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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?

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(c) is the very statement of the Baire Category Theorem. As a consequence, (a) is also true.

(b) need not be true: If $V_n=\Bbb R\setminus\{\frac1n\}$, then the intersection of all $V_n$ is not open.

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  • $\begingroup$ Baire theorem states that any complete metric space is not the union of a countable nowhere dense subset. Suppose on contrary $V=\emptyset$,$X=X\setminus V= X\setminus \bigcap_{n\in \mathbb N} V_n=X\setminus \bigcap_{n\in \mathbb N} V_n$ $\endgroup$ – Math geek Jun 6 at 14:39
  • $\begingroup$ that is $X=\bigcup_{n\in \mathbb N}X \setminus V_n$ $\endgroup$ – Math geek Jun 6 at 14:41
  • $\begingroup$ Can you help me to complete the proof of (a)? $\endgroup$ – Math geek Jun 6 at 14:45
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    $\begingroup$ @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). $\endgroup$ – Henno Brandsma Jun 6 at 16:36
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Jun 6 at 16:39

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