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I have to prove that any uncountable $B\subseteq \mathbb{R}$, where $(\mathbb{R},\epsilon^1)$ is euclidean topology and topology on B is relative, is separable. And I know it's true because every subset of separable metric space is separable.

But what if we are given separable space $(X,\tau)$, $X$ uncountable, and $A \subseteq X$ uncountable subset with relative topology. Is $(A,\tau_A)$ separable and if it is, how to prove it?

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3 Answers 3

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Consider the Niemytzki (or Moore) plane. This space is separable (the family of points with both coordinates rational is dense), but the $x$-axis $A = \{ \langle x , 0 \rangle : x \in \mathbb{R} \}$ is an uncountable closed discrete subset (and so $A$ with the subspace topology is discrete, and is therefore not separable).

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Thank you for answer, link and counterexample =) You helped a lot. –  Mare May 14 '13 at 8:32
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Another example is this;

Let $ \omega$ denote infinite countable cardinal and $\mathcal E \subset [\omega]^\omega$ is a maximal almost disjoint family, where $[\omega]^\omega = \{ A \subset \omega: |A| = \omega \}$. Let $\Psi(\mathcal E)$ denote the topological space whose point-set is $\omega \cup \mathcal E$, with the topology generated by isolating each $\alpha \in \omega$, and the basic nbhds about $E \in \mathcal E$ are all sets of the form $\{E\}\cup (E\setminus F)$, where $F \in [E]^{< \omega}$. This we can called $\Psi$ space.

Claim: It is separable. However the subspace $\mathcal E$ of $\Psi$ is uncountable closed discrete, and hence it is not separeblae.

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The Sorgenfrey Plane is another example. It's a separable space, but the antidiagonal line {⟨x,-x⟩:x∈R} is discrete and uncountable.

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