Supremum of $(-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$

Question

Let $S:=(-\sqrt{2},\sqrt{2})\cap \mathbb{Q}$. It is clear that $\sup(S) = \sqrt{2}$, but I don't know how to prove this.

Suppose that $\sup(S)<\sqrt{2}$, then $\forall a\in S$ and $\forall\varepsilon>0$, $\sup(S)-\varepsilon<a\le\sup(S)<\sqrt{2}$. But what to do from this point on?

Answer 1

It is easy to see that $\sqrt{2}$ is an upper bound of $S$.

Now use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$, one can show that $\sqrt{2}$ must be the least upper bound of $S$ also:

To show that $\sqrt{2}$ is the least upper bound of $S$, suppose otherwise there exists $s<\sqrt{2}$ as an upper bound of $S$. By the density of $\mathbb{Q}$, there exists $r\in S$ such that $s<r<\sqrt{2}$, which is a contradiction.

Answer 2

Since it is obvious that $x<\sqrt 2$ for all $x$ in the interval, and it is easy to construct a sequence in the interval which converges to $\sqrt 2$ (namely, its truncated decimal expansión), we have the result.