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I have encountered this well-known result when reading about separable spaces.

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Let $(E, \tau)$ be a separable topological space. Let $X \in \tau$ and $\tau_X$ its subspace topology. Then $(X, \tau_X)$ is separable.

Could you verify if my below proof is fine?

Let $D$ be a countable dense subset of $E$. This implies $\overline{D}^{\tau} = E$, or equivalently $$ \forall x \in E, \forall \text{ nbh } V \text{ of } x \text{ in } \tau, V\cap D \neq \emptyset. $$

Notice that if $x \in X$ and $V$ is a nbh of $x$ in $\tau$, then $V \cap X$ is also a nbh of $x$ in $\tau$. This implies $$ \forall x \in X, \forall \text{ nbh } V \text{ of } x \text{ in } \tau, (V \cap X)\cap (D \cap X) \neq \emptyset. $$

Hence $$ \forall x \in X, \forall \text{ nbh } V \text{ of } x \text{ in } \tau_X, V\cap (D \cap X) \neq \emptyset. $$

It follows that $D\cap X$ is countable and dense in $X$, i.e., $\overline{D}^{\tau_X} = X$.

Sambo
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Akira
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  • I think it's fine. But I also think it would get more understandable if you write your conclusion that $D\cap X$ is dense in $X$ at the begining. – Yuz Apr 27 '22 at 16:12

2 Answers2

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The logic in your proof looks right to me! However, in my opinion, the way you structured it makes it a bit hard to follow. I think it would read better if you went backwards from how you wrote it.

More specifically, you're trying to prove $D \cap X$ is dense in $X$, which, as you noted, translates to: $$ \forall x \in X, \forall \text{ nbh } U \text{ of } x \text{ in } \tau_X, \; U \cap (D \cap X) \neq \varnothing $$ Try proving this statement directly: choose $x$ and $U$, write $U = V \cap X$, and then use your argument about $V \cap X$ being a neighborhood of $x$ in $\tau$ to reach the conclusion you want.

Sambo
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$(E, \tau) $ separable and $X\in \tau$.

Claim: $(X, \tau_X) $ is separable.

Proof: $(E, \tau) $ separable. Let, $D\subset E$ be countable and $\tau$-dense .

Claim: $X\cap D$ is countable and $\tau_Y$-dense.

$X\cap D\subset D \implies D$ is countable.

Choose, $U\in \tau_X$ .

We have to show $U\cap(X\cap D) \neq \emptyset$

$U\in\tau_X\implies U=X\cap V $ for some $V\in \tau$

Then, $X\cap V\in \tau$ and $D$ is dense implies $(X\cap V) \cap D\neq\emptyset$

Hence, $(X\cap V)\cap(Y\cap D)\neq \emptyset$

$\implies U\cap(X\cap D) \neq \emptyset$

Hence, $X\cap D $ is $\tau_Y$ -dense.

SoG
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