Let $A$ be a subset of $B$, where $A$ is nonempty. How can I show that $\sup A \leq \sup B$?
My attempt
I said let element $a$ be in $A$, which means $a$ is in $B$ because $A$ is a subset of $B$. I also stated that $\sup A$ is an upper bound of $A$, so for all elements $a \in A$, $a \leq \sup A$. Also if $\sup B$ is an upper bound of $B$, $b \leq \sup B$ for all elements $b$. How can use these to arrive at the result I desire?
Since $A \subset B$, $\sup B$ is an upper bound for $A$. Since $\sup A$ is the least upper bound for $A$ by definition, it must be less than or equal $\sup B$.