Proof ambiguity regarding $\mathbb{Q}$ as an intersection of open sets

Question

This proof is to show that $\mathbb{Q}$ is not the countable intersection of open sets. I am having a hard time seeing why $\bigcap\limits_{k\in\omega}W_k = \varnothing$? (see the bolded part in the proof below) Because Every $_$ contains $_$ except for the corresponding ${_}$, and each open $_$ covers all of $\mathbb{Q}$, so there are a lot of things in the intersection $\bigcap\limits_{k\in\omega}W_k$

Here is the proof:

Suppose that $\mathbb{Q} = \bigcap\limits_{k\in\omega}V_k$, where each $V_k$ is open in the usual topology on $\mathbb{R}$. Clearly each $V_k$ is dense in $\mathbb{R}$. Let $\mathbb{Q}=\{q_k:k\in\omega\}$ be an enumeration of the rationals, and for each $k\in\omega$ let $W_k=V_k\setminus \{q_k\}$; clearly each $W_k$ is dense and open in $\mathbb{R}$, and $\bigcap\limits_{k\in\omega}W_k = \varnothing$

Let $(a_0,b_0)$ be any non-empty open interval such that $[a_0,b_0]\subseteq W_0$. Given a non-empty open interval $(a_k,b_k)$, let $r_k=\frac14(b_k-a_k)$; clearly $a_k<a_k+r_k<b_k-r_k<b_k$. Since $W_{k+1}$ is dense and open, there is a non-empty open interval $(a_{k+1},b_{k+1})$ such that $$(a_{k+1},b_{k+1}) \subseteq [a_{k+1},b_{k+1}] \subseteq W_{k+1}\cap (a_k+r_k,b_k-r_k),$$ and the construction can continue.

For $k\in\omega$ let $J_k = [a_k,b_k] \subseteq W_k$. For each $k \in \omega$ we have $J_k \supseteq J_{k+1}$, so $\{J_k:k\in\omega\}$ is a decreasing nest of non-empty closed intervals. Let $J = \bigcap\limits_{k\in\omega}J_k$; $J\subseteq J_k \subseteq W_k$ for each $k\in\omega$, so $J \subseteq \bigcap\limits_{k\in\omega}W_k = \varnothing$. But the nested intervals theorem guarantees that $J \ne \varnothing$, so we have a contradiction. Thus, $\mathbb{Q}$ cannot be a $G_\delta$-set in $\mathbb{R}$.

Answer 1

The key facts (you can fill in the details):

• Firstly, we see that the $W_k$ are each dense in $\mathbb{R}$ by the simple fact that they contain every rational as $V_k$ does, except for one, $q_k$. $W_k$ will have all of the other rationals, and hence you can find a sequence of rationals $r_n \to q_k$ as $n\to \infty$. Thus, $q_k \in \overline{W_k}$; this should be the only troublesome element to argue that is in the closure, seeing as each $V_k$ is dense, and thus the closure of each is $\mathbb{R}$, giving density.

• Since the $V_k$ are open, they may be written as the countable union of open balls for each $k$. An open ball (open interval) in $\mathbb{R}$ minus any point inside of it just gives you two more intervals: $(a,b) \setminus\ \{c\} = (a,c) \cup (c,b)$ provided $a<c<b$. We may extrapolate this to the $V_k$, and use it on the open ball(s) in $V_k$ (when we write it as such a union) that contain $q_k$. We thus easily have $W_k$ as a union of these open balls (some split, but nonetheless countably many since the "splitting" adds at most as many balls as there were originally).

• For the intersection, suppose otherwise, with $x \in W_k$ for all $k$. But then (since each $W_k$ excludes a rational), $x$ would have to be irrational. Thus, $x \in V_k$ for all $k$. But then $x \in \bigcap V_k$ contradicting the initial assumption.

Answer 2

If the given intersection was non-empty, then there would exist an $x \in W_k$ for every $k \in \omega$. In particular, $x$ could be none of the rationals. So, it must be irrational. But if $x$ is irrational and belongs to each $V_k$, then $\bigcap_{k \in \omega} V_k \neq \mathbb{Q}$ and that is a contradiction.

Answer 3

Clearly $\cap W_k \subseteq \cap V_k = \mathbb{Q}$. But take any $q \in \mathbb{Q}$; then $q = q_k$ for some $k \in \omega$. Notice that $q_k \not \in W_k$ by construction, which means that $q_k \not \in \cap W_k$. Since $q_k$ from $\mathbb{Q}$ was chosen arbitrarily, this means that $q \not \in \cap W_k$ for every $q \in \mathbb{Q}$. So $\cap W_k \cap \mathbb{Q} = \emptyset$ but $\cap W_k \subseteq \mathbb{Q}$ which means that $\cap W_k = \emptyset$