# Assume $X$ and $Y$ be nonempty subsets of $R$ such that $x<y$ for every $x \in X$ and $y \in Y$. Prove that $\sup X \leq \inf Y$. Is my proof correct?

We have that both sets are nonempty, by the completeness axiom, both sup and inf exist for both sets. Since $x < y$, $y$ is an upper bound for $X$ and hence by def. of sup, sup$X \le y$. So $\sup X$ is a lower bound for $Y$ and thus by def. of inf, $\sup X \leq \inf Y$.

• When you say $x < y$, you're considering a single $y$ and all $x \in X$ to conclude that sup $X$ $\le y$.
• Because sup $X$ $\le y$ is true for all $y \in Y$, we can also conclude that sup $X$ is a lower bound for $Y$. It is not a lower bound for $y$: we say that sets have bounds, but numbers do not.
Actually it's enough that, given $x \in X$ there exists an $y_x \in Y$ such that $x \leq y_x$. In fact in this case we have $supY\geq y_x \geq x \ \forall x \in X$, which implies $supY \geq sup X$.