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

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 $\sup(A) \le \inf(B)$

• You are essentially saying: if $m\le n$, $p\le q$, and $m\le q$, then $n\le p$. This is not true. – Joey Zou Oct 10 '15 at 19:47
• Nope, you haven't proved that $sup(A) \le inf(B)$, the second to last line doesn't follow. First show that for any $a \in A$, $a \le inf(B)$. ... – BrianO Oct 10 '15 at 19:49

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 $\sup A \leq \inf B$.
• You don't really have to worry about $A$ or $B$ being empty, the hypotheses say they're "bounded", as in, have (finite) sup and inf respectively. – BrianO Oct 10 '15 at 19:52
• But "$A$ is bounded" could be interpreted as meaning that there exist $c_1, c_2 \in \mathbb{R}$ such that for all $a \in A$, $c_1 \leq a \leq c_2$. If $A = \emptyset$ then any $c_1, c_2 \in \mathbb{R}$ satisfy this condition. Therefore the least upper bound is $- \infty$ and the greatest lower bound is $+ \infty$. – Simon Oct 10 '15 at 19:57
• You may well be right, but I'm not at all sure its a universal convention that bounded means having a lub and glb. The supremum axiom for the real numbers is usually given as Every nonempty set of real numbers having an upper bound has a least upper bound" for precisely this reason. – Simon Oct 10 '15 at 20:13