## 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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### 2 Responses to Any subspace of a separable metric space is separable

1. XiangYu says:

Reblogged this on XiangYu's Blog.

2. Peter De Cupis says:

More in general, tuo can extend a non-separable space X by adding a monadic set including a single element y, with y non-belonging to X; then in such extended space Xy you impose that the closure of such added monadic set is the whole space Xy (it is trivially verified that in this way yhe original topology in X is correctly obtained as sub-space topology from Xy, i.e. X is an authentic topological subspace of a topological “super-space” Xy). Tautologically Xy will result as the closure of a set with cardinality=1