Separable metric spaces and closed cover

Question

When I read a proof of theorem, there is a statement

"Since $X$ is a separable metric space, then we can find a closed cover $\{F_n^k\}_{k\in\mathbb{N}}$ of $X$ such that diam$\{F_n^k\}<\frac{1}{n}$ for each $k$."

I never read about separable metric space which using closed cover before.
Then, I wonder, does the statement above have relation with Lindelöf, i.e. if $X$ a separable metric space then $X$ is Lindelöf ?

Answer 1

You can prove it directly:

Assume $Q=\{q_1,q_2\ldots\}$ is a dense subset. Denote $F_n^k=D(q_k,1/2n)$. Then $\{F_n^k\}_{k\in\Bbb N}$ is a closed cover (here $D$ stands for a closed ball). To verify it, consider any $x\in X$. Since $B(x,1/3n)\cap Q\neq \emptyset$, there is an index $k$ such that $d(q_k,x)<1/3n$. This gives $x\in B(q_k,1/3n)\subset F_n^k$.

If you want to use the Lindelöf property, you can consider the cover $$\left\{B\left(x,\frac 1{3n}\right)\,|\,x\in X\right\}$$ and find a countable subcover $$\left\{B\left(x_i,\frac 1{3n}\right)\,|\,i\in \Bbb N\right\}.$$ The closures of these sets define a desired closed cover.

Remark:

• $\mathrm{diam}\overline A = \mathrm{diam} A$;
• $\mathrm{diam}B(x,r)\leq \mathrm{diam}D(x,r)\leq 2r$.