1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ You are essentially saying: if $m\le n$, $p\le q$, and $m\le q$, then $n\le p$. This is not true. $\endgroup$ – Joey Zou Oct 10 '15 at 19:47
  • $\begingroup$ Nope, you haven't proved that $sup(A) \le inf(B)$, the second to last line doesn't follow. First show that for any $a \in A$, $a \le inf(B)$. ... $\endgroup$ – BrianO Oct 10 '15 at 19:49
5
$\begingroup$

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 $\sup A \leq \inf B$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – BrianO Oct 10 '15 at 19:52
  • 1
    $\begingroup$ 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 $+ \infty$. $\endgroup$ – Simon Oct 10 '15 at 19:57
  • $\begingroup$ It could, in general, but not in the presence of "Assume A and B are bounded sets. This means they have a least upper bound and a greatest lower bound." I assume this means finite lub and glb. (Anyway, not a big deal.) $\endgroup$ – BrianO Oct 10 '15 at 20:10
  • $\begingroup$ 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. $\endgroup$ – Simon Oct 10 '15 at 20:13
  • 1
    $\begingroup$ As a universal convention, no -- you're right about that. I just meant that in this case, that's apparently so. In any event, emptiness has to be addressed or dismisssed, & you did.. $\endgroup$ – BrianO Oct 10 '15 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.