Proving that if $A \subset B$ then $\sup A \leq \sup B$

Question

Note: $\sup a$ being supremum.

I am trying to prove that if $A \subset B$ then $\sup A \leq \sup B$, with $A \subset \mathbb{R}$ and $B \subset \mathbb{R}$. Here's is my attempt.

Let's reason by contradiction. Suppose that $\sup A > \sup B $. Let then $\epsilon \in \mathbb{R}$ such that $\epsilon = \sup A - \sup B$, we have $\epsilon > 0$.

Let then $\delta \in \mathbb{R}$ such that $0 < \delta < \epsilon$.

Let $y= \sup A$. Then we have $y-\delta \in A$ and $y-\delta > y-\epsilon$, thus $y-\delta \not \in B$. But this is absurde, as $A \subset B$

Thus we have $\sup(A) \leq \sup(B) $.

Is my proof correct?

Answer 1

The statement $y-\delta \in A$ is not correct. What is true is that there exists an element $a \in A$ such that $a > y-\delta$. [Otherwise $y-\delta$ would be a smaller upper bound on $A$.] Then the rest of the proof is good: then we have $a>y-\delta > y-\epsilon = \sup B$, which implies $a \notin B$, contradicting $A \subset B$.

Answer 2

A direct proof is the simplest here:

$\sup B$ is an upper bound for $B$, hence also for $A$. Because $\sup A$ is the least upper bound, $\sup A\leq \sup B$.

It also has the advantage that it works for all partially ordered sets.