Compact sets are closed.

Question

I am a beginner in studying functional analysis, so please pardon me if my argument seems absurd. I am having hard time, coming in terms with the proof of "Compact subsets of $\mathbb{R}$, are closed". Now lets say that I have a set $ A = [1,2) $, and follow the proof. If I take the fixed point $x$ as $ {2} \in \mathbb{R} \backslash A $, then for each point $ y \in A$, I $\mathbf{can}$ find small disjoint neighborhoods for $x$ and $y$. I am not sure if these small neighborhoods of $y$ cover $A$, but it is given that $A$ is compact. Or is it that I have to prove that these neighborhoods of $y$ doesn't cover $A$? Thanks

Answer 1

$A$ is not compact:

Let $\alpha=\{(-1,1.1),(1,2-\frac{1}{n}) \mid \forall n \in \mathbb{N}\}$ be an open cover of $A$.

There is not any finite subcover of $\alpha$.

Moreover, in metric spaces, any compact set is closed and bounded, and Heine–Borel theorem says that in $\mathbb{R}^n$, any bounded and closed set is compact.

Edit: I tried to comment this but it is too long:

@Prashanth If I'm right, the proof's claim you sent is: A compact set $K$ in a Hausdorff topological space $X$ is closed. Which means that the complement of any compact set $K$ in a $T_2$ topological space $X$ is open. To see that, you have to show that $\forall x \in K^c$, there is an open neighbourhood $V$ such that $x \in V \subseteq K^c$ , (The definition of open set). and to show that, try to use the definition of compactness; the fact that any finite intersection of open sets is an open set; and the fact that $X \in T_2$ ($X$ is Hausdorff).

Hint: given a point $x \in K^c$ , because $X$ is Hausdorff, try to look at the (maybe infinite) union of the open sets $U_k\subseteq K$ such that $k\in U_k \subseteq K$ such that $U_k \cap V =\emptyset$, when $x\in V\subseteq K^c$, and $V$ is an open set. Do this for all $x\in K^c$, for all $x\in K^c$ use the fact that $K$ is compact and see what you got.