Question : Let $A$ and $B$ $\subset \mathbb{R}$ nonempty and bounded above. Prove that $A \subset B$ implies $\sup A \le \sup B$.
This is a pretty basic analysis question. I am presenting my solution here so that anybody who has more experience in mathematics can help me to check if there is any flaw in the argument. So here it goes:
From the hypothesis, $A$ and $B$ nonempty and bounded above, $\alpha := \sup A$ and $\gamma := \sup B$ exist by the least upper bound property of $\mathbb{R}$. Goal is to prove $\alpha \le \gamma$.
$\alpha \ge a \quad \forall a \in A$. Since if $a \in A \Rightarrow a \in B$, we have $a \le \gamma \quad \forall a \in A$. $\gamma$ is an upper bound for $A$ by definition of upper bound. Again by definition of supremum, $\alpha \le \gamma$ since $\alpha$ is the least upper bound of A. $\Box$
Just a side note: The definition of supremum I am using has the following component: If $s$ is $\sup A$ for some subset $A$ in $\mathbb{R}$, $s \le u \quad \forall u$ the upper bounds of $A$.