1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Your proposed "lemma" is usually taken to be the DEFINITION of compactness. $\qquad$ $\endgroup$ – Michael Hardy May 5 '16 at 23:11
3
$\begingroup$

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 $X$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – John Trail May 5 '16 at 23:15
  • $\begingroup$ @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. $\qquad$ $\endgroup$ – Michael Hardy May 5 '16 at 23:21
  • $\begingroup$ Oh, i get what it's meant now...thanks a lot Michael! $\endgroup$ – John Trail May 5 '16 at 23:24
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.