Prove that $x$ is an accumulation point of a set $S$ iff there exists a sequence $(s_n)$ of points in $S\setminus \{x\}$ such that $(s_n)$ converges to $x$.


Since this is a show-me-what-you've-got-so-far community, I'll give you what I've come up with so far even though it's incomplete:

Suppose that $x$ is an accumulation point of a set $S$. By the definition of accumulation point, for all $\epsilon>0$ we have that $N^*(x;\epsilon)\cap S\neq \varnothing$. Thus, we can conclude that $N^*(x;\frac{1}{n})\cap S\neq \varnothing$ by setting $\epsilon_n=\frac{1}{n}$ for $n\in\mathbb{N}$, and so for all $n\in\mathbb{N}$ we can select the desired sequence $s_n\in N^*(x;\frac{1}{n})\cap S$. Now, for all $n\in\mathbb{N}$ we have $s_n\neq x$, and $s_n\in S$ since $s_n\in N^*(x;\frac{1}{n})\cap S$---for all $n\in\mathbb{N}$. To show that for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that if $n>N$, then $\lvert s_n-s\rvert <\epsilon$ let $\epsilon>0$; by the Archimedian property there exists $N\in\mathbb{N}$ such that $\frac{1}{N}<\epsilon$, and by design $s_n\in N^*(x;\frac{1}{n})$ for all $n$. Thus $\lvert s_n-x\rvert <\frac{1}{n}$ for all $n$, and so $n>N$ implies $\lvert s_n-x\rvert <\frac{1}{n}<\frac{1}{N}<\epsilon$ . . . still working . . .

  • $\begingroup$ I don't know what is considered missing here. It seems complete to me. (I'm assuming $N^*(x; \epsilon)$ does not contain $x$.) $\endgroup$
    – Tunococ
    Sep 9 '13 at 23:57
  • $\begingroup$ @Tunococ Well, I'll tell you. This is the -> direction only, and this is an iff condition. Yes, $N^*$ is the deleted $\epsilon$-neighborhood. $\endgroup$ Sep 10 '13 at 2:17
  • $\begingroup$ I was under the impression that you had already got the other direction because I think the direction you proved is more difficult. $\endgroup$
    – Tunococ
    Sep 10 '13 at 7:27
  • 1
    $\begingroup$ @Tunococ (and Loie Benedicte): I have edited out a slightly inflammatory remark from L.B.'s comment above, and also a direct response to said remark from T.'s comment. Keep it civil, folks! $\endgroup$
    – user642796
    Sep 12 '13 at 7:39

Sequence implies accumulation

Let $X$ be an arbitrary topological space.

Let $S\subseteq X$ and let $x\in S$.

Let $(s_n)$ be a sequence in $S\setminus\{x\}$ converging to $x$.

Let $U$ be an open set containing $x$. Then by the definition of convergence, there is an $N$ such that whenever $n \ge N$, $s_n \in U$. In particular, $s_N\in U$. Since $(s_n)$ is a sequence in $S\setminus \{x\}$, $s_N \in S\cap U$ and $s_N \ne x$. That is, every open neighborhood of $x$ contains an element of $S$ not equal to $x$, so $x$ is an accumulation point of $S$.

Accumulation implies sequence in a metric space:

Let $(X,d)$ be a metric space, let $S\subseteq X$, and let $x$ be an accumulation point of $S$.

For each positive integer $n$, let $B_n$ be the open ball about $x$ with radius $\frac1n$. Since $x$ is an accumulation point of $S$, $S\cap B\setminus\{x\}$ is non-empty for each $n$. Thus by the axiom of countable choice, there is a sequence $(s_n)$ in $S$ such that $s_n\in B_n$ for each $n$. Then since the set of all these open balls form a neighborhood basis at $x$, $(s_n)$ converges to $x$.

Accumulation does not imply sequence in a general topological space:

Let $S=\omega_1$ be the first uncountable ordinal. Let $X=\omega_1^+$ be considered under the order topology. Then $\omega_1$ is clearly an accumulation point of $S$. Suppose $(s_n)$ is a sequence in $S$. Then $(s_n)$ is bounded above by its union, which is a countable union of countable ordinals, and hence a countable ordinal. Thus $\bigcup_n s_n<\omega_1$, but $s_n$ never exceeds it, so $(s_n)$ does not converge to $\omega_1$.

Edit: The fact that a countable union of countable sets is countable is called the countable union condition, and is a form of the axiom of choice weaker than the axiom of countable choice but stronger than the axiom of countable choice for collections of finite sets.


sufficiency: suppose that $\{s_n\}$ converges to x, for every m, there is an N such that for all $n>N$, $|x-s_n| < 1/m$. Put $1/m$ less than any given $\epsilon > 0$. We see that $x$ is an accumulation point since $\{s_n\} \in S$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.