Let $S \subseteq \mathbb{R}$.
To prove that every neighborhood of $x$ contains infinitely many points of $S$ whenever $x$ is an accumulation point of $S$, we will suppose to the contrary that there exists a neighborhood $N$ of $x$ such that
$$N \cap S = \{s_1, \cdots, s_n\}$$
Since $x$ is an accumulation point of $S$, then every deleted neighborhood $N^*_{\delta} \cap S \neq \varnothing$. We then select a particular $\delta = \min\{|x-s_i| \, : \, s_i \neq x\}$. Then $\delta > 0$, but we have
$$N^*_{\delta} \cap \, (N \cap S) = \varnothing$$
which is a contradiction, so every neighborhood of $x$ contains infinitely many points of $S$.
Are there any problems with this proof?
Best Answer
Let $B_\alpha$ be a neighborhood of $x$. That is:
$$ B_\alpha = \{z : |x - z| < \alpha\},$$
for $\alpha > 0$.
Take the sequence $$a_n = x + \frac{1}{n}.$$ This converges to $x$ as $n$ approaches to $+\infty$. This means that $x$ is accumulation point for some sets that contain $s$. Moreover:
$$\forall \varepsilon > 0 \exists m \in \mathbb{N} : n > m \Rightarrow a_n \in B_\varepsilon. $$
Now, consider the sets $S = B_\varepsilon$, $N = B_\alpha$ and $S \cap N$. It's clear that you can always find a $m'$ such that
$$n > m' \Rightarrow a_n \in S \cap N,$$
and the $a_n$'s with $n > m'$ are, of course, infinite.