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
K1⊂⋃ni=1Gαi [so far so good]. But this means that
K1∩Kα1∩…Kαn
is empty, in contradiction to our hypothesis.
The area enclosed by the green is , by the fuschia and together they make up the set The blue lines indicate the area of . And 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
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?