Clarification of Baire category theorem: a (counter-?)example

Question

I am trying to understand the statement of Baire's category theorem.

Why is $\bigcap_{n \in \mathbb{N}} A_n := \bigcap_{n \in \mathbb{N}} \big[(n, n+\frac{1}{2}) \cap \mathbb{Q}\big] = \emptyset$ dense in $\mathbb{R}$? As I understand $\mathbb{R}$ is a complete metric space and for each $n$, $A_n$ is an open dense set in $\mathbb{R}$, which would yield the claim by Baire's theorem. But it does not seem in accordance with any conceptual idea I have of a dense set.

Answer 1

It seems like the issue here is confusion over the definition of a dense set.

A subset $S$ of $X$ is dense $X$ if every point in $X$ is either an element of $X$ or a limit point of $S$.

It should be clear here that none of the $A_n$ here are dense in $\mathbb R$, because $-1$ is not in any $A_n$, nor is a limit point of any $A_n$.

Answer 2

$[n,n+\frac 1 2)\cap \mathbb Q$ is not dense in $\mathbb R$. It is only dense in $[n,n+\frac 1 2)$. It is also not open.

Answer 3

$A_n=(n,n+\frac12)\cap \Bbb Q$ is neither open nor dense in $\Bbb R$, therefore $\bigcap_{n\in\Bbb N}A_n$ not being dense is perfectly consistent with Baire category theorem.