A question about sup and inf

Question

My professor last week told us about max, min, sup and inf of a set, and while talking about sup and inf, he stated that they always do exist. For example if $A = (2, 4)$ then $\sup (A) = 4$ and $\inf(A) = 2$, even if they do not belong to the set.

In the same way, we can have $\sup(G) = \pm \infty$, where $G$ is some set.

However, I thought about this set:

$$B = \{ x\in\mathbb{Q}:\ 2< x^2 < 3\}$$

And I got doubts and got stuck: this set is the set of all the rationals between $\sqrt{2}$ and $\sqrt{3}$, hence it's like to write $(\sqrt{2}, \sqrt{3})$ while working in $\mathbb{Q}$.

What are $\sup(B)$ and $\inf(B)$ in this case? For I don't think they do exist, because the square roots here are not rational numbers. Am I wrong?

Thank you!

Answer 1

This is one of the defining distinctions between the rational numbers and the real numbers. In a sense, the real numbers fill in the "holes between all the rational numbers", in such a way that suprema and infima always exist. Both classic constructions of $\Bbb R$ from $\Bbb Q$, either by Cauchy sequences or by Dedekind cuts, seem to be constructed to directly address this deficiency in the rational number line.

In $\Bbb Q$, there is no supremum for $B$. Just as you say. However, $B$ can also be seen as a subset of the real numbers (even though it happens to only contain rational numbers), and in that sense it does have a supremum: $\sqrt3$.

Answer 2

You make a good observation: $\Bbb Q$ is not complete. But when your professor states all suprema exist, they are correct... because they meant to say that all suprema exist in $\overline{\Bbb R}$! The supremum of $B$ is $\sqrt{3}\in\Bbb R$ and this exists just fine. In fact, a common way to rigorously prove $\sqrt{3}$ exists is by defining it as a suitable supremum.

It is correct and well-known and completely fine that not all suprema exist in $\Bbb Q$.