separable space in nLab
To be explicit, XX is separable if there exists an infinite sequence a:ℕ→Xa\colon \mathbb{N} \to X such that, given any point bb in XX and any neighbourhood UU of bb, we have a i∈Ua_i \in U for some ii.
A second-countable space is separable (and trivially it is first-countable). However, first-countability plus separability do not imply second-countability; a counterexample is ℝ\mathbb{R} with the half-open topology (see Munkres, page 192), denoted ℝ l\mathbb{R}_l.
An arbitrary product of separable spaces need not be separable, but a product of as many as a continuum number of separable spaces is separable (with the product computed in Top); a proof of the more general Hewitt-Marczewski-Pondiczery theorem can be found here. In particular, the space ℝ ℝ\mathbb{R}^\mathbb{R} of all functions ℝ→ℝ\mathbb{R} \to \mathbb{R} under pointwise convergence is separable, but is not even first-countable (and thus not second-countable either). A first-countable space need not be separable; a simple example of that is a discrete space of uncountable cardinality.
Subspaces of separable spaces need not be separable. Example: the product ℝ l×ℝ l\mathbb{R}_l \times \mathbb{R}_l, also called the Sorgenfrey plane, is separable, but the subspace defined by the equation y=−xy = -x is uncountable and discrete and therefore not separable. However, open subspaces of separable spaces are separable.
Proof
The second property is implied by the first: given any dense subset {x n∣n∈ℕ}\{x_n \mid n\in \mathbb{N}\}, form the countable system of sets {B 1/m(x n)∣n,m∈ℕ}\{B_{1/m}(x_n) \mid n,m\in\mathbb{N}\}. To see that this is indeed a base for the topology of XX, take any open set U⊂XU\subset X, a point x∈Ux\in U and a radius 1/k1/k such that B 1/k(x)⊂UB_{1/k}(x)\subset U. By separability there is some nn such that x n∈B 1/(2k)(x)x_n\in B_{1/(2k)}(x) and therefore x∈B 1/(2k)(x i)⊂Ux\in B_{1/(2k)}(x_i)\subset U.
To show (2) implies (3), let {U n}\{U_n\} be a countable base of the topology. Given any open cover {V λ}\{V_\lambda\} of XX, we can form the index set I⊂ℕI\subset \mathbb{N} of those nn that are contained in some V λV_\lambda. By assumption ⋃ i∈IU i=⋃ λV λ=X\bigcup_{i\in I} U_{i} = \bigcup_\lambda V_\lambda = X. The axiom of countable choice provides now a section of ⨆ i∈I{λ∣U i⊂V λ}→I\bigsqcup_{i\in I} \{\lambda \mid U_i \subset V_\lambda\}\to I.
Finally, we prove that (3) implies (1). Consider the open covers {B 1/1(x)∣x∈X}\{B_{1/1}(x) \mid x\in X\}, {B 1/2(x)∣x∈X}\{B_{1/2}(x) \mid x\in X\}, … From each extract a countable subcover corresponding to collection of centers A 1,A 2,…A_1, A_2, \ldots. We claim that that the union A 1∪A 2∪…A_1\cup A_2\cup\ldots forms a dense set. Indeed, given any y∈Xy\in X and nn the point xx has to be contained in some B 1/n(x)B_{1/n}(x) for some x∈A nx\in A_n.
Similar in spirit to (1)⇔\Leftrightarrow(2) but less well-known is the following.
Proof
One direction is not hard: if x ix_i is a countable dense subset of XX and r jr_j is an enumeration of the rationals, then according to the proof of Theorem , the balls B r j(x i)B_{r_j}(x_i) form a countable base (XX is a second-countable space). Hence every open set is a union of a family of such balls that is at most countable.
The other direction is trickier. (This is based on a MathOverflow discussion, which for the moment we record with little adaptation.) Suppose XX is not separable; construct by recursion a sequence x βx_\beta of length ω 1\omega_1 such that no x βx_\beta lies in the closure of the set of its predecessors x αx_\alpha (each such set is countable and therefore not dense, so such x βx_\beta outside its closure can be found at each stage).
