How to prove that $\mathbb{Q}$ is not the intersection of a countable collection of open sets WITHOUT Baire's Category Theorem.

Question

I am looking for a way to prove that $\mathbb{Q}$ is not the intersection of a countable collection of open sets, but without Baire's Category Theorem. I am not sure if that is even possible, but if you have any idea it would help me a lot. Thanks.

Answer 1

Assume for a contradiction that $\mathbb Q=\bigcap_{n=1}^\infty D_n$ where each $D_n$ is open and of course dense in $\mathbb R$.

Fix an enumeration $\mathbb Q=\{r_n:n\in\mathbb N\}$.

Construct an infinite sequence of nested closed intervals $I_1\supseteq I_2\supseteq I_3\supseteq\cdots$ so that $I_n\subseteq D_n\setminus\{r_n\}$.

By the nested intervals theorem, $\bigcap_{n=1}^\infty I_n\ne\emptyset$. Choose a point $x\in\bigcap_{n=1}^\infty I_n$. Then $x\in\bigcap_{n=1}^\infty D_n$ but $x\notin\mathbb Q$, contradicting our assumption that $\mathbb Q=\bigcap_{n=1}^\infty D_n$.

Answer 2

One argument is by using the perfect set property.

Assume towards a contradiction that $\mathbb{Q} = \bigcap_n G_n$ where $G_n$ form a decreasing sequence of open sets.

Take arbitrary rationals $r_0 < r_1$. There are open intervals $I_0$ and $I_1$ with disjoint closures around those points, and included in $G_1$. Chose new rationals $r_{00}<r_{01}$ in $I_0$ and $r_{10}<r_{11}$ in $I_1$ and then find open intervals $I_{ij}$ ($i,j=0,1$) with $\overline{I_{ij}} \subseteq I_i \cap G_2$ and $\overline{I_{i0}}\cap \overline{I_{i1}}=\emptyset$. Proceed inductively to define $r_s$ and $I_s\subseteq G_n$ for every binary sequence $s$ of length $n$. You may arrange to make the length of the intervals go to zero with increasing $n$.

Now the map $2^{\mathbb{N}} \to \mathbb{Q}$ given by $x \mapsto \lim_{n\to \infty} r_{x_0x_1\dots x_n}$ is an injection; the map is well defined since the $r_s$ belong to a nested family of closed intervals, and the intersection of this family is a subset of all $G_n$. But this is a contradiction, since $2^{\mathbb{N}}$ is uncountable.

Note that the previous argument shows something stronger: Every dense $G_\delta$ set of reals is equinumerous with $\mathbb{R}$.