I have looked through the many questions about this theorem on this website, but none that I could see adressed exactly what I am wondering about. For completeness, I will state the theorem below and the part of the proof I am wondering about.
2.41 Theorem If a set $E$ in $\mathbb{R}^{k}$ has one of the following three properties, then it has the other two:
- $E$ is closed and bounded.
- $E$ is compact.
- Every infinite subset of $E$ has a limit point in $E$.
The part of the proof I am wondering about concerns proving that (3) implies that $E$ is closed. The proof is set up by proving the contrapositive.
Proof If $E$ is not closed, then there is a point $\mathbf{x_{0}}\in R^{k}$ which is a limit point of $E$ but not a point of $E$. For $n=1,2,3,...,$ there are points $\mathbf{x_{n}}\in E$ such that $|\mathbf{x_{n}} - \mathbf{x_{0}}| < 1/n$. Let $S$ be the set of the points $\mathbf{x_{n}}$.
The proof continues, but the part I am wondering about is the last part, namely "let $S$ be the set of points $\mathbf{x_{n}}$." I understand that because $\mathbf{x_{0}}$ is a limit point, there must exist such points for all $n$. However, is just knowing that such points exist enough to form a set consisting of such points? For each $n\in \mathbb{N}\setminus \{0\}$ there must exist infinitely many such points since $\mathbf{x_{0}}$ is a limit point, must there not? And if so (since such points are not unique), if we want to form the set $S$ described above, must we not either specify some procedure describing these points for each $n$, or appeal to the axiom of (countable?) choice to choose one such point for every $n$?
Unfortunately my instructor has deemed set theory "boring" every time I have tried to talk to him about the potential usage of the axiom of choice. Thank you very much in advance to anyone that took the time to read this post. Any answer is much appreciated.
The first two are equivalent without the axiom of choice (see here and here), but the third is not: by Theorem $\bf{3.27}$ of Horst Herrlich, Axiom of Choice, Springer-Verlag, $2006$, the equivalence of $(2)$ and $(3)$ for arbitrary pseudometric spaces is equivalent to the axiom of countable choice. In Disaster $\bf{4.53}$ he also notes that it is consistent with $\mathsf{ZF}$ that $\Bbb R$ have a subset $A$ with an accumulation point $x$ that is not the limit of any sequence in $A$.