Sequentially compactness

Question

Let (X,d) be a metric space . If A is sequentially compact set in X then A is closed . My approach :- let x in A be any arbitrary point . Then in order to show A is closed I have to construct a sequence (Xn) such that (Xn) converges to x so that I can conclude that x is limit point . But how to construct the sequence I am not getting .

Answer 1

Maybe try to prove the contrapositive. Suppose that A is not closed (does not contain all of its limit points) and then show that it cannot be sequentially compact. Perhaps you can construct a sequence which does not have any sub sequences which converge in A.

Answer 2

I don't think your direction is quite right. A subset $A$ of a metric space is closed if it contains its limit points, if and only if every convergent sequence comprised of elements in $A$ has a limit in $A$. The bottom line is that you don't have to construct a converging sequence; you assume that you have one.

Let $(x_n)_{n = 1}^{\infty} \subset A$ be a sequence, and suppose $x_n \to x$. Since $A$ is sequentially compact, there exists a subsequence $(x_{n_k})_{k=1}^{\infty}$ of $(x_n)_{n=1}^{\infty}$ which converges to $x' \in A$. On the other hand, since $(x_{n_k})_{k=1}^{\infty}$ is a subsequence of the convergent sequence $(x_n)_{n=1}^{\infty}$, it must also converge to $x$. Since limits are unique in metric spaces, it follows that $x' = x$, and therefore $x \in A$. This proves that $A$ is closed.