The only assumption that I have is the space being metric and total boundedness, defined as:

For all $\varepsilon>0$ there exist a net of finite elements $x_1, \cdots, x_n$ such that the balls with radius $\varepsilon$ centered at $x_i$ cover the space.

  • 1
    $\begingroup$ HINT: Use the total boundedness to show that the space is separable, then use that to show that it has a countable base, and then use that to get the desired result. $\endgroup$ Oct 21 '20 at 3:36

You can do the following: Let $(A_\lambda)_{\lambda\in L}$ be a cover of your metric space $M$. Since $M$ is totally bounded, for each $n$ you can find $x_1,\dots,x_k\in M$ s.t. $\,\bigcup_i B(x_i, 1/n) = M$.

Define $B$ as the family of all such balls with every possible radius $1/n$. Clearly $B$ is countable, since it is a countable union of finite sets (you can think of $B$ as $\bigcup_n \{B(x_1,1/n),\cdots B(x_k,1/n)\}$).

Then take $C$, a subfamily of $B$, s.t. every element of $C$ is a ball that is contained in some $A_\lambda$.

I say that $C$ is a cover of $M$. In fact, if $p\in M$ then $p\in A_\mu$ for some $\mu\in L$, since $A_\mu$ is open, there exists $\varepsilon>0$ s.t. $\,B(p,\varepsilon)\subset A_\mu$.

Let $n$ be an integer s.t. $1/n<\varepsilon/3$. Note that there is a element of $B$ with radius $1/n$, say $B(x,1/n)$, s.t. $\,p\in B(x,1/n)$, and actually, by the definition of $n$, we have that $B(x,1/n)\subset B(p,\varepsilon)\subset A_\mu$.

Therefore $B(x,1/n)\in C$ and $p\in B(x,1/n)$, which implies that $C$ is a cover of $M$.

To conclude observe the following:

  1. $C$ is countable, and therefore we can write $C=\{C_1,\cdots, C_n,\cdots\}$
  2. For each $j\in \mathbb N$ you may choose one element of the family $(A_\lambda)_{\lambda\in L}$, say $A_{\lambda_j}$, s.t. $C_j\subset A_{\lambda_j}$.
  3. The family $(A_{\lambda_j})_{j\in \mathbb N}$ is a countable sub cover of $(A_\lambda)_{\lambda\in L}$.
  • $\begingroup$ I am not sure, I had a similar argument, but the choice of your Cj depend on x, when x vary in the space, so does your covers A_u, or I am missing something here $\endgroup$
    – Omar
    Oct 21 '20 at 14:49
  • $\begingroup$ What do you mean? the cover $A_\lambda$ is set. When you define the $A_{\lambda_j}$ you could have more than one $A_\lambda$ that contains $C_j$ in this case choose whichever, and is all set. $\endgroup$ Oct 21 '20 at 14:56
  • $\begingroup$ Remember that $C$ is defined as the subfamily of $B$ s.t. it's elements are entirely contained in at least one $A_\lambda$, there is no $x$ in this definition. $\endgroup$ Oct 21 '20 at 15:06

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