# Proof by contradiction that compact sets are closed

I have started on the following attempt that if a set $$E \subseteq \mathbb{R}^{k}$$ is compact, then it is closed (I have yet to cover my second case). Am I on the right track?

Suppose $$E$$ is compact, but it is open. Then, for each $$x\in E$$, $$\exists$$ $$r_{x}>0$$ s.t. $$B(x, r_{x})\subseteq E$$. $$E \subseteq \bigcup_{x\in E}\{B(x,r_x)\}$$, and so $$\{B(x, r_{x})\}_{x\in E}$$ is an open cover of $$E$$. We now show that, from $$\{B(x, r_{x})\}_{x\in E}$$, can be derived an open cover of $$E$$ that does not have a finite subcover.

Consider all $$y\in E$$ which are such that there does not exist an $$x\in E$$, $$x\neq y$$, such that $$y\in B(x, r_x)$$. Let $$\Phi$$ be the set containing all balls with such centres.

Now, remove some arbitrarily chosen ball from $$\Phi$$. Call this ball $$B(x_{1}, r_{x_1})$$. Note that $$\{B(x, r_{x})\}_{x\in E \setminus \{x_{1}\}}$$ is an open cover of $$E$$. Now, choose some ball $$B(x_{2}, r_{x_{2}})$$ in $$\Phi \setminus \{B(x_{1}, r_{x_1})\}$$ which satisfies the following two conditions:

1. It is not the only set in $$\{B(x, r_{x})\}_{x\in E \setminus \{x_1\} }$$ that contains $$x_{1}$$

2. $$\exists B(a, r_{a})$$ such that $$a \neq x_{2}$$, and $$B(a, r_{a}) \in \{B(x, r_x)\}_{x\in E \setminus \{x_{1}\}}$$, such that $$x_{2} \in B(a, r_{a})$$

First, consider the case in which there does not exist any such $$B(x_{2}, r_{x_{2}})$$ in $$\Phi \setminus \{B(x_{1}, r_{x_1})\}$$. Then, each ball in $$\{B(x, r_{x})\}_{x\in E \setminus \{x_1\} }$$ is of at least one of the following types:

1. It is the only ball in that collection which contains $$x_{1}$$.

2. It is the only ball in that collection which contains the point which is at its centre.

A ball of the first type cannot be removed to form a subcover of $$\{B(x, r_x)\}_{x\in E \setminus \{x_{1}\}}$$, because then the presumptive subcover would not include $$x_{1}$$.

A ball of the second type cannot be removed to form a subcover of $$\{B(x, r_x)\}_{x\in E \setminus \{x_{1}\}}$$, because then the presumptive subcover would not include any such ball's central point.

So, if there does not exist a ball in $$\Phi \setminus \{B(x_{1}, r_{x_1})\}$$ that satisfies both condition (1) and (2), then no balls in $$\Phi \setminus \{B(x_{1}, r_{x_1})\}$$ can be removed to form a finite subcover of $$\{B(x, r_{x})\}_{x\in E \setminus \{x_{1}\}}$$. But, neither can any balls in $$\{B(x, r_x)\}\setminus \Phi$$. So $$\{B(x, r_{x})\}_{x\in E \setminus \{x_{1}\}}$$ does not have a finite subcover, contradicting the compactness of $$\{B(x, r_{x})\}_{x\in E}$$

Second, consider the case in which there exists at least one ball in $$\Phi \setminus \{B(x_{1}, r_{x_1})\}$$ that satisfies conditions (1) and (2).

• "Suppose E is compact, but it is open." Do you know that if a set isn't open, that doesn't mean that it's closed? Honestly, I stopped reading there, because if you assumed that, then the rest of your argument would be built on a false assumption. Jul 27, 2022 at 23:45
• @JonathanZsupportsMonicaC Apologies for not recognising that. Thanks for the reminder. Jul 27, 2022 at 23:57
• I think it's much easier to prove that the complement of $E$ is open, which you do by choosing an arbitrary point in the complement and proving that there's an open ball around it that is disjoint from $E$. Jul 28, 2022 at 0:21
• @Charles - No need to apologize. So many people make the same mistake that there are jokes based on it. Jul 28, 2022 at 3:00
• So, are you planning to repair this serious flaw? Jul 28, 2022 at 20:30

Let $$E$$ be compact. If $$E^c$$ is empty, then $$E$$ is closed because the empty set is open and the complement of an open set is closed.

If $$E^c$$ is not empty then there exists some $$p\in E^c$$. Let $$\{U_i\}$$ be a family of open sets defined by $$U_i=\{x\in\mathbb R^k: |x-p|\gt\frac{1}{i}\}$$, for $$i\in\mathbb N$$. Note that I am defining the natural numbers as the counting numbers so division by zero is not an issue.

Every element $$e\in E$$ is some positive distance from $$p$$ and we can always find $$i$$ such that $$\frac{1}{i}\lt |e-p|$$. Therefore, the $$\{U_i\}$$ form an open cover for $$E$$. But $$E$$ is compact so there exists a finite subcover $$\{U_{i_1},...,U_{i_n}\}\subset\{U_i\}$$ of $$E$$. Since this collection is finite, there is an index $$i_r$$ for which $$\frac{1}{i_{r}}$$ is minimum.

Then $$p$$ is contained in an open ball of radius $$\frac{1}{2I_r}$$ which does not intersect $$E$$. Since this construction can be carried out for each $$p\in E^c$$, $$E^c$$ is open and, therefore, $$E$$ is closed.

Let $$(\mathbb{R}^{n},d)$$ be the Euclidean metric space.

We say a subset $$X\subseteq\mathbb{R}^{n}$$ is compact iff every sequence in $$X$$ admits a convergent subsequence in $$X$$.

Suppose by contradiction that $$X$$ is not closed. This means $$X\not\supseteq\partial X$$. Consider then that $$x\in\partial X\backslash X$$.

Since $$x$$ is an adherent point of $$X$$, there is a sequence of points in $$(x_{n})_{n\in\mathbb{N}}$$ in $$X$$ which converges to $$x$$.

Due to the compactness of $$X$$, we conclude that there is a subsequence $$(x_{s(n)})_{n\in\mathbb{N}}$$ which converges to $$x\in X$$. But we also know that a sequence of points in a metric space converges to some point iff every subsequence converges to the same point. Thus we deduce that $$x_{n}\to x\in X$$, which is a contradiction.

Hopefully this helps!

• The OP's definition of compactness is the general one (existence of a finite subcover for any open cover), not the particular sequential one. Sep 11, 2022 at 11:03
• Dear @AnneBauval, in the context of metric spaces (which is the case), both are equivalent. Sep 11, 2022 at 19:58
• Yes, I know, but I doubt the OP does. Sep 11, 2022 at 20:45
• @AnneBauval do you have any suggestion then? Sep 11, 2022 at 21:33
• @AnneBauval I will keep it as it may be useful for future reference. Sep 11, 2022 at 21:51