1 (x (0, 1)) of I contains a rational number qx , and we get an injection x qx of (x1 , x2 ) into Q, which is absurd. This space is compact by Theorem 27.1, and Hausdorff because every order topology is Hausdorff. That it is first countable is Problem 21.4. If D is any dense subset of I o2 , D meets x (0, 1) for any x I , so π1 maps D onto I . Therefore I o2 is not seperable. Finally, any neighborhood of 0 1 contains some set [a, b] I , which is homeomorphic to I o2 , and hence Lindel¨ of but not second countable, and therefore not metrizable. Thus I o2 is not locally metrizable. (e) R This space is totally disconnected; for x < y the separation
∩ ×
∈
⊆
∪{ }
§
∈
×
×
×
×
§
×
∈
×
→
× ×
×
×
×
∈
×
{(−∞, y), [y, ∞)} shows that x and y are in different components. Hence R is neither connected nor locally connected by (2) and (3). The subset Z has no limit point, so R is not limit point compact. Being Hausdorff, if it is locally compact, there is some > 0 such that [0, ) = [0, ) is compact. But [0, δ) 0 < δ < ) is an open cover of [0, ) with no finite subcover. By Example 2 of 31, R is normal, and by Example 3 of 30 it is first countable, Lindel¨ of and separable, but not second countable. Any non-empty open subset of R contains an interval (a, b), which in the subspace topology is homeomorphic to R . Since (a, b) is separable but not second countable, it is not metrizable, and so R is not locally metrizable.
{
|
§
§
(f) R2 This space is not normal by Example 3 of 31, and it is not limit point compact because Z2 has no limit point. Since R is neither connected, locally connected, locally compact nor locally metrizable, the same is true of R2 ; since R is completely regular, first countable and separable, so is R2 .
§
12
(g) Rω in the product topology This space is path connected and locally path connected since it is a product of path connected and locally path connected spaces. No set containing a neighborhood of the origin is limit point compact, since it contains an infinite set 0 0 Z 0 0 with no limit point. This ω implies that R is neither limit point compact nor locally compact. It is second countable by Theorem 30.2, and metrizable by Theorem 20.5. (h) Rω in the uniform topology This space is not connected by Problem 23.8. For any x Rω and > 0, we show that Bρ¯(x, ) is path connected, so this space is locally path connected. Rω by f (t) = For any y Bρ¯(x, ), let d = ρ¯(x, y) < and define f : I (f n (t)) = ((1 t)xn + tyn ). For s, t I we have f n (s) f n (t) ds t, and hence ρ¯(f (s), f (t)) d s t . Therefore f is continuous, and taking t = 0 shows that f maps into Bρ¯(x, ); it is a path from x to y, so the proof is complete. 1 Rω For any r with 0 < r 2 , let Ar be the set of all a = (an ) with an = r for all n. For a = b Ar , ρ¯(a, b) = 2r. Suppose that x is a limit point of Ar . Take a Ar with ρ¯(x, a) < r and a = x. Then Bρ¯(x, r) Ar = a , and Bρ¯(x, ρ¯(x, a)) Ar = , a contradiction. It follows that this space is neither limit point compact nor locally compact. If D is any dense set, for each a A1/2 choose f (a) D Bρ¯(a, 1/2). Then f is an injection from A1/2 to D, so Rω is not separable. It is metrizable by definition of the uniform topology. (i) Rω in the box topology This space is not connected by Example 6 of 23; recall from this example that A, B is a separation, where A is the set of bounded sequences and B is the set of unbounded sequences. For any a Rω with an = 0 for all Rω given by ha (x) = (an xn ) is easily seen to n, the function ha : Rω be a homeomorphism. Any neighborhood U of 0 contains a point x with xn = 0 for all n. Let an = xn /n. The homeomorphism ha takes 0 to 0 and the sequence (n) to x, so 0 and x lie on opposite sides of the separation ha (A), ha (B) , and this space is not locally connected. Let D = 0, 1 and consider the set Dω of all sequences of 0’s and 1’s. For any x Rω , there is a neighborhood U n of xn with U n D xn for ω each n. Now U = U n is a box neighborhood of x with U D x . ω Thus D has no limit points, and this space is not limit point compact. For any neighborhood U of 0, take a U with an = 0 for all n. The ω homeomorphism ha takes D into U , so this space is not locally compact. It is completely regular by Exercise 33.9; it is unknown whether it is normal: see Exercise 32.5. It is not first countable by Example 1 of 21, and it is neither Lindel¨of nor separable because the coarser uniform topology has neither of these properties. (j) RI in the product topology, where I = [0, 1] This space is path connected and locally path connected since it is a product of path connected and locally path connected spaces. No set containing a neighborhood of the origin is limit point compact, since it contains an infinite set s∈I As , where some As0 is Z and As = 0 for s = s0 , with no limit point. This implies that RI is neither limit point compact nor locally
× · ·· × × × × ·· · × × ·· ·
∈
∈
−
≤ | −| ≤
±
∩
{}
∈
|
∈ ∈ ∩
∈
{
−
→
|≤ | − | ∈
∅ ∈ ∩
§
}
∈
→
{
}
∈
{ }
∩ ⊆{ } ∩ ⊆{ }
∈
§
{}
13
compact. It is completely regular because it is a product of completely regular spaces, and not normal by Example 1 of 32 (see the supplementary notes). It is not first countable by Example 2 of 21, and is separable by Exercise 30.16.
