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?

  • 2
    $\begingroup$ The claim is not true in general. For example in a finite space with indiscrete topolgy, every point is an accumulation point of ecevery nonempty subset. Does the use of $|\cdot|$ mean that the problem is specifically about $S\subseteq \Bbb R$? $\endgroup$ – Hagen von Eitzen May 18 '16 at 20:03
  • $\begingroup$ yes, $S \subseteq \mathbb{R}$ $\endgroup$ – Jeremiah Dunivin May 18 '16 at 20:05
  • $\begingroup$ Looks good. You should point out that $N_{\delta}^* \cap S = \emptyset$ rather than $N_{\delta}^* \cap(N\cap S)$ though. $\endgroup$ – Alex G. May 18 '16 at 20:12
  • $\begingroup$ But is it clear that $N^*_{\delta} \cap S = \varnothing$? $\endgroup$ – Jeremiah Dunivin May 18 '16 at 20:31
  • $\begingroup$ @AlexG. I don't think we any information that $N^*_{\delta} \cap S = \varnothing$, since $S$ is an arbitrary subset of the reals. The only fact that we do have is that $N^*_{\delta} \cap (N\cap S)$ is empty by definition of $\delta$. Please correct if I'm wrong. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 11:13

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.

  • $\begingroup$ Nice direct proof $\endgroup$ – Jeremiah Dunivin May 18 '16 at 21:36

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