The rationals are an intersection of open set-contradiction to Baire theorem

Question

Prove or contradict:

There are infinite open sets $U_1,U_2,...\in \mathbb{R}$ such that : $\mathbb{Q}=\bigcap^\infty _{i=1} U_i $

So I saw the following answer:

No, becuase if it was true,$\mathbb{Q}=\bigcap^\infty _{i=1} U_i $ ,then:

$\phi = \bigcap_{q\in \mathbb{Q}}\mathbb{R} \setminus \{q\} \ \cap \bigcap^\infty_{i=1}U_i$

which contradicts $Baire$ theorem.

Why does it contradicts it?

Answer 1

Baire's theorem says that the intersection of countably many open sets with a certain property is never empty.

• What is that property?

• Do you see why the sets of the form $\mathbb{R}\setminus\{q\}$ and $U_i$ have this property? (HINT: the first is easy; for the second, what can you say about the sets $U_i$ and $\mathbb{Q}$?)

Side note: "$\mathbb{R}\setminus q$" is incorrect notation. "$A\setminus B$" only makes sense if $A, B$ are both sets.

Answer 2

$\mathbb{R}\setminus\{q\}$ is open as $\{q\}$ is closed, and dense as $\{q\}$ is not isolated.

If $\mathbb{Q} =\cap_n U_n$, where all $U_n$ are open, $\mathbb{Q} \subset U_n$ for all $n$, so every $U_n$ is dense as it contains a dense set: $\mathbb{R} =\overline{\mathbb{Q}} \subset \overline{U_n} \subset \mathbb{R}$.

The intersection of all these (countably many, as $\mathbb{Q}$ is countable) dense open sets is empty, so far from dense. This contradicts Baireness of the reals, which follows either from its local compactness or it's metric completeness, depending on your taste or background.