Is the union of open neighbourhoods of all rational numbers always $\mathbb{R}$?

Question

I have the following problem:

For every rational number $q\in \mathbb{Q}$ is $U_q\subseteq \mathbb{R}$ an open neighbourhood of $q$. Is the following true? $$\mathbb{R}=\bigcup_{q\in\mathbb{Q}}U_q$$

Intuitively I'd say that it is not true.
Every open neighbourhood of a rational number $q$ is an interval $]a,b[\subset \mathbb{R}$ with $a< q < b$. We define $\epsilon_q:=\max\{q-a,b-q\}$.

What if there is an $r\in \mathbb{R}\backslash\mathbb{Q}$ with $|r-q|=e>\epsilon_q$ $\forall q\in \mathbb{Q} $? This would be possible because a single point $r$ in $\mathbb{R}$ is closed, which means that $\mathbb{R}\setminus\{ r\}$ is open. So you can find such a neighbourhood for every $q\in \mathbb{Q}\setminus\{r\}$. Is this correct? Is this already a counterexample to the statement above?

Thanks in advance.

Answer 1

For each rational $q$, define$$U_q=\begin{cases}\left(\sqrt2,+\infty\right)&\text{ if }q>\sqrt2\\\left(-\infty,\sqrt2\right)&\text{ otherwise.}\end{cases}$$Then $\bigcup_{q\in\mathbb{Q}}U_q=\mathbb{R}\setminus\left\{\sqrt2\right\}.$

Answer 2

No. A couple of ideas: if you enumerate the rationals $q_1$, $q_2,\ldots$ and put an interval of length $\epsilon/2^n$ about $q_n$ you get a set of Lebesgue measure $\le\epsilon$ overall so it can't equal $\Bbb R$.

Or, pick your favourite irrational, say $\sqrt2$. For a given rational $q$, make sure your open interval about $q$ is small enough to avoid $\sqrt2$.