This question refers to How to prove closure of $\mathbb{Q}$ is $\mathbb{R}$
I want to prove that the closure of $\mathbb{Q}$ is $\mathbb{R}$.
I am trying to understand the accepted answer, but when it comes to "This shows that the complement of $\mathbb{Q}$ has empty interior, so the closure of $\mathbb{Q}$ is all of $\mathbb{R}$.", I get stuck.
What does it mean intuitively that a set has empty interior?
What I know is that a point is interior to a set when it is the center of some open ball inside that set. In that sense, for a set to have empty interior would mean that it has no interior points. Furthermore, the set of all its adherent points – a point $m$ is adherent to the set of rationals when there is a sequence of rational points $x_k$ such that $limx_k = m.$
Being so, the complement of rationals would have no interior points.
How does it follow from this that the closure of $\mathbb{Q}$ is $\mathbb{R}$?
Thanks in advance!
Best Answer
Take $x \in \mathbb{R}$. Since $\mathbb{R}\ - \mathbb{Q}$ has empty interior, it follows that $x \notin int(\mathbb{R} - \mathbb{Q})$. It means that, for all $r > 0$, the ball $B(x,r)$ centred in $x$ with radius $r$ is not contained in $\mathbb{R}\ - \mathbb{Q}$. So, there is a $q \in \mathbb{Q}$ such that $q \in B(x,r)$. In other words, $x$ is in the closure of $\mathbb{Q}$. Since $x$ is arbritary, $\overline{\mathbb{Q}} = \mathbb{R}$.