§
§
(k) RK The open set (0, ) inherits its usual topology, in which it is connected, so (0, ) = [0, ) is connected. The open set R K also inherits its usual topology, which has a number of consequences. One component of R K is ( , 0], which is not open in RK , so RK is not locally connected, but as the union of the connected subsets ( , 0] and [0, ) it is connected. As the union of the open metrizable subsets (0 , ) and R K , it is locally metrizable. Since the identity, considered as a map from RK to R, is continuous, any connected subspace of RK is connected in the usual topology, and hence RK is a path with convex. Suppose, for a contradiction, that f : I f (0) 0 and f (1) > 0. If f (s) < 0, there is some t > s with f (t) = 0, since f ([s, 1]) is connected. Since f −1 (( , 0]) is closed, it has a greatest element s0 < 1, and f ((s0 , 1]) (0, ). Since R K is a neighborhood of 0, there is some s > s0 with f ([s0 , s]) R K . But f ([s0 , s]) is connected, so [0, f (s)] f ([s0 , s]), a contradiction. Hence RK is not path connected. That RK is Hausdorff but not regular is Example 1 of 31. No subset of K has a limit point in RK , so no set containing infinitely many points of K (in particular RK ) is limit point compact. If RK were locally compact, ¯ then since it is Hausdorff there would be a neighborhood U of 0 with U ¯ contains infinitely many points of K , a contradiction. compact. But U Finally RK is second countable since it has a basis consisting of all sets (a, b) and (a, b) K with a and b rational.
∞ −∞
∞
∞
−
−∞
−
∞
∞
−
→
≤
−∞ ⊆ ∞ ⊆ −
⊆
−
§
−
The answers to Problems 4S.2–4S.6 are summarized in Table 2. The column headings 4A, 4C and 4O refer to arbitrary, closed and open subspaces in Problem 4S.4, respectively, and 5A, 5C and 5F to arbitrary, countable and finite products in Problem 4S.5. In the columns for Problems 4S.2 and 4S.3, the use of and indicates implications within the column as before; in the columns for Problems 4S.4 and 4S.5, it indicates the trivial implications between the parts of each problem. There are 83 entries in need of justification.
◦
−
Problem 4S.2. The one is obvious, and from Problem 4S.1, Rω in the uniform topology gives all × entries except for local connectedness. The space K = 0 1 Z+ R is metrizable but not locally connected. n n
•
{ }∪
{ | ∈ }⊆
Problem 4S.3. The entries.
• entries are obvious, and from Problem 4S.1, S ¯Ω gives all
×
Problem 4S.4. The space R is path-connected and locally path-connected, while its closed subspace K is neither connected nor locally connected, and its open subspace R 0 is not connected. This gives the negative entries in the first 4 rows. That an open subset of a locally (path) connected space is locally (path) connected follows trivially from the definitions. The space [0, 1] is compact (hence limit-point compact), but its open subspace (0, 1) is not limit-point compact (hence not compact), and the subspace Q of the
−{ }
14
2 connected path connected locally connected locally path connected compact limit point compact locally compact T 2 Hausdorff regular completely regular normal first-countable second-countable Lindel¨ of separable locally metrizable metrizable Table 2.
3
4A 4C 4O 5A 5C 5F
− − − − − − − − − • − • ◦ − • ◦ • ◦ • • ◦ ◦ ◦ • ◦ ◦ ◦ • ◦ ◦ ◦ − • ◦ • ◦ − − • ◦ ◦ − • − ◦ − • ◦ • − • ◦
×
×
×
×
×
×
×
× ×
×
× ×
×
×
×
×
× ×
×
◦ ◦ • ◦ ◦ • • • • • ◦ ◦ • − • • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ − ◦ • ◦ ◦ • ◦ − − • • • ◦ • ◦ − • ◦ • ◦
×
• • − − • − − • • • −
×
×
×
×
×
×
×
× × × ×
×
×
×
×
×
×
×
×
6
×
×
×
×
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Answers to Problems 4S.2–4S.6
locally compact Hausdorff space R is not locally compact (see the answer to Problem 29.3). That a closed subset of a compact space is compact is Theorem 26.2, and that a closed or open subset of a locally compact Hausdorff space is locally compact is Corollary 29.3. It is easy to see that a closed subset A of a limit point compact space X is limit point compact: if B is an infinite subset of A and x is a limit point of B in X , then x A and hence is a limit point of B in A. That a subspace of a Hausdorff, regular or completely regular space is of the same kind is Theorem 31.2 or 33.2, and that a closed subspace of a normal space is normal is easy. Examples 1 and 2 of 32 both give non-normal open subsets of normal spaces. That a subspace of a first or second countable space is of the same kind is Theorem 30.2, and is easy to see that a closed subspace of a Lindel¨of space is Lindel¨ of, and that an open subspace of a separable space is seperable. Example 5 of 30 gives an open subspace of a Lindel¨of space that is not Lindel¨o f. The Sorgenfrey plane R2 is separable, but the closed subspace x ( x) x R is uncountable and discrete, so it is not separable. Finally, it is easy to see that a subspace of a metrizable space is metrizable (this is Problem 21.1), and it follows that a subspace of a locally metrizable space is locally metrizable.
