I believe the following are equivalent, but I am wanting to show why $(1)\implies (2)$ in regards to below. I came across a proof on this site as to why accumulation/limit points are related to Hausdorff to try to start the proof, but I got stuck on the part that discussing $N-1$ elements. After getting stuck, I wasn't sure even if this was an appropriate route to take to prove why $(1)$ implies $(2)$. Any help would be greatly appreciated!

  • For the following, $X$ is a metric space, $A\subseteq X$, and the notation $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$ denotes the sequence converging to the limit $x\in X$.

$\textbf{(1)}$ $\forall$ sequence $\lbrace a_n \rbrace_{n\in \mathbb{N}}$ that is a subset of $A$, [if $\lbrace a_n \rbrace_{n\in \mathbb{N}}\rightarrow x$, then $x\in A$].

$\textbf{(2)}$ the set $A$ is closed.

I have previously shown the following are equivalent definitions of a closed set concerning the problem.

\begin{align*} A \text{ is a closed set} &\leftrightarrow X-A \text{ is open }\\ &\leftrightarrow closure(A)=A\\ &\leftrightarrow boundry(A)\subseteq A\\ &\leftrightarrow \forall \text{ limit point } c \text{ of } A, c\in A.\\ \end{align*}

  • $\begingroup$ As its name suggests, a limit point $x$ (in a metric space) is a limit point precisely because there exists a sequence $x_n$ such that $x_n\to x$ as $n\to\infty$. Just think of open balls with radius $1/n$ to construct such a sequence. $\endgroup$ – Clayton Jul 10 '18 at 18:07
  • $\begingroup$ I am using the following as a definition of a limit point: $\endgroup$ – W. G. Jul 10 '18 at 18:09
  • $\begingroup$ Let $A\subseteq X$. Let $b\in X$. We say $b$ is a limit point of $A$ iff $\forall B_x$ (an open ball with center $x$), $(B_x-\lbrace x \rbrace)\cap A \neq \phi$. $\endgroup$ – W. G. Jul 10 '18 at 18:10
  • $\begingroup$ Is that the best way to solve this problem? Could I see more how to solve it then? $\endgroup$ – W. G. Jul 10 '18 at 18:20

This is true. Let $x$ be a limit point in a metric space $(X,d)$. We claim $X$ is closed if and only if for every convergent sequence, it converges in $X$.

$\Rightarrow$ Assume $X$ is closed, e.g. it contains all of its limit points. Now let $\{x_n\}$ be a convergent sequence contained in $X$ that converges to $x$. Then for any $r>0$, $B_r(x) \cap X \neq \emptyset$, so $x$ is a limit point, therefore it's in $X$.

$\Leftarrow$ Assume for every $\{x_n\}$ be a convergent sequence that is contained in $X$, it converges in $X$. Let $x$ be a limit point, then take $r_n = \frac{1}{n}$ to get a convergent sequence. Then by assumption $x \in X$, so $X$ has all of its limit point, so by def it's closed.

  • $\begingroup$ Can you explain more what you mean by taking $r_n=\frac{1}{n}$ to get a convergent sequence? Reading the first comment, I see you do something with open balls. Is the convergent sequence something like $B(x, r_n)$ (I am not sure)? I see your using the last definition of the closed set to prove this but not sure how you go about finding a convergent sequence. $\endgroup$ – W. G. Jul 10 '18 at 18:40
  • $\begingroup$ To construct a sequence, let $x_1$ in $B_{1}(x) \cap X$ and similarity for $x_n \in B_{1/n}(x) \cap X$ $\endgroup$ – user518441 Jul 10 '18 at 18:44
  • $\begingroup$ I just am thinking of the reasons as to why that would be well defined in the first place.Why would $\forall n\in \mathbb{N}$, $x_n \in B_r(x)$ when $r=\frac{1}{n}$ hold true? $\endgroup$ – W. G. Jul 10 '18 at 19:01
  • $\begingroup$ I'm not sure if you are getting it from the limit point definition or what. I appreciate your help. $\endgroup$ – W. G. Jul 10 '18 at 19:09
  • $\begingroup$ It comes from the definition. $x$ is a limit point in $(X,d)$ if and only if for all $r > 0$, $B_r(x) \cap X \neq \emptyset$. Taking $r = \frac{1}{n}$, we see $B_{1/n}(x) \cap X \neq \emptyset$. As it's non-empty, we can choose a point in the ball and choose that to be our $x_n$ $\endgroup$ – user518441 Jul 10 '18 at 19:30

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