The problem says:
Show that if $x_0\in A^´$, then there exists some sequence $\{x_n\}$ in $A$ of distinct points such that
$$\lim_{n\rightarrow\infty}x_n=x_0$$.
I've seen proofs of a similar result that omits the requirement of the points being distinct.
The proof that I came up with is, in shorter words.
First, construct a sequence $\{x_n\}$ of points such that $x_n\in\ (V_{\frac{1}{n}}(x_0)\cap A)\setminus\{x_0,x_1,…,x_{n-1}\}$ (the sequence starting at $x_1$). This can be done because $V_{\frac{1}{n}}(x_0)\cap A$ is an infinite set and $\{x_0,x_1,…,x_{n-1}\}$ is finite.
Then, it´s easy to show that this sequence converges to $x_0$.
My question is if there is an error by defining a sequence this way. My doubt comes from the fact that all of the proofs that I've seen of the other result make use of the Axiom of choice, and I don't know if I'm using it when constructing the sequence or if the sequence is not valid.
Best Answer
The method is fine, it's a standard recursion on $\Bbb N$, where you indeed use so-called dependent choice (a relatively mild consequence of AC) that isn't controversial (like full AC is for some). Its use cannot be avoided, as I think this results is itself (almost?) equivalent to DC.