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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 $A$,

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

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