Proof. Let $A$ be nonempty and bounded from above. Let $s_{1}$, $s_{2}$ be two supremums of A. Since $s_{1}$ is an upper bound and $s_{2}$ is less than equal to any upper bound. We have $s_{2}\leq s_{1}$. Similarly, $s_{1}\leq s_{2}$. Therefore, $s_{1}=s_{2}$.
My question is: in proof, ''Since $s_{1}$ is an upper bound and $s_{2}$ is less than equal to any upper bound. We have $s_{2}\leq s_{1}$'' So, why?