Therefore, for each x βx_\beta we may choose a rational radius r βr_\beta such that the ball B r β(x β)B_{r_\beta}(x_\beta) contains no predecessor x αx_\alpha. There are uncountably many β\beta, so some rational radius rr was used uncountably many times. The collection of x βx_\beta for which r β=rr_\beta = r forms another ω 1\omega_1-sequence which we now use instead of the first; since all the radii are the same, this sequence has the property that each B r(x β)B_r(x_\beta) contains no other x αx_\alpha, no matter whether α\alpha appears before or after β\beta in the sequence.
Provided that there uncountably many such x αx_\alpha that are non-isolated in XX, we can construct an open set UU that is not a countable union of balls. For around each such x αx_\alpha we can find a point p α≠x αp_\alpha \neq x_\alpha within distance r/4r/4 of x αx_\alpha; put U=⋃ αB d(x α,p α)(x α)U = \bigcup_\alpha B_{d(x_\alpha, p_\alpha)}(x_\alpha). Notice that p α∉Up_\alpha \notin U. If B s(x)B_s(x) is any ball contained in UU, then d(x,x γ)<r/4d(x, x_\gamma) \lt r/4 for some γ\gamma. Supposing we have distinct x α,x β∈B s(x)x_\alpha, x_\beta \in B_s(x), then r<d(x α,x β)≤d(x α,x)+d(x,x β)<s+sr \lt d(x_\alpha, x_\beta) \leq d(x_\alpha, x) + d(x, x_\beta) \lt s + s, so r/2<sr/2 \lt s. But then from d(x γ,p γ)<r/4d(x_\gamma, p_\gamma) \lt r/4 and d(x,x γ)<r/4d(x, x_\gamma) \lt r/4, we have d(x,p γ)<r/2<sd(x, p_\gamma) \lt r/2 \lt s so that p γ∈B s(x)⊂Up_\gamma \in B_s(x) \subset U, a contradiction. We conclude that any ball contained in UU contains at most one x αx_\alpha, and so UU cannot be covered by countably many balls.
We are therefore left to deal with the case where there are at most countably many non-isolated x βx_\beta. Discard these, so without loss of generality we may suppose all the x βx_\beta are isolated points of XX and that for some fixed rr the ball B r(x β)B_r(x_\beta) contains no other x αx_\alpha. Let ZZ be this set of x βx_\beta. For each β\beta, let t βt_\beta be the supremum over all tt such that B t(x β)∩ZB_t(x_\beta) \cap Z is countable. It follows that B t β(x β)∩ZB_{t_\beta}(x_\beta) \cap Z is itself countable, as is {α:α<β}\{\alpha: \alpha \lt \beta\}. At each stage β\beta, there is a countable set C β⊂ZC_\beta \subset Z disjoint from B t β(x β)∪{x α:α<β}B_{t_\beta}(x_\beta) \cup \{x_\alpha: \alpha \lt \beta\} such that for each s>t βs \gt t_\beta, there exists y∈C βy \in C_\beta with d(x β,y)<sd(x_\beta, y) \lt s. By transfinite induction, we can construct a cofinal subset II of ω 1\omega_1 such that x β∉C αx_\beta \notin C_\alpha whenever α,β∈I\alpha, \beta \in I and α<β\alpha \lt \beta. The set Y={x α:α∈I}Y = \{x_\alpha: \alpha \in I\} is open in XX, and we claim that it is not a countable union of balls. For suppose otherwise. Let FF be such a countable family of balls; then there is some minimal α\alpha for which B t(x α)∈FB_t(x_\alpha) \in F is uncountable, so that t>t αt \gt t_\alpha. By construction of C αC_\alpha, there exists z∈C α∩B t(x α)z \in C_\alpha \cap B_t(x_\alpha). But since Y=⋃FY = \bigcup F, we have that z=x βz = x_\beta for some β∈I\beta \in I, and this contradicts our condition on II.