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*}