Any subspace of a separable metric space is separable

Let X be a separable, metric space, A \subset X, and let Q \subset X be countable and dense.  For each q \in Q, n \in \mathbb{N}, choose a_n^q \in A \cap B_{1/n}(q) if such intersection is nonempty.  Then, A^* = \bigcup_{n,q} \{ a_n^q \} is a countable subset of A.

Let a \in A, 0 < \epsilon < 1. Then choose q \in B_{\epsilon / 4}(a) \cap Q.  Now, choose N such that \epsilon / 4 \le 1 / N \le \epsilon / 2.  Then, B_{1/N}(q) \cap A \neq \emptyset, so there exists a^* \in A^*, such that d(a^*, q) < 1/N.

Now d(a^*,a) \le d(a^*,q) + d(q,a) < \epsilon / 2 + \epsilon / 4 < \epsilon.  So a^* \in B_\epsilon(a) \cap A.  Hence, the closure of A^* in A is equal to A.

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This entry was posted in Metric Spaces, Separability, Topology. Bookmark the permalink.

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