Question:

Let $\displaystyle A \subset \mathbb{R}$, then prove that:

$\displaystyle A$ is compact $\displaystyle \Rightarrow A$ is closed and bounded.

approach:

$\displaystyle A$ is compact then $\displaystyle A$ is bounded:

The open intervals $\displaystyle (-n,n), n \in \mathbb{Z}^+$ have union $\displaystyle \mathbb{R}$, so they definitely cover $\displaystyle A$. since $\displaystyle A$ is compact and the above covering is open so it admits a finite subcovering which means $\displaystyle A \subset (-N,N)$ for some $\displaystyle N \in \mathbb{R}$ and hence $\displaystyle A$ is bounded.

now to show that A is closed too.

here is the part where i am stuck:

Let $\displaystyle (x_n)$ be a sequence in $\displaystyle A$ such that $\displaystyle x_n \rightarrow x$. I must show that $\displaystyle x \in A$. i tried to use contradiction method, that is, i assumed that there exists a sequence $\displaystyle (x_n)$ in $\displaystyle A$ with $\displaystyle x_n \rightarrow x$ such that $\displaystyle x \notin A$. now i must exhibit an open covering of $\displaystyle A$ which does NOT admit a finite subcovering in order to bring up a contradiction but i am not able to go further.