Is A is open/closed?

Question

let $\{V_n\}_{n=1}^{\infty}$ be sequence of open and dense subsets of $\mathbb{R}^{\mathbb{N}}$. set $A= \cap_{n=1}^{\infty}V_n$ then A is

choose the correct option

$1.)$ open

$2.)$ closed

$3.)$ countable

$4.)$ uncountable

My attempt : I take $n = 1.$

Since $\mathbb{R}$ is a complete metric space, the Baire category theorem implies that $A$ is dense, in particular it is nonempty. We also claim that $A$ has to be uncountable. Indeed, suppose that $A=\{x_1, x_2, \ldots \}$ was countable, and define the sets $W_n := \mathbb{R} \smallsetminus \{x_n\}$. Then clearly each $W_n$ is an open dense subset of $\mathbb{R}$, and thus applying the Baire category theorem to the countably many sets $V_n,W_n$ yields that $\bigcap_{n \geq 1} (V_n \cap W_n)$ is dense in $\mathbb{R}$, in particular it is nonempty. On the other hand $$ \bigcap_{n \geq 1} (V_n \cap W_n) = \left(\bigcap_{n \geq 1} V_n \right) \cap \left( \bigcap_{n \geq 1} W_n \right) = A \cap (\mathbb{R} \smallsetminus \cup_{n \geq 1} \{ x_n \} ) = A \cap (\mathbb{R} \smallsetminus A) = \emptyset.$$ This is a contradiction, so $A$ is uncountable.

Im confusing that A must be open?

Any hints/solution

thanks u

Answer 1

No, $A$ doesn't have to be open. Let $\{q_n\,|\,n\in\mathbb{N}\}$ be an enumeration of $\mathbb Q$. For each $n\in\mathbb N$, let $V_n=\mathbb{R}\setminus\{q_n\}$. Then $V_n$ is open and dense. But $\bigcap_{n\in\mathbb N}V_n=\mathbb{R}\setminus\mathbb Q$, which is not open.

Answer 2

Consider $V_n =(-1/n,1/n)$, $n \in \mathbb{Z^+}$, open and dense subsets.

$A:=\displaystyle{\cap}_{n \in \mathbb {Z^+}} V_n= ${$0 $}.

Hence $A$ is not open.

Rule out option $1)$.

Note: In this particular example $A$ is closed.

This does not!!! Imply that option $2)$ is correct (cf.Carlos' example).