Can we find $V_n$ in such a way that $\bigcap\limits_{n=1}^{\infty} V_n$ is countable?

Question

If $V_n$ is open, dense subset of $\mathbb R$ for each $n\in \mathbb N$ then by Baire's Category theorem we know that $\displaystyle \cap_{n=1}^{\infty} V_n$ is dense in $\mathbb R$.

My question is that : Can we find $V_n$ in such a way that $\displaystyle \bigcap_{n=1}^{\infty} V_n$ is countable set?

Any help is appreciated. Thank you.

Answer 1

No, we cannot. Suppose otherwise. Then $\bigcap_{n\in\mathbb N}V_n=\{r_1,r_2,r_3,\ldots\}$. Now, let $W_n=V_n\setminus\{r_n\}$. Then $W_n$ is also an open dense subset of $\mathbb R$. But now $\bigcap_{n\in\mathbb N}W_n=\emptyset$, which is impossible, by Baire's theorem.

Answer 2

No. If $S = \int_{n=1}^\infty V_n$ were countable, you could make it empty by intersecting with another countable sequence of dense open sets (the complements of each member of $S$), and thereby violate Baire.