QUESTION:
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*}
Best Answer
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.