# 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

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
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