Rudin' PMA: theorem 2.36 confusion about the proof

Question

This theorem is fully presented in this question:Confusion in Theorem 2.36 Baby Rudin My question is different from this existing question, but I have understood where is the contradiction of this theorem. Thank you for the help.

but I have a new question now. if all K_a do not contain any points in K_1, does this mean that the intersection of K_1 with all K_a is empty? and does this mean that the intersection of K_1 with any finite collection of K_a is also empty?

I am thinking of using picture to explain. suppose each K_a is a circle. Then if all circles of K_a have no overlaps with K_1, then any finite collection of K_a also have no overlaps with K_1's circle. Is this right?

Answer 1

No, if $K_1$ has no point that belongs to every $K_\alpha$ doesn't by itself mean that intersection of $K_1$ with any collection of $K_\alpha$-s is empty. For example, consider $K_i = \{n \in \mathbb N | n > i\}$.

Contradiction is achieved by assuming $K_1 \cap \left(\bigcap\limits_\alpha K_\alpha\right) = \varnothing$ and deducing from it that $K_1 \cap \left(\bigcap\limits_{\alpha \in A}K_\alpha \right) = \varnothing$ for some finite $A$.

Note that statements "$K_1$ has no points contained in any other $K_\alpha$" and "no point of $K_1$ belongs to every $K_\alpha$" are different: the former means that for every point in $K_1$ and any other set $K_\alpha$, the point doesn't belong to the set, while the latter means that for every point in $K_1$ there exists set $K_\alpha$ s.t. the point doesn't belong to the set.

Answer 2

The point is that we assume no point of $K_1$ belongs to each and every $K_\alpha$. Well, if it doesn't belong to $K_\alpha$, it belongs to $G_\alpha=K_\alpha^c$. But the $G_\alpha$'s are an open cover of $K_1$. So $K_1$ is a subset of a union of finitely many of these, and thus $K_1$'s intersection with $\bigcap_{j=1}^n G_{\alpha_j}^c = \bigcap_{j=1}^n K_{\alpha_j}$ is empty! That's the contratiction.