# Prove all closed subspace of a compact space are compact: Redundancy?

I see a redundancy in the following proof of the statement. First, we have a lemma that this proof uses

A subspace $A \subseteq X$ is compact if and only if every open cover of $A$ by open subsets of $X$ has a finite subcover.

Now the proof

Let $X$ be compact and $A \subseteq X$ be closed. Let $(U_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. Then $(U_i)$ together with $X \setminus A$ is an open cover of $X$. Since $X$ is compact, it has some finite subcover, $J \subseteq I$ that is finite such that

$$(X\setminus A) \cup \left(\bigcup_{j \in J} U_j\right)=X$$

Then $A \subseteq \bigcup_{j \in J} U_j$ and by the lemma, we have $A$ to be compact.

Now, I think I can shorten this as follows

Let $X$ be compact and $A \subseteq X$ be closed. Let $(U_i)_{i \in I}$ be a cover of $A$ by open subsets of $X$. Since $X$ is compact, there exists some finite $J \subseteq I$ such that $\bigcup U_j=X$. Since $A \subseteq X$, we clearly have $A \subseteq \bigcup_{j \in J} U_j$ as required.

Well..? Why do we need to consider the complement $X\setminus A$? Why put that in the finite open cover that we already have? Shouldn't $\bigcup U_j$ suffice on its own as a finite open cover of $A$ anyway? I do't see the role of $X\setminus A$ at all in the proof. can someone explain?

• Your proposed "lemma" is usually taken to be the DEFINITION of compactness. $\qquad$ – Michael Hardy May 5 '16 at 23:11

## 2 Answers

Since $X$ is compact, there exists some finite $J\subseteq I$ such that $\bigcup U_j=X$.

That is the mistake: The open cover that covers $A$ was not assumed to cover all of $X$, so it's not an open cover of $X$. If it doesn't cover $X$ then it can't have a finite subset that covers $X$.

• Ah hang on, how about if I write $(U_i)_{i \in I}$ such that $\cup U_i=X$? Such a family of open cover with finite subcover should exist since $X$ is compact, right? Then the subcover surely contains $A$? So instead of the cover of $A$ by subsets of $X$, but considering the open cover of $X$ itself. And the finite subcover that must exist. Which must have $A$ in it. – John Trail May 5 '16 at 23:15
• @JohnTrail : Yes, but that doesn't rule out the possibility that some covers of $A$ that don't cover all of $X$ lack finite subcovers. One must rule that out in order to prove what was to be proved. $\qquad$ – Michael Hardy May 5 '16 at 23:21
• Oh, i get what it's meant now...thanks a lot Michael! – John Trail May 5 '16 at 23:24

There's no reason why you would have $\bigcup_i U_i = X$. usually it's false.

In your argument, you don't use the fact that $A$ is closed, so if it was correct, it would work with any $A$. And $]0,1[$ is not compact while $[0,1]$ is