1st December 2004

Munkres §30

Ex. 30.3 (Morten Poulsen). Let X be second-countable and let A be an uncountable subset of X. Suppose only countably many points of A are limit points of A and let A_{0} ⊂ A be the countable set of limit points.

For each x ∈ A − A_{0} there exists a basis element U_{x} such that x ∈ U_{x} and U_{x} ∩ A = {x}. Hence if a and b are distinct points of A − A_{0} then U_{a} ≠ U_{b}, since U_{a} ∩ A = {a} ≠ {b} = U_{b} ∩ A. It follows that there uncountably many basis elements, contradicting that X is second-countable.

Note that it also follows that the set of points of A that are not limit points of A are countable.

Ex. 30.4 (Morten Poulsen).

Theorem 1. Every compact metrizable space is second-countable.

Proof. Let X be a compact metrizable space, and let d be a metric on X that induces the topology on X.

For each n ∈ Z_{+} let A^{n} be an open covering of X with 1/n-balls. By compactness of X there exists a finite subcovering A_{n}.

Now B = ⋃_{n∈Z_{+}} A_{n} is countable, being a countable union of finite sets.

B is a basis: Let U be an open set in X and x ∈ U. By definition of the metric topology there exists ε > 0 such that B_{d}(x, ε) ⊂ U. Choose N ∈ Z_{+} such that 2/N < ε. Since A_{N} covers X there exists B_{d}(y, 1/N) containing x. If z ∈ B_{d}(y, 1/N) then

d(x, z) ≤ d(x, y) + d(y, z) ≤ 1/N + 1/N = 2/N < ε,

i.e. z ∈ B_{d}(x, ε), hence B_{d}(y, 1/N) ⊂ B_{d}(x, ε) ⊂ U. It follows that B is a basis. □

Ex. 30.5. Let X be a metrizable topological space.

Suppose that X has a countable dense subset A. The collection {B(a, r) | a ∈ A, r ∈ Q_{+}} of balls centered at points in A and with a rational radius is a countable basis for the topology: It suffices to show that for any y ∈ B(x, ε) there are a ∈ A and r ∈ Q_{+} such that y ∈ B(a, r) ⊂ B(x, ε). Let r be a positive rational number such that 2r < ε−d(x, y) and let a ∈ A∩B(y, r). Then y ∈ B(a, r), of course, and B(a, r) ⊂ B(x, ε) for if d(a, z) < r then d(x, z) ≤ d(x, y) + d(y, z) ≤ d(x, y) + d(y, a) + d(a, z) < d(x, y) + 2r < ε.

Suppose that X is Lindelöf. For each positive rational number r, let A_{r} be a countable subset of X such that X = ⋃_{a∈A_{r}} B(a, r). Then A = ⋃_{r∈Q_{+}} A_{r} is a dense countable subset: Any open ball B(x, ε) contains a point of A_{r} when 0 < r < ε, r ∈ Q.

We now have an extended version of Thm 30.3:

Theorem 2. Let X be a topological space. Then

\xymatrix{X has a countable dense subset & X is 2nd countable \ar@{=>}[l] \ar@{=>}[r] \ar@{=>}[d] & X is Lindelöf \\  & X is 1st countable}

If X is metrizable, the three conditions of the top line equivalent.

Ex. 30.6. R_{ℓ} has a countable dense subset and is not 2nd countable. According to [Ex 30.5] such a space is not metrizable.

The ordered square I_{o}^{2} is compact and not second countable. Any basis for the topology has uncountably many members because there are uncountably many disjoint open sets (x × 0, x × 1), x ∈ I, and each of them contains a basis open set. (Alternatively, note that I_{o}^{2} contains the uncountable discrete subspace \{x × \frac{1}{2} | x ∈ I\} so it can not be second countable by [Example 2 p 190].) According to [Ex 30.4] or [30.5(b)] a compact space with no countable basis is not metrizable.

Ex. 30.7. (Open ordinal space and closed ordinal space) Sets of the form (α, β), −∞ ≤ α < β ≤ +∞, form bases for the topologies on the open ordinal space S_{Ω} = [0, Ω) and the closed ordinal space \overline{S}_{Ω} = [0, Ω] [§14, Thm 16.4]. The sets (α, β] = (α, β + 1) = [α + 1, β] are closed and open. Let n denote the nth immediate successor of the first element, 0.

[0, Ω) is first countable: {0} = [0, 1) is open so clearly [0, Ω) is first countable at the point 0. For any other element, α > 0, we can use the collection of neighborhoods of the form (β, α] for β < α.

[0, Ω) does not have a countable dense subset: The complement of any countable subset contains [Thm 10.3] an interval of the form (α, Ω) (which is nonempty, even uncountable [Lemma 10.2]).

[0, Ω) is not second countable: If it were, there would be a countable dense subset [Thm

30.3]. [0, Ω) is not Lindelöf: The open covering consisting of the sets [0, α), α < Ω, does not

contain a countable subcovering. [0, Ω] is not first countable at Ω: This is a consequence of [Lemma 21.2] in that Ω is a limit point of [0, Ω) but not the limit point of any sequence in [0, Ω) for all such sequences are bounded [Example 3, p. 181].

[0, Ω] does not have a dense countable subset: for the same reason as for [0, Ω).

[0, Ω] is not second countable: It is not even first countable.

[0, Ω] is Lindelöf: It is even compact [Thm 27.1]. S_{Ω} = [0, Ω) is limit point compact but not compact [Example 2, p. 179] so it can not be

metrizable [Thm 28.2]. S_{Ω} is first countable and limit point compact so it is also sequentially compact [Thm 28.2].

