Let $(E,d)$ be a metric space. Then $X$ is second-countable if and only if $X$ is Lindelöf if and only if $X$ is separable

Question

In proving every subspace of a separable metric space is separable, I need below result. Could you check if my proof is fine?

Theorem: Let $(E,d)$ be a metric space. Then $X$ is second-countable if and only if $X$ is Lindelöf if and only if $X$ is separable.

Proof: Let $X$ be second-countable, i.e., $X$ has a countable base $(O_i)_{i \in \mathbb N}$. Let $(C_j)_{j \in J}$ be an open cover of $X$. Then for each $j \in J$, there is $I_j \subseteq \mathbb N$ such that $C_j = \bigcup_{i \in I_j} O_i$. Let $I := \bigcup_{j \in J} I_j \subseteq \mathbb N$. For each $i \in I$, let $\varphi(i)$ be one of those $j$ such that $i \in I_j$. Then $(C_{\varphi(i)})_{i \in I}$ is a countable subcover of $(C_j)_{j \in J}$.

Let $X$ be Lindelöf. For each $n \in \mathbb N^+$, the collection of open balls $\mathcal U_n :=\{\mathbb B(x, 1/n) \mid x \in X\}$ is an open cover of $X$. Then for each $n \in \mathbb N^+$, there exists a countable subcover $\mathcal U_n' \subseteq \mathcal U_n$. Clearly, $\mathcal U := \bigcup_n \mathcal U'_n$ is a countable union of countable sets and thus countable. It's easy to verify that $\mathcal U$ is a dense countable subset of $X$.

Let $X$ be separable, i.e., $X$ has a countable dense subset $E \subseteq X$. For each $x \in X$ there is a sequence $(e_{x,n}, 1/n)_n$ in $E \times \mathbb N^*$ such that $x \in \mathbb B(e_{x, n}, 1/n)$. Let $\mathcal U$ be the collection of all balls of the form $\mathbb B(e_{x, n}, 1/n)$ where $x \in E$ and $n \in \mathbb N$. Clearly, $\mathcal U$ is countable. Let $O \subseteq X$ be open. For each $x \in O$, there is $r_x >0$ such that $\mathbb B(x, r_x) \subseteq O$. Let $n_x$ be the least $n$ such that $1/n < r_x / 2$. It follows that $x \in \mathbb B(e_{x, n_x}, 1/n_x) \subseteq O$. Hence $O = \bigcup_{x \in O} \mathbb B(e_{x, n_x}, 1/n_x)$. Hence $\mathcal U$ is a countable base of $X$.

Answer 1

I treat a more general fact here, but some comments on your proof: second countable implying Lindelöf is a quite general fact for all spaces (as you can see from your proof, which could be clear IMO e.g. see here which makes the dependence on ACC more clear.

For Lindelöf to separable make it more clear that the dense set will be the centres of the balls from the subcovers and don't handwave with "it's easy to verify" but do that verification for real.

In the final separable to second countable proof I'd just use the dense set as centres with all rational radii as a direct countable base, I don't see the need for the extra $e_{x,n_x}$ points that converge to a dense subset. Simplify!

Answer 2

FYI. You can show directly that if $(X,d)$ is a separable metric space and $Y\subset X,$ then $Y$ is separable, as follows:

Let $E$ be a countable dense subset of $X$. For $n\in \Bbb N$ let $E_n=\{e\in E: B_d(e,1/n)\cap Y\ne \emptyset\}.$ And for $e\in E_n$ take $\psi_n(e)\in B_d(e,1/n)\cap Y.$

Now let $D=\cup_{n\in \Bbb N}\{\psi_n(e):e\in E_n\}.$ Then $D$ is a countable subset of $Y$.

For $y\in Y$ and $r\in \Bbb R^+,$ we show that $B_d(y,r)\cap D\ne\emptyset$ as follows:

Take $n\in \Bbb N$ with $1/n<r/2.$ Now $E$ is dense in $X$ so take $e\in E\cap B_d(y,1/n).$. Then $e\in E_n$ because $y\in B_d(e,1/n)\cap Y.$ So $$d(y,\psi_n(e))\le d(y,e)+d(e,\psi_n(e))<2/n<r.$$ So $\psi_n(e)\in B_d(y,r)\cap D$.