2
$\begingroup$

Let $G$ be a profinite group:

Theorem 1. Let $C_{1},C_{2},...$ be a countably infinite set of nonempty closed subsets of $G$ having empty interior. Then $$G \neq \bigcup_{n}^{\infty}C_{i}.$$

Theorem 2. Let $(C_{n}\mid n\in \mathbb{N})$ be a family of closed subsets of $G$ such that $\bigcup_{n}C_{n}$ contains a nonempty open set, then some $C_{n}$ contains a nonempty open set.

Both theorem are versions of Baire's Category Theorem for Profinite Groups. I know a proof for theorem 1, but I'm interested in the statement of theorem 2. My question is: how to show that theorem 1 implies theorem 2? Maybe it's a simple question, but I cannot see.


According to Tsemo's answer, can someone give me a hint for prove the theorem 2? I can only use results of profinite groups.

$\endgroup$
1
$\begingroup$

Theorem $2$ is a consequence of Baire theorem. If every $C_n$ has an empty interior, so is $\cup_nC_n$.

I dont believe that 2 is a consequence of $1$ since it is a weaker consequence of Baire, as to show 1, one uses the fact that if every $C_n$ has a empty interior, Baire implies that $\cup_nC_n$ has an empty interior, so it cannot be $G$. It is the fact that $\cup_n C_n$ has a non empty interior interior that is relevant to deduce 2 also.

$\endgroup$

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.