Quote by Damidami
I'm working on some topology in [itex] \mathbb{R}^n [/itex] problem, and I run across this:
Given [itex]\{F_n\}[/itex] a family of subsets of [itex] \mathbb{R}^n [/itex], then if [itex]x[/itex] is a point in the clausure of the union of the family, then
[itex]x \in \overline{\cup F_n}[/itex]
wich means that for every [itex]\delta > 0[/itex] one has
[itex]B(x,\delta) \cap (\cup F_n) \neq \emptyset [/itex]
Now, if I say that because the intersection is a nonempty set, I have a point [itex]y \in \mathbb{R}^k [/itex] such that
[itex] d(x,y) < \delta \wedge y \in (\cup F_n)[/itex]
did I use the axiom of choice?
Because I think it this way, I choose an element [itex] y [/itex] from an infinite (nonumerable) indexed family parametrized by [itex]\delta[/itex] (for each value of [itex]\delta[/itex] I have another intersection).
So If I don't want to use the AC I can't just say that I can choose an element from that nonempty intersection?
I'm a little confused :?
Thanks!

You did not use the axiom of choice. In fact, you've made only
one choice. The axiom of choice is only necessary if you want to make an infinite number of choices (and if you can't specify the choices you make).