# If $A, B$ are two nonempty sets s.t. $\forall_{x \in A} \exists_{y \in B} \; x \leq y$, then $\sup A \leq \sup B$ and $\inf A \leq \inf B$

This is a remark from my course notes:

If $$A, B \subseteq \mathbb{R}$$ are two nonempty subsets such that $$\forall_{x \in A} \exists_{y \in B} \; x \leq y$$, then $$\sup A \leq \sup B$$ and $$\inf A \leq \inf B$$. My understanding for the first part:

Assume $$\forall_{x \in A} \exists_{y \in B} \; x \leq y$$. Suppose $$\sup A > \sup B$$. Then for all $$\epsilon > 0$$ there exists some $$x \in A$$ such that $$\sup A - \epsilon < x < \sup A$$. If you set $$\epsilon = \sup A - \sup B$$ then this implies that there exists some $$x \in A$$ such that $$x > \sup B$$. Then $$x$$ is greater than all the elements in $$B$$, which is a contradiction, so $$\sup A \leq \sup B$$.

For the second part, I'm not sure how to do it. If I assume that $$\inf A > \inf B$$, I am not sure how I can get a contradiction as there still can be an element in $$B$$ greater than all elements in $$A$$,

The second part is wrong. Have a look at $$A=\{0\}$$ and $$B=\{-1,1\}$$.