2

I am trying to show that $\mathbb{Q}$ is not a not a $G_{\delta}$. However, I am confused with the following argument:

Clear for all $r \in \mathbb{Q}$, $(r-\frac{1}{n} ,r+\frac{1}{n})$ is an open set

For each n, I am going to define the open set
$ V_n= \cup_{r \in \mathbb{Q}} (r-\frac{1}{n} ,r+\frac{1}{n}) $

Clearly, $V_n$ is open, since it is the union of open sets.

Now isn't $\mathbb{Q}= \cap_{n} V_n$

I am confused since I want to show that Q is not a $G_{\delta}$ set

Demha
  • 21

2 Answers2

8

The point that you are missing is that for each and every $n\in\mathbb{N}$ the set $V_n=\mathbb{R}$ (this is because $\mathbb{Q}$ is dense).

Hence $\bigcap V_n = \mathbb{R}$ which is not what you wanted.

Here you can find a proof for the statement.

Yanko
  • 13,758
2

For each fixed $n$ we have that $V_n = \mathbb{R}$, because if $x \in \mathbb{R}$, $(x-\frac{1}{n}, x+\frac{1}{n})$ is open and non-empty so contains a rational number $r'$,so that $|x-r'| < \frac{1}{n}$, and then we have that $x \in (r'-\frac{1}{n}, r'+\frac{1}{n}) \subseteq V_n$. So $\bigcap_n V_n = \mathbb{R}$ also.

Henno Brandsma
  • 242,131