# A question about the proof of closed and bounded $\Rightarrow$ compactness

A subset $$S$$ of $$\mathbb{R}$$ is compact iff it is bounded and closed. I am going over the proof of this theorem. To show that bounded and closed implies compactness, the argument follows the following strategy:

1. Let $$\mathcal{F}$$ be an open cover of $$S$$.
2. Define $$S_x = S \cap (-\infty, x]$$
3. $$B = \{ x : S_x \text{ is covered by a finite subcover of } \mathcal{F} \}$$
4. If $$B$$ is bounded, then it will yield a contradiction
5. Therefore, $$B$$ is not bounded and it follows that $$S$$ is compact.

My question is why $$B$$ not being bounded implies that $$S$$ is compact.

My guess is that if $$x \in B$$, then $$S_x$$ is covered by a finite subcover of $$\mathcal{F}$$. $$S \subset S_x$$ if $$x > \sup S$$, so if $$B$$ is not bounded then all the points of $$S$$ are covered by a finite subcover of $$\mathcal{F}$$, which implies that $$S$$ is also covered by a finite subcover of $$\mathcal{F}$$. One thing that is unsatisfactory with my observation is that according to this view, it would be enough to show there exists $$y$$ such that $$\sup S < y$$ and $$y \in B$$.

• en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem Commented Oct 28, 2022 at 22:25
• That is to say that your equivalence does not hold in general. Commented Oct 28, 2022 at 22:25
• Apologies, maybe it is implicit in your question, but you should specify that $S$ is a subset of $\mathbb{R}$. Commented Oct 28, 2022 at 22:29
• I will be very honest with you. I did not understand that proof. I think we have to use both closedness and boundness somehow. the proof in the wikipedia link I posted is slightly different. do you have a reference for yours? Commented Oct 28, 2022 at 23:09
• Yes, you understand the proof correctly. Step 4 shows that $B$ is unbounded, hence there is $y \in B$ s.t. $\sup S < y$. So, $S=S \cap (-\infty, y]=S_y$ is covered by finitely many members of $\mathcal{F}$, thus $S$ is compact.
– Vit
Commented Oct 29, 2022 at 2:48

Let $$S \subseteq \mathbb{R}$$ be closed and bounded. Assume for contradiction that $$B$$ is bounded above. Then $$q:= \sup B< \infty$$.
Case 1: $$q \notin S$$. Since $$S$$ is closed, its complement is open and we can pick $$r>0$$ such that $$S$$ doesn't intersect the interval $$(q-r,q+r)$$. By definition of supremum, $$S \cap (- \infty, q-r/2]$$ is covered by a finite subcover of $$\mathcal{F}$$. But then the same must be true of $$S \cap (- \infty, q+r/2]$$, which contradicts that $$q$$ is the supremum.
Case 2: $$q \in S$$. Then there is an open set $$Q \in \mathcal{F}$$ and $$r>0$$ such that $$(q-r,q+r) \subseteq Q$$. By definition of supremum, $$S \cap (- \infty, q-r/2]$$ is covered by a finite subcover of $$\mathcal{F}$$. But then that same subcover with $$Q$$ included if not already must cover $$S \cap (- \infty, q+r/2]$$, which contradicts that $$q$$ is the supremum.
In either case, we have a contradiction, so we conclude that $$B$$ isn't bounded above. Since $$S$$ is bounded, there is $$N \in \mathbb{R}$$ such that $$S \cap (- \infty,N]=S$$. Pick $$M \in B$$ such that $$M \geq N$$. Then we have a finite subcover of $$S \cap (- \infty,M]=S$$.
While it seems we didn't use that $$S$$ is bounded below, this is in fact needed to show that $$B$$ is nonempty, before the argument above can be used