Let $X,Y⊆ℝ$ be two non-empty sets. Prove that if $\sup Y$ exists and $\forall x \in X \exists y \in Y$ s.t. $x \le y$, then $\sup X$ also exists and $\sup X \le \sup Y$.
For this question, I proved $\sup X$ exists but I don't know how to show second part.
For all $x \in X$ :
There is a $y_0 \in Y$ s.t.
$x\le y_0 \le \sup Y$, since $\sup Y$ is an (the least) upper bound for $Y$.
$y_0$ is an upper bound for $X$;
$\sup X$ exists, and
$\sup X \le y_0 \le \sup Y$, since
$\sup X$ is the least upper bound for $X$.
Assume for a contradiction that $\sup X > \sup Y$.
I claim that in this case there is some $q \in X$ that lies in the interval $\left(\sup Y, \sup X\right]$. If there was no such $q$, then $\sup Y$ would be an upper bound of $X$, so the least upper bound of $X$, $\sup X$, would be smaller than $\sup Y$, contradicting our assumption.
But for any $x \in X$ there is some $y \in Y$ with $x \leq y$. Take such a $y$ corresponding to $x=q$. Then $\sup Y < q \leq y$, so $\sup Y$ is smaller than some element of $Y$, a contradiction.
Since assuming $\sup X > \sup Y$ leads to a contradiction, we have $\sup X \leq \sup Y$.
