# Random stuff 2026 week 8
Some stream of thought about compactness in $\mathbb{R}$.
At the core, one should exploit the (Dedekind) completeness of $\mathbb{R}$.
In particular, let us show Heine-Borel holds in $\mathbb{R}$, namely for a subset $A \subset \mathbb{R}$, $A$ is compact if and only if $A$ is closed and bounded.
Note generally, in a topological space, a compact set need not even be closed.
First let show a lemma: If $A = [a,b]$ is a closed interval, then $A$ is compact.
There are many ways to prove this. Such as via nested interval property, Konig's lemma, etc. But we will prove it using directly the least upper bound property of $\mathbb{R}$. And a useful philosophy of constructing some set $B$ whose supremum is a desired value.
Let ${\mathscr U} = \set{U_{\lambda}}_{\lambda\in \Lambda}$ be an open cover for $A$.
Consider $I_{x} = [a,x]$ for each $x$ in $A$. Note $I_{a} = [a,a] = \set{a}$. Denote $B = \set{x \in [a,b] : I_{x} \text{ is finite subcoverable by } U_{\lambda}}$.
Note since $a$ is finite subcoverable, it is in one of the open sets $U$. Which makes $a$ an interior point of said $U$. This means some $[a, a+ \delta] \subset U$.
Note, as $\set{U_{\lambda}}_{\lambda \in \Lambda}$ is an open cover for $A$, the point $a$ is in one of $U_{\lambda}$, for some $\lambda \in\Lambda$. So the set $I_{a} = [a,a]$ is finite sub-coverable. So $a \in B$, namely $B$ is not empty. In fact, above showed that some $[a, a + \delta] \subset B$.
Also, by construction, $B$ is bounded above by $b$. So by completeness axiom of the reals, $\sup B$ exists, say $\beta = \sup B$. Note $\beta \le b$.
If $\beta = b$, then this means $[a,b]$ is finite coverable. Done.
If $\beta < b$, then note as $a < a + \delta \le \beta$, we have $\beta \in (a,b)$.
We claim $I_{\beta}$ is finite coverable, namely $\beta \in B$.
Since $\beta \in [a,b]$, the point $\beta \in U$ for some $U \in \set{U_{\lambda}}_{\lambda\in\Lambda}$.
Since $U$ is open and $\beta$ is an interior point of $U$, the point $\beta - \epsilon \in U$ for some positive $\epsilon > 0$.