Does the set of rational numbers between 0 and 2 have the least upper bound property?

Question

Let $A = \{ a \in Q : 0 < a < 2\}$

Does A have the least upper bound property?

Definition: $A$ has the least upper bound property if $\forall$ nonempty $B \subseteq A$, if $B$ has an upper bound in $A$, then it has a least upper bound in $A$.

My claim: $A$ does not have the least upper bound property.

My attempt at a proof:

Take $B \subseteq A$. Now since there are infinitely rationals between any two rationals in $(0,2)$, $B$ must have an upper bound in $A$. That is, if $b \in B$ is the largest element in $B$, there must be an $a \in A$ such that $b < a < 2$.

This is the part I'm not sure about...

I need to show that $B$ does not have a least upper bound in $A$. So let $a\in A$ be an upper bound for $B$. Since the greatest element in $B$ is a rational and $a$ is a rational, I can find an $a' \in A$ such that $b<a'<a$. I can also find a rational between $b$ and $a'$, etc. So there does not exist a least upper bound for $B$ in $A$.

Does that look correct?

Answer 1

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 $A$.

Answer 2

Let $B = \{ q \in A | q < \sqrt{2} \}$. Then $B$ has an upper bound in $A$ (say $\frac{3}{2}$), but no least upper bound in $A$.