# From open cover to ball cover - role of AC

Let $$X$$ be a metric space and assume that, for every $$\varepsilon>0$$ there is a countable open cover $$(A_i)$$ of $$X$$ with $$diam(A_i)\le \varepsilon$$ for each $$i$$.

Of course I can assume the cover is made only of basic open sets: if $$\varepsilon$$ is fixed then, for each $$i$$, we can pick $$x_i\in A_i$$ and consider the ball $$B_i$$ with center $$x_i$$ and diameter $$2\varepsilon$$ to obtain a $$2\varepsilon$$-cover of $$X$$.

The question is: can I assume the open cover is made of open balls even without $$AC$$ (or, in this case, without $$AC(\omega)$$)? Is there a way to build a ball cover from a generic open cover that is not dependent on the choice of a point for each set in the cover?

• I don't get your question. What is the statement you want to prove without choice? – Asaf Karagila Oct 6 '18 at 17:12
• Passing from a generic open cover to a ball cover. Let me rephrase it in a clearer way – Manlio Oct 7 '18 at 9:59

In Cohen's first model there is an infinite Dedekind-finite set of reals which is dense in $$\Bbb R$$. This is a set which is infinite, but has no countably infinite subset. And as a set of reals, it is of course a metric space, and by being dense, it is also not bounded.
Now, cover $$A$$ with rational intervals (or their intersection with $$A$$) with diameter $$\varepsilon$$, this is a countable cover. But if you want these to be open balls of diameter $$\varepsilon$$, you need to choose centers for these balls, which would necessitate a countably infinite subset of $$A$$.