# Set $V =\bigcap_{n\in \mathbb N} V_n$. Which of the following statements are true?

Let $$\{V_n\}$$ be a sequence of open and dense subsets of $$\mathbb R^N$$ . Set $$V =\bigcap_{n\in \mathbb N} V_n$$. Which of the following statements are true?

a. $$V \neq ∅.$$

b. $$V$$ is an open set.

c.$$V$$ is dense in $$\mathbb R^N$$

My Try:- (a)We have $$V_n^o=V_n$$ and $$\overline {V_n}=X ,\forall n\in \mathbb N$$. Consider $$V =\bigcap_{n\in \mathbb N} V_n$$. It is enough to prove that $$X\setminus V\neq X.$$

$$X\setminus V= X\setminus \bigcap_{n\in \mathbb N} V_n=X\setminus \bigcap_{n\in \mathbb N} V_n^o=\bigcup_{n\in \mathbb N}\overline{X \setminus V_n}$$

(b)Enough to prove that $$X\setminus V$$ is closed $$\overline{X\setminus V}=\overline{X\setminus \bigcap_{n\in \mathbb N} V_n}=\overline{X\setminus \bigcap_{n\in \mathbb N} V_n^o}=\overline{\bigcup_{n\in \mathbb N}\overline{X \setminus V_n}}=\bigcup_{n\in \mathbb N}\overline{\overline{X \setminus V_n}}=\bigcup_{n\in \mathbb N}{X \setminus V_n}=X\setminus V \implies X\setminus V$$ is a closed set. So, $$V$$ is open. ($$\because$$ Closure of the union = Union of closures)

(c)$$\overline{V}= \overline{\bigcap_{n\in \mathbb N} V_n}\subseteq \bigcap_{n\in \mathbb N}\overline V_n=X.$$ How do I prove the reverse inclusion?

(b) need not be true: If $$V_n=\Bbb R\setminus\{\frac1n\}$$, then the intersection of all $$V_n$$ is not open.
• Baire theorem states that any complete metric space is not the union of a countable nowhere dense subset. Suppose on contrary $V=\emptyset$,$X=X\setminus V= X\setminus \bigcap_{n\in \mathbb N} V_n=X\setminus \bigcap_{n\in \mathbb N} V_n$ – Math geek Jun 6 at 14:39
• that is $X=\bigcup_{n\in \mathbb N}X \setminus V_n$ – Math geek Jun 6 at 14:41
• @Mathgeek this does not need an extra proof once you know c). It's false in a general space, so you need something specific of $\mathbb{R}^n$ (like the fact that Baire holds). – Henno Brandsma Jun 6 at 16:36
• @Mathgeek if $V_n$ is open and dense, its complement is nowhere dense. Hence the link with your version of Baire. And Baire holds inside every non-empty open set too, hence the denseness of the intersection. – Henno Brandsma Jun 6 at 16:39