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Problem 4S.5. That a product of connected spaces is connected is Problem 23.10, and the corresponding result for path connected spaces is Problem 24.8(a). That a finite product of locally (path) connected spaces is locally (path) connected follows, since in the finite case any product of open sets is open. Let X be a space that is locally connected but not connected, and let J be an infinite set. If U is any non-empty open subset of X J then U contains U α where U α = X for all but 15
finitely many α. For any α with U α = X , the projection πα maps U onto X , so U is not connected. The same argument works for local path-connectedness. That a product of compact spaces is compact is Tychonoff’s Theorem. In Exercise 28.3, Munkres refers you to Example 112 of Steen and Seebach for a product of two limit point compact spaces that is not limit point compact, and I can do no ˇ better. (The example starts from the Stone- Cech compactification of the integers, and it gets worse from there.) That a finite product of locally compact spaces is compact follows from (the finite case of) Tychonoff because in the finite case any product of open sets is open. Let X be a space that is locally compact Hausdorff but not compact, and let J be an infinite set. If U is any non-empty open subset of X J then U contains U α where U α = X for all but finitely many α. For any α ¯ onto X , so U ¯ is not compact. with U α = X , the projection πα maps U That a product of Hausdorff, regular or completely regular spaces is of the same kind is Theorem 31.2 or 33.2, while Example 2 of 32 gives a product of two normal spaces that is not normal. A countable product of first or second countable spaces is of the same kind by Theorem 30.2, while R is second countable but RI is not first countable (problem 4S.1(j)). A product of two Lindel¨of spaces need not be Lindel¨ of by Example 4 of 30. That a product of separable spaces indexed by I (and hence any countable product) is seperable is an easy adaptation of the solution to Problem 30.16(a), while part (b) of that problem shows that this is not the case for arbitrary products. That a countable product of metrizable spaces is metrizable was proved in Theorem N20.1 in the notes, and it follows that a finite product of locally metrizable spaces is locally metrizable by an argument seen twice already. An uncountable product of metrizable spaces need not be metrizable by Example 2 of 21. Let X be a space that is locally metrizable but not metrizable (for instance, S Ω or RK by problem 4S.1), and let J be an infinite set. Any non-empty open subset of X J contains a subspace homeomorphic to X , so U is not metrizable.
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Problem 4S.6. That connectedness, path connectedness and compactness are preserved by continuous maps is Theorem 23.5, Problem 24.8(a) and Theorem 26.5, and the proof for the Lindel¨of property is virtually the same as for compactness. That the continuous image of a separable space is separable follows from the rela¯ tion f (A) f (A) for continuous f . Most of the negative entries can be seen by the following cheap trick. Suppose some property is posessed by all discrete spaces, and that X is a space not posessing it. If Y is the set X with the discrete topology, the identity considered as a map Y X shows that the property is not preseved by continuous maps. Any discrete space is trivially locally connected, locally path connected and locally compact. It is also metrizable, and hence locally metrizable, first countable, normal, completely regular, regular and Hausdorff. The same trick will work for second countability provided we can find a countable space X that is not second countable. The following is Example 26 of Steen and Seebach, who call it Arens-Fort space. As a set, X = Z2 . The open sets are of two kinds.
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(1) Any set not containing 0 0. (2) A set which, for all but finitely many integers m, contains all but finitely many points of the column m Z.
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This is easily seen to be a topology, and clearly 0 0 is in the closure of X 0 0 . We show that no sequence (xn ) in X 0 0 converges to 0 0, so X is not even first countable. If there are finitely many columns containing all the xn , then 16
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X xn n Z+ is a neighborhood of 0 0 showing that (xn ) does not converge to 0 0. Otherwise, we can take a subsequence (xn ) such that each column contains at most one term, and then X xn k Z+ is a neighborhood of 0 0 showing that (xn ), and hence (xn ), does not converge to 0 0. Finally, it is observed in Example 1 of 28 that X = Z+ Y is limit point compact, where Y is an indiscrete two-point space, and projection on the first factor is a continuous map onto the non-compact space Z+ . (It is easy to see that countable compactness is preserved by continuous maps, so X has to be non-T 1 here.) k
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