Let $A$ be a set, $X$ a metric space, $x$ an accumulation point of $A$ that is, every nbhood of $x$ contains a point $a \in A$, $a \neq x$.
I wrote a proof of the fact "every nbhood of $x$ contains infinitely many points of $A$". But for one I don't know if it is correct and also if it is correct whether it can be improved. Here is my proof:
The open ball $B(x,\varepsilon)$ is a neighbd of $x$ and by assumption contains $a \in A, a \neq x$. Since $B(x, \varepsilon)$ is open there is $\delta$ such that $B(x, \delta) \subseteq B(x, \varepsilon)$ and $a \notin B(x, \delta)$. Since $B(x, \delta)$ is a nbhood of $x_0$ by assumption there is $a' \in A$ with $a' \in B(x, \delta)$.
Then I wrote: "Repeat process to obtain infinitely many points $a \in A$ in $B(x, \varepsilon)$".
Is this valid? And if it is not: how to finish the proof? And if it is correct: is there a neater way to write this proof? Thank you.
Best Answer
The idea is sort of OK, but can be made more formally correct.
Assume there is a neighbourhood $U = B(x_0, \epsilon)$ such that $U \cap A$ is a finite set, (and we leave out $x_0$ if $x_0 \in A$) say it equals $\{ a_1,\ldots, a_k\}$ for some finite $k$. We have to show this leads to a contradiction. We now apply the idea you also use: take a ball around $x_0$ with radius $r < \min(d(x_0,a_1), d(x_0,a_1),\ldots,d(x_0, a_k))$. Then $B(x_0,r) \subset U$ and by construction the $a_i$ are not in this ball, but there must be some $a \neq x_0$ with $a \in A$ in this ball, as $x_0$ is an accumulation point. This shows that $A \cap U$ could not have been this finite set (we have found a new point). Contradiction.
We can avoid radii and metric, by noting that a finite set is closed (so we need $T_1$ in a general topology setting) and note that $U \setminus (A \cap U)$ is also open and should contain some new point of $A$.