\overline{S}_{Ω} = [0, Ω] is not metrizable since it is not first countable.

Ex. 30.9. A space X is Lindelöf if and only if any collection of closed subsets of X with empty intersection contains a countable subcollection with empty intersection. Since closed subsets of closed subsets are closed, it follows immediately that closed subspaces of Lindelöf spaces are Lindelöf.

The anti-diagonal L ⊂ R_{ℓ} × R_{ℓ} is a closed discrete uncountable subspace [Example 4 p 193]. Thus the closed subset L does not have a countable dense subset even though R_{ℓ} × R_{ℓ} has a countable dense subset.

Ex. 30.12. Let f: X → Y be an open continuous map.

Let B be a neighborhood basis at the point x ∈ X. Let f(B) be the collection of images f(B) ⊂ f(X) of members B of the collection B. The sets in f(B) are open in Y, and hence also in f(X), since f is an open map. Let f(x) be a point in f(X). Any neighborhood of f(x) has the form V ∩ f(X) for some neighborhood V ⊂ Y of f(x). Since p^{−1}(V) is a neighborhood of x there is a set B in the collection B such that x ∈ B ⊂ f^{−1}(V). Then x ∈ f(B) ⊂ V ∩ f(X). This shows that f(B) is a neighborhood basis at f(x) ∈ f(X).

Let B be a basis for the topology on X. Let f(B) be the collection of images f(B) ⊂ f(X) of members B of the collection B. The sets in f(B) are open in Y, and hence also in f(X), since f is an open map. Since B ia a covering of X, f(B) is a covering of f(X). Suppose that f(x) ∈ f(B_{1}) ∩ f(B_{2}) where x ∈ X and B_{1}, B_{2} are basis sets. Choose a basis set B_{3} such that x ∈ B_{3} ⊂ f^{−1}(f(B_{1}) ∩ f(B_{2})). Then f(x) ∈ f(B_{3}) ⊂ f(B_{1}) ∩ f(B_{2}). This shows that f(B) is a basis for a topology T_{f(B)} on f(X). This topology is coarser than the topology on f(X) since the basis elements are open in f(X). Conversely, let f(x) ∈ V ∩ f(X) where V is open in Y. Choose a basis element B such that x ∈ B ⊂ f^{−1}(V). Then f(x) ∈ f(B) ⊂ V ∩ f(X). This shows that all open subsets of f(X) are in T_{f(B)}. We conclude that f(B) is a basis for the topology on f(X).

We conclude that continuous open maps preserve 1st and 2nd countability.

Ex. 30.13. Let D be a countable dense subset and U a collection of open disjoint subsets. Pick a member of D inside each of the open open sets in U. This gives an injective map U → D. Since D is countable also U is countable.

Ex. 30.16. For each natural number k ∈ Z_{+}, let D_{k} be the set of all finite sequences

(I_{1}, …, I_{k}, x_{1}, …, x_{k})

where I_{1}, …, I_{k} ⊂ I are disjoint closed subintervals of I with rational endpoints and x_{1}, …, x_{k} ∈ Q are rational numbers. Since D_{k} is a subset of a countable set,

D_{k} ↪ (\overbrace{Q × Q) × ⋯ × (Q × Q)}^{k} × \overbrace{Q × ⋯ × Q}^{k} = Q^{3k},

D_{k} itself is countable [Cor 7.3]. Put D = ⋃_{k∈Z_{+}} D_{k}. As a countable union of countable sets, D is countable [Thm 7.5].

For each element (I_{1}, …, I_{k}, x_{1}, …, x_{k}) ∈ D_{k}, let x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) ∈ R^{I} be the element given by

π_{t}x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) = \begin{cases}x_{j} & t ∈ I_{j} for some j ∈ \{1, …, k\} \\ 0 & t ∉ I_{1} ∪ ⋯ ∪ I_{k} \end{cases}

where π_{t}: R^{I} → R, t ∈ I, is the projection map. This defines a map x: D → R^{I}.

(a). The basis open sets in R^{I} are finite intersections ⋂_{j=1}^{k} π_{i_{j}}^{−1}(U_{i_{j}}) where i_{1}, …, i_{k} are k distinct points in I and U_{i_{1}}, …, U_{i_{k}} are k open subsets of R. Choose disjoint closed subintervals I_{j} such that i_{j} ∈ I_{j} and choose x_{j} ∈ U_{i_{j}} ∩ Q, j = 1, …, k. Then x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) ∈ ⋂_{j=1}^{k} π_{i_{j}}^{−1}(U_{i_{j}}) for π_{i_{j}}x(I_{1}, …, I_{k}, x_{1}, …, x_{k}) = x_{j} ∈ U_{i_{j}} for all j = 1, …, k. This shows that any (basis) open set contains an element of x(D), ie that the countable set x(D) is dense in R^{I}.

(b). Let D be a dense subset of R^{J} for some set J. Let f: J → P(D) be the map from the index set J to the power set P(D) of D given by f(j) = D∩π_{j}^{−1}(2003, 2004). Let j and k be two distinct points of J. Then f(j) ≠ f(k) for

f(j) − f(k) = (π_{j}^{−1}(2003, 2004) − π_{k}^{−1}(2003, 2004)) ∩ D

⊃ (π_{j}^{−1}(2003, 2004) ∩ π_{k}^{−1}(2002, 2003)) ∩ D ≠ ∅

since D is dense. This shows that f is injective. Thus card J ≤ card P(D).

[figure: R_{k}; R_{j}; _{2003 2004}; {2004 \atop 2003}]

References