nLab separable space

Separable spaces



topology (point-set topology, point-free topology)

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Separable spaces


A topological space is separable if it has a countable dense subset.

To be explicit, XX is separable if there exists an infinite sequence a:Xa\colon \mathbb{N} \to X such that, given any point bb in XX and any neighbourhood UU of bb, we have a iUa_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 \mathbb{R} with the half-open topology (see Munkres, page 192), denoted 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 \mathbb{R}^\mathbb{R} of all functions \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.


Although an arbitrary product of separable spaces need not be separable, an arbitrary product of separable spaces does satisfy the countable chain condition (see there for more discussion). This is somewhat remarkable, since it is not necessarily true that even a finite product of spaces satisfying the countable chain condition also satisfies the countable chain condition (whether it does is independent of ZFC).

Subspaces of separable spaces need not be separable. Example: the product l× 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.

Many results in analysis are easiest for separable spaces. This is particularly true if one wishes to avoid using strong forms of the axiom of choice, or to use arguments predicative over the natural numbers. For example, the Hahn-Banach theorem for separable Banach spaces can be established using only mild forms of choice, e.g., dependent choice. More precisely, it can be established in the weak subsystem WKL 0WKL_0 of second-order arithmetic; see Brown-Simpson.

Separable metric spaces

A classical fact is


Using countable choice, then:

For a metric space XX the following are equivalent

  1. XX is separable;
  2. XX is second-countable;
  3. XX is Lindelöf, i.e. any open cover admits a countable subcover.

The second property is implied by the first: given any dense subset {x nn}\{x_n \mid n\in \mathbb{N}\}, form the countable system of sets {B 1/m(x n)n,m}\{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 UXU\subset X, a point xUx\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 nB 1/(2k)(x)x_n\in B_{1/(2k)}(x) and therefore xB 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 II\subset \mathbb{N} of those nn that are contained in some V λV_\lambda. By assumption iIU 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 iI{λU iV 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)xX}\{B_{1/1}(x) \mid x\in X\}, {B 1/2(x)xX}\{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 1A 2A_1\cup A_2\cup\ldots forms a dense set. Indeed, given any yXy\in X and nn the point xx has to be contained in some B 1/n(x)B_{1/n}(x) for some xA nx\in A_n.


In the proof any variant of the axiom of choice is only used for the implication (2)(3)(2)\Rightarrow(3). On the other hand, assuming countable choice, this implication holds in every topological space. The implication (2)(1)(2)\Rightarrow(1) holds in any topological space as well (see Example )

Similar in spirit to (1)\Leftrightarrow(2) but less well-known is the following.


A metric space XX is separable iff every open set is a countable union of balls.


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 αx 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 αUp_\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 βB 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 γB 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 βZC_\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 yC 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 βC αx_\beta \notin C_\alpha whenever α,βI\alpha, \beta \in I and α<β\alpha \lt \beta. The set Y={x α:αI}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 zC αB 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 βI\beta \in I, and this contradicts our condition on II.



Every second-countable topological space is separable. To see this take a countable cover and select a point in each member of the cover (using countable choice).

Axioms: axiom of choice (AC), countable choice (CC).




  • James Munkres, Topology, a first course, Prentice-Hall (1975).

  • D. K. Brown and S. G. Simpson, Which set existence axioms are needed to prove the separable Hahn-Banach theorem?, Annals of Pure and Applied Logic 31 (1986), pp. 123-144.

  • Henno Brandsma, Thread on Hewitt-Marczewski theorem, Ask a Topologist (2010) (link)

Last revised on June 2, 2022 at 01:43:43. See the history of this page for a list of all contributions to it.