compactness and seq.compact of metric space

Question

prove that a sequentially compact subspace of a metric space $X$ is closed in $X$?

I wil solve this question from defination of sequentially compact but I dont know how?.(I dont want to solve it from compactness)

Answer 1

Suppose Y is a sequentially compact subset of a metric space X, so for any sequence $(x_n)_{n\geq 1} $ there is a convergent subsequence $(x_n)_{n_j\geq 1} \to s \in $ Y . Therefore, given $y_0$ a limite point of Y, i.e, $(x_n)_{n\geq 1} \subset Y$ such that $x_n \to y_0$, by definiton of Y there is a convergente subsequence $(x_n)_{n_j\geq 1} \to s $ in Y. But X is a metric space, therefore it is Housdorff, so $y_0=s \in$ Y, then Y is closed.

Answer 2

Consider a sequence $(x_n), n∈\mathbb N$ in $Y$ that converges in $X$, $ Y\subset X $, to a limit $x_0∈X$. It suffices to show that $x_0∈Y$. (Why? This follows from a basic theorem on closed subspaces of metric spaces.) As $Y$ is sequentially compact,there exists a subsequence $(x_{n_k}),k∈\mathbb N$ that converges in $Y$.Let $y_0∈Y$ be its limit.The same subsequence also converges to $ x_0 $,and by the uniqueness of limits, $x_0=y_0∈Y$. Q.E.D.