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.


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.


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