separable space

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 first-countable, but the converse need not (see Steen Seebach Example 51 ).

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.

Separable metric spaces


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


One direction is easy: if x ix_i is a countable dense subset of XX and r jr_j is an enumeration of the rationals, then the balls B r j(x i)B_{r_j}(x_i) is 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 easily construct an open set UU that is not a countable union of balls. For around each such x αx_\alpha we can find a small ball B s α(x α)B_{s_\alpha}(x_\alpha) of radius at most r/2r/2, such that B r(x α)\B s α(x α)B_r(x_\alpha) \backslash B_{s_\alpha}(x_\alpha) is inhabited by a point p αp_\alpha. Put U= αB s α(x α)U = \bigcup_\alpha B_{s_\alpha}(x_\alpha). No ball B s(x α)B_s(x_\alpha) interior to UU can contain another x βx_\beta, since for that we would need s>rs \gt r and such B s(x α)B_s(x_\alpha) would then include p αp_\alpha, which is excluded from UU by design. Thus, since there are uncountably many x βx_\beta, we need uncountably many balls to cover UU.

We may therefore assume that 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. For each β\beta, let t βt_\beta be the supremum over all tt such that B t(x β)B_t(x_\beta) is countable. It follows that B t β(x β)B_{t_\beta}(x_\beta) is itself countable, as is {α:α<β}\{\alpha: \alpha \lt \beta\}. At each stage β\beta, there is a countable set C βC_\beta disjoint from B t β(x β){α:α<β}B_{t_\beta}(x_\beta) \cup \{\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<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 yC αB t(x α)y \in C_\alpha \cap B_t(x_\alpha). But since Y=FY = \bigcup F, we have that y=x βy = x_\beta for some βI\beta \in I, and this contradicts our condition on II.

Revised on September 19, 2014 20:50:28 by Todd Trimble (