# Two versions of Baire Category Theorem for Profinite Groups

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.