Let $B=\{x\in A: x^2<2\}$. $B$ has upper bound 1.5, but its least upper bound is not rational, so not in $A$.
The proof that $\mathbb{R}$ does indeed have the Least Upper Bound property really depends upon how you're defining the real numbers; for example, if $\mathbb{R}$ is constructed using Dedekind cuts, then the proof is rather straight-forward and easy. If you're constructing $\mathbb{R}$ using equivalence classes of Cauchy sequences, then it's involved (at least the proofs I've seen/done).
Thus, I'll use this answer to address the concerns of why the Least Upper Bound property needs to be mentioned at all, and why it doesn't trivially hold true for everything.
Firstly, the Least Upper Bound property is essentially the reason calculus can be done; as we shall see, there are "gaps" in the rational numbers. The ability to take limits, which is central to everything done in Real Analysis, is closely related to the Least Upper Bound property. To show why this is the case, I'll quote some equivalences:
- Least Upper Bound Property
- Bolzano-Weierstrass Theorem (all cauchy sequences are convergent) and the Archimedean property
- Monotone Convergence Theorem and the Archimedean property
- Nested Intervals Theorem
All of the above are equivalent, and all are central to Real Analysis.
As far as the Least Upper Bound property not holding more generally in, for example, $\mathbb{Q}$, consider the following set: $$\{ q\in \mathbb{Q} \mid q>0 \wedge q^2<2 \}$$ It is clearly bounded above; 2 is an upper bound, for example. Does this set have a least upper bound? In $\mathbb{R}$ it certainly would, and would be $\sqrt{2}$; since $\mathbb{Q}$ is dense in $\mathbb{R}$, if there were a least upper bound in $\mathbb{Q}$ for this set, then it would also be a least upper bound of the set in $\mathbb{R}$. But we know that $\sqrt{2}$ is irrational, so this cannot be the case. Thus, $\mathbb{Q}$ cannot have the Least Upper Bound property. This is why the Real numbers are necessary.
Best Answer
$\sup S$ is shorthand for the least upper bound of $S$ in some set $T\supset S$ with respect to an order $\leq$ on $T$. However, we often omit $T$ and $\leq$ when they are obvious from the context.
Consider the case when $T=\mathbb{Q}$ and $\leq$ is the usual order. Since $\sup\left\{ x\in\mathbb{Q}:x^{2}<2\right\} $ has an upper bound but no least upper bound (in $T=\mathbb{Q}$), this supremum does not exist. Hence, $\mathbb{Q}$ does not satisfy the least upper bound property (a.k.a. Dedekind completeness).
However, if instead we had $T=\mathbb{R}$, the above supremum is $\sqrt{2}$.
Another example is given by @Hagen von Eitzen, where we look for $\sup \emptyset$. Regardless of whether $T=\mathbb{Q}$ or $T=\mathbb{R}$, in both of these spaces, $0$ is an upper bound of $\emptyset$. Furthermore, if $x$ is an upper bound of the empty set, so too is $x-1$. Therefore, there exists no least upper bound. We can write this as $\sup\emptyset = -\infty$ (in fact, if $T=\overline{\mathbb{R}}$, the extended reals with the obvious order, this is a precise statement).
Note that this does not contradict Dedekind completeness of the reals because Dedekind completeness only requires bounded nonempty sets to have supremums.