Why is $\mathbb{Q}$ dense in $\mathbb{R}$?

Question

I've heard this fact numerous times in the past but I've never fully understood its proof. So I'm hoping to do it right this time:

Take the usual topology on $\mathbb{R}$ and take the subspace topology for $\mathbb{Q}$.

To prove that $\mathbb{Q}$ is dense in $\mathbb{R}$ it suffices to show that $\mathbb{R}\subseteq\overline{\mathbb{Q}}$.

The closure of $\mathbb{Q}$ can be written as $\overline{\mathbb{Q}}=\{x\in\mathbb{R}\mid \text{ if } x\in (a,b) \text{ then }(a,b) \cap \mathbb{Q} \neq \emptyset\}$

Let $x\in\mathbb{R}$, and let $(a,b)$ be an arbitrary open interval in $\mathbb{R}$ containing $x$. It suffices to show that there exist $p\in\mathbb{Q}$ such that $p\in (a,b)\cap\mathbb{Q}$, or equivalently, $a<p<b$.

How can I proceed from here?

Answer 1

There is some confusion here. You don't need to consider the subspace topology on $\mathbb Q$. You only need to see $\mathbb Q$ as a subset of $\mathbb R$. And $\mathbb R\subset\overline{\mathbb Q}$ looks weird; here, the whole universe is $\mathbb R$ and therefore what you want to prove is that $\mathbb R=\overline{\mathbb Q}$.

Now, if $a,b\in\mathbb R$ with $a<b$, just consider the sequence $\left(2^{-n}\left\lfloor2^{n-1}(a+b)\right\rfloor\right)_{n\in\mathbb N}$, which is a sequence of rational numbers smaller than or equal to $\frac{a+b}2$ which converges to $\frac{a+b}2$. Therefore, if $n$ is big enough, $2^{-n}\left\lfloor2^{n-1}(a+b)\right\rfloor\in(a,b)$.

Answer 2

It appears that the problem is to prove that for every $a,b\in\mathbb R$ with $a<b,$ there exists $p\in\mathbb Q$ such that $a<p<b.$

The set $\mathbb N = \{1,2,3,\ldots\}$ cannot have an upper bound $\mathbb R,$ since then it would have a smallest upper bound $m\in\mathbb R,$ so that $m-1$ is not an upper bound. That would imply $m-1< n$ for some $n\in\mathbb N,$ so that $m<n+1\in\mathbb N,$ so that $m$ is not an upper bound of $\mathbb N.$

Thus for some $n\in\mathbb N$ we have $n > \dfrac 1 {b-a}.$

Consequently $0<\dfrac 1 n< b-a.$

Thus some member of $\left\{ \dfrac 0 n, \pm \dfrac 1 n, \pm\dfrac 2 n, \pm\dfrac 3 n, \ldots \right\}$ is in the interval $(a,b).$

Answer 3

It's true by construction, really. What are the reals? We can construct them as the metric completion of the rationals (using equivalence classes of Cauchy sequences), or as the order completion of the rationals, using Dedekind cuts. These turn out to give homeomorphic/isomorphic results, but in both cases $\mathbb{Q}$ being dense in $\mathbb{R}$ is true by construction.