How do I prove that a set $A$ can have at most one least upper bound?

Question

So of course if $A$ is null then there are no least upper bounds. And so then if we take $A=[x:x^2<10]$ then we have $A$ has 1 least upper bound: namely $\sqrt{10}$. So then I'm thinking assume we have a non empty set $A$ which has more than one least upper bound and finding a contradiction.

Answer 1

If you have two least upper bounds $u_1$ and $u_2$ then we have $u_1\le u_2$ and $u_2 \le u_1$

Thus they are the same.

Answer 2

By the definition of supremum: if we have 1 supremum $\alpha$ and the other $\beta$ then

$$\alpha\leq\beta$$ and $$\beta\leq\alpha$$ For these to both hold, $$\alpha=\beta$$ Thus we have at most 1 supremum for any set $A$