Intersection of two compact sets in $\mathbb R$ is compact (without Heine-Borel theorem)

Question

I refer to this question here.

The answer here used Heine-Borel theorem since it used the fact, that every compact set is closed.

Without using it, how can I show that union of two compact sets in $\mathbb R$ is compact?

Answer 1

In the title you ask about the intersection, while in the body of the question you ask about the union. The union is literally trivial from the definitions, since an open cover of $K_1\cup K_2$ is an open cover of $K_1$ and also an open cover of $K_2$.

The intersection is simple as well: If $O$ is an open cover of $K_1\cap K_2$ then $O\cup\{X\setminus(K_1\cap K_2)\}$ is an open cover of $K_1$. Except for that to work we need to kow that compact sets are closed. No, that doesn't require Heine-Borel, it's true in much greater generality:

If $X$ is a Hausdorff space and $K\subset X$ is compact then $K$ is closed in $X$.

Proof: Suppose $p\in X\setminus K$. For every $x\in K$ choose open sets $U_x$ and $V_x$ with $x\in U_x$, $p\in V_x$ and $U_x\cap V_x=\emptyset$. Since $K$ is compact there exist $x_1,\dots,x_n\in K$ with $K\subset\bigcup_{j=1}^n U_{x_j}$. So if $V=\bigcap_{j=1}^n V_{x_j}$ then $V$ is open, $p\in V$ and $V\cap K=\emptyset$, which is to say $V\subset X\setminus K$.

Note regarding the Obvious Questions that arise here: It's trivial to give an example of a non-Hausdorff space with a compact but non-closed subset, simply because finite sets are compact. It's perhaps not quite so obvious how to give an example of two compact sets with non-compact intersection:

Let $X=[0,1]\cup\{-1\}$. Let $\tau$ be the standard topology on $[0,1]$. Then $$B=\tau\cup\{(V\setminus\{0\})\cup\{-1\}:V\in\tau, 0\in V\}$$is a basis for a topology on $X$. If $K_1=[0,1]$ and $K_2=(0,1]\cup\{-1\}$ then $K_1$ and $K_2$ are compact but $K_1\cap K_2$ is not compact.