If $K \subset \mathbb{R}^2$ is compact, then $K$ is closed?

Question

If $K \subset \mathbb{R}^2$ is compact, then $K$ is closed? How I can prove it? Thanks in advance.

Answer 1

Any compact subset $K$ of a Hausdorff $X$ space is closed.

We can easily show that $X\setminus K$ is open. Let $x\in X\setminus K$. For each $k\in K$, there exist disjoint open sets $V_k$ and $W_k$ with $x\in V_k$ and $k\in W_k$. So the $\{W_k\}$, indexed by $k\in K$, form an open cover of $K$. As $K$ is compact, there is a finite subcover: there exist $k_1,\ldots,k_n$ such that $K\subset \bigcup_1^n W_{k_j}$.

Let $V=\bigcap_1^n V_{k_j}$. This is open, since it is a finite intersection of open. We have $x\in V$, since $x\in V_{k_j}$ for all $j$, and $V\subset X\setminus K$, since $V$ is disjoint with $W_{k_1},\ldots,W_{k_n}$.

Answer 2

Yes. Every compact subset of $\mathbb{R}^2$ is closed. To prove it, take any converging sequence in your compact set. You want to show that the limit belongs to your set. By compactness, there is a convergent subsequence to an element of your set. So the limit belongs to your set. Hence it is closed.

Answer 3

Let P be a point adherent to X. One may construct a sequence of points of X each lying within 1/n (n is a positive integer) of P. P is then the limit of that sequence, because P lies within 1/N for all elements with n > N. Because X is compact, there is a subsequence of the constructed one that belongs to X that converges to Q in X. Since all subsequences of a convergent sequence have the same limit, P and Q are identical and P belongs to X.