One of the definition I am using is a set is closed if its complement is open.
Definition of open set is that for every element in the set, there exists a positive $\delta$ such that the open $\delta$-ball around the element is a subset of the original set.
The other definition is that a set is closed if for every convergent sequence $\{a_n\}\in \mathbb{original \space set}$, the limit is also a member of the original set.
How do I show the equivalence?
Best Answer
Let $S$ be open in metric space $X$. Let $C=X\setminus S$, i.e., the complement. Let $d$ be a limit point of $C$. If $d\notin C$, then $d\in S$, which is a contradiction since there is no neighborhood of $d$ that solely contains points of $S$ by the definition of a limit point. Hence $C$ is closed.
To prove the converse, we let $C$ be closed by your second definition in $X$. Hence, $C$ contains all of its limit points. Suppose $S=X\setminus C$ is not open. This means that for some $x\in S$, $\forall \epsilon>0$, there is a point in $C$ that is within $\epsilon$ of $x$.
Let $a_n$ be the sequence of points such that $a_n\in C$ and the distance between $x$ and $a_n$ is less than $\frac1n$. This is a convergent subsequence, so $x$ is a limit point of $C$, a contradiction.