# Thread: Easy compactness proof, What is wrong with my picture/logic?

1. ## Easy compactness proof, What is wrong with my picture/logic?

Theorem. If {Kα} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {Kα} is nonempty, then Kα is nonempty.
Proof. Fix a member K1 of {Kα} and put Gα=Kcα [this denotes the complement of Kα]. Assume that no point of K1 belongs to every Kα. Then the sets Gα form an open cover of K1 [this took me a bit but that's by the last assumption]; and since K1 is compact, there are finitely many indices α1,,αn such that K1ni=1Gαi [so far so good]. But this means that
K1Kα1Kαn
is empty, in contradiction to our hypothesis.

The area enclosed by the green is $\displaystyle K_1$, by the fuschia $\displaystyle K_2$ and together they make up the set $\displaystyle \{K_\alpha \}$ The blue lines indicate the area of $\displaystyle K_2^c$. And $\displaystyle G_\alpha$ is the union of the blue shaded area and the part of $K_2$ not occupied by $K_1$ (which I didn't draw out as it looked too messy for one picture)

Line for line of the proof;

Fix a member of $K_1$ of $\{K_\alpha\}$...

I think I'm good on this line. It just means call the green blob's insides $K_1$

Assume that no point of $K_1$ belongs to every $K_\alpha$.

This means that $K_1 \cap K_2$ is empty. So the bit of the green that overlaps the fuschia is empty.

Then the sets $G_\alpha$ form an open cover of $K_1$;

As blue area and the compliment to $K_1$ definitively covers everything but $K_1 \cap K_2$, but that's empty, so this looks acceptable to me.

and since $K_1$ is compact, there are finitely many indices $\displaystyle \alpha_1,...,\alpha_n$ s.t. $K_1 \subset G_{\alpha_1} \cup ... \cup G_{\alpha_n}$

I think this means that there is a collection of open sets which cover $K_1$. This is only true because we've removed that chunk with the; "Assume that no point of $K_1$ belongs to every $K_\alpha$.".

But this means that $K_1 \cap K_2 \cap ... \cap K_{\alpha_n}$ is empty, in contradiction to our hypothesis.

I don't see how this contradicts anything. He assumed the intersection of all K's was empty and then showed that the intersection of all K's is empty. It looks like he's written $A \land A$ conclude $\perp$ which is nonsense.

Where are my errors?

2. ## Re: Easy compactness proof, What is wrong with my picture/logic?

Hi,
It appears to me that your main problem is unfamiliarity with some basic set operations. I have rewritten your theorem and proof slightly. If you have question about what I have posted, post again and I will try to explain.