nLab
separable space

Separable spaces

Definitions

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.

Properties

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.

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 be 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 that metric space is separable if and only if it is second-countable. Similar in spirit but less well-known is the following (whose proof subsumes the aforementioned classical fact).

Theorem

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

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 by playing around with the triangle inequality, one can show that the balls B r j(x i)B_{r_j}(x_i) form a countable base (XX is a second-countable space), and in particular 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.

References

  • 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)

Revised on September 29, 2014 16:25:22 by Todd Trimble (67.81.95.215)