Does the Compactness Property Hold for Finite Subcollections of a Metric Space?

[mynameisfunk]

Hey guys, sorry for practically flooding the forum today but I have an analysis exam and nobody is more helpful than phys forum folk.

I am having trouble understanding a line in Rudin. Thm 2.36:
If {$$K_{\alpha}$$} is a collection of compact subsets of a metric space X s.t. the intersection of every finite subcollection of {$$K_{\alpha}$$} is nonempty, then $$\bigcap K_{\alpha}$$ is nonempty.

pf:
Fix a member $K_1$ of {$K_{\alpha}$} and put $G_{\alpha}=K^{c}_{\alpha}$. Assume that no point of $K_1$ belongs to every $K_{\alpha}$. Then the sets $G_{\alpha}$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,...,\alpha_n$ such that $K_1 \subset G_{\alpha_{1}} \cup... \cup G_{\alpha_{n}}$. But this means that $K_1 \cap K_{\alpha_{2}} \cap... \cap K_{\alpha_{n}}$ is empty, in contradiction to our hypothesis.



I do not see how this proves anything. We are assuming the opposite of the theorem is true and we arrive at the opposite of our result, but that shouldn't really constitute a proof should it?

 

[awkward]

Hey guys, sorry for practically flooding the forum today but I have an analysis exam and nobody is more helpful than phys forum folk.

I am having trouble understanding a line in Rudin. Thm 2.36:
If {$$K_{\alpha}$$} is a collection of compact subsets of a metric space X s.t. the intersection of every finite subcollection of {$$K_{\alpha}$$} is nonempty, then $$\bigcap K_{\alpha}$$ is nonempty.

pf:
Fix a member $K_1$ of {$K_{\alpha}$} and put $G_{\alpha}=K^{c}_{\alpha}$. Assume that no point of $K_1$ belongs to every $K_{\alpha}$. Then the sets $G_{\alpha}$ form an open cover of $K_1$; and since $K_1$ is compact, there are finitely many indices $\alpha_1,...,\alpha_n$ such that $K_1 \subset G_{\alpha_{1}} \cup... \cup G_{\alpha_{n}}$. But this means that $K_1 \cap K_{\alpha_{2}} \cap... \cap K_{\alpha_{n}}$ is empty, in contradiction to our hypothesis.



I do not see how this proves anything. We are assuming the opposite of the theorem is true and we arrive at the opposite of our result, but that shouldn't really constitute a proof should it?

Yes, that is a valid proof.

The method is indirect proof, or proof by contradiction. You want to prove a statement S. You start by assuming S is false. This leads to a contradiction. The conclusion is that your assumption that S is false is incorrect, i.e. S is true.

 

[mynameisfunk]

I am kind of confused by the form of proving if X is true then Y is true by saying If Y isn't true then X isn't true. It seems like it is not disproving that if X is true then Y may still not be true.

 

[losiu99]

"It seems like it is not disproving that if X is true then Y may still not be true. "

But since we proved "If Y isn't true then X isn't true", such situation clearly cannot happen.

 

[HallsofIvy]

The "contrapositive" of the statement "if P then Q" is "if not Q then not P". Notice that we have not only changed each part to "not", we have swapped hypotheses and conclusion.

If the hypotheses cannot be true when the conclusion is false then knowing that the hypothesis is true tells us that the conclusion is true. A statement is true if and only if its contrapositive is true.