proving if $a \le b$ then $\sup(A) \le \inf(B)$

Question

Let $A$ and $B$ be bounded sets of real numbers such that $a\le b$ for all $a \in A$ and for all $b \in B$. Show that $\sup(A) \le \inf(B)$

Pf: Assume A and B are bounded sets. This means they have a least upper bound and a greatest lower bound.

let $x \le \sup(A)$ for some $x \in A$ and $\inf(B) \le y $ for some $y \in B$

Since we know that $a \le b $ for $\forall a\in A $ and for $\forall b \in B$, $x \le y$

Therefore, $x \le \sup(A) \le \inf(B) \le y$

Thus proving that $\sup(A) \le \inf(B)$

Answer 1

If $A$ is empty then $\sup A = -\infty$ (Exercise: why ?) and we are done. Similarly if $B$ is empty then $\inf B = \infty$ (Exercise: why ?) and we are done.

So suppose that $A$ and $B$ are non-empty. Let $a \in A$ and $b \in B$. By hypothesis, $a \leq b$. Since $a \in A$ is arbitrary, it follows that for all $a \in A$, $a \leq b$. Thus $b$ is an upper bound for $A$, and by definition of $\sup$, we conclude that $\sup A \leq b$.

However, $b \in B$ is also arbitrary, so it follows that for all $b \in B$, $\sup A \leq b$. Thus $\sup A$ is a lower bound for $B$, and by definition of $\inf$, we conclude that $\sup A \leq \inf B$.