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I'm trying to prove the following statement:

Every subspace of a separable metric space is separable.

Could you please verify if I correctly apply the concept relative topology? Thank you so much!


$\textbf{My attempt}$

First, I need the following lemma:

Let $X$ be a metric space. The following statements are equivalent:

  • $X$ satisfies the second countability axiom.

  • $X$ is a Lindelöf space.

  • $X$ is separable.

Let $(X,d)$ be a separable metric space, $Y \subseteq X$, and $d_Y$ the induced metric of $d$ on $Y$. By Lemma, $X$ satisfies the second countability axiom. Then the topology induced by $d$ has a countable basis $\mathcal M$. Let $\mathcal N = \{A \cap Y \mid A \in \mathcal M\}$. By relative topology, $\mathcal N$ is a countable basis of the topology induced by $d_Y$. Hence $(Y,d_Y)$ satisfies the second countability axiom. Thus $(Y,d_Y)$ is separable by Lemma.


Update: From @Henno Brandsma's answer, I include the proof that $\mathcal N$ is a basis of the topology induced by $d_Y$.

Assume $A \subseteq Y$ is open in $Y$. Then $A = O \cap Y$ for some $O \subseteq X$ that is open in $X$. It follows from $O$ is open in $X$ that there is $\mathcal M' \subseteq \mathcal M$ such that $O = \bigcup \mathcal M'$. Let $\mathcal N' = \{Y \cap B \mid B \in \mathcal M'\}$ Then $\mathcal N' \subseteq \mathcal N$ and $A = \bigcup \mathcal N'$. Hence $\mathcal N$ is a basis of the topology induced by $d_Y$.

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In this answer I show a bigger equivalence, with all proofs.

The fact that second countable implies that all subspaces are separable you have shown correctly. You need a small lemma that when $\mathcal{B}$ is a base for $X$, its intersections with $Y$ form a base for $Y$ too, but I assume you already showed that before.

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  • $\begingroup$ I've added a part that $\mathcal N$ is a basis of the topology induced by $d_Y$. Please have a check on my update! $\endgroup$ – LAD Feb 5 at 23:16
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    $\begingroup$ @AbstractAnalysis yes that seems fine too. I supposed it would have been part of you studied theory. $\endgroup$ – Henno Brandsma Feb 5 at 23:22
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    $\begingroup$ Thank you so much for your dedicated support ;) ^o^ $\endgroup$ – LAD Feb 5 at 23:25

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