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Let $A$ and $B$ be two nonempty bounded sets of nonnegative real numbers. Define the set $C:= \{ab: a\in A, b \in B\}$. Show that $C$ is a bounded set and that $\sup (C) = \sup (A)\sup (B)$ and that $\inf (C) = \inf (A)\inf (B)$.

I have asked the mathematical assistant center at my school but no one knows how to solve this problem. I've come here as a last result. Can someone please help me?

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4 Answers 4

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Let $\bar a=\sup A$ and $\bar b=\sup B$. To prove that $\bar c=\bar a \bar b$ is the supremum of $C$, you must prove two things: $\bar c$ is an upper bound for $C$ and $\bar c - \epsilon$ is not an upper bound for $C$ for any $\epsilon > 0$.

First off, it is clear that if any of $\bar a$ or $\bar b$ equals $0$, then one of the sets $A, B$ contains only $0$ and therefore $C=\{0\}$ and $\sup C = 0 = \bar a \bar b$.

Now we focus on the case where $\bar a,\bar b > 0$.

  • Take any $c\in C$. You know that $c=ab$ for some $a\in A, b\in B$. Because $\bar a$ is an upper bound for $A$ and $\bar b$ for $B$ (because they are supremums), you know that $\bar a\geq a\geq 0$ and $\bar b \geq b\geq 0$, you know that $\bar c = \bar a\bar b \geq ab$, so $\bar c$ is indeed an upper bound.
  • Take any $\epsilon>0$. You know that, for any $\min\{\bar a, \bar b\}>\delta>0$, because $\bar a-\delta$ is not a upper bound for $A$ (because it is a supremum),that there exists $a\in A$ such that $a>\bar a - \delta$. In the same way, you get $b\in B$ such that $b>\bar b - \delta$. You know that $ab\in C$ and also that $$ab>(\bar a - \delta)(\bar b - \delta) = \bar c - (\bar a + \bar b)\delta + \delta ^2.$$ Choosing $\delta$ small enough that $(\bar a + \bar b)\delta - \delta ^2<\epsilon,$ the equation means that $ab>\bar c - \epsilon$, proving that $\bar c = \sup C$.
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  • If $a\in A$ and $b\in B$ then $ab \le \sup A \sup B$ so $\sup C\le \sup A\sup B$.

There's no need of sequences for the other inequality. In fact the following approach uses only the definition of the least upper bound property.

  • If $a\in A^*$ and $b\in B^*$ (WLOG),then $ab \leq \sup C$

Hence $a\leq \frac{\sup C}{b}$

Hence $ \forall b \in B^*, \sup A \leq \frac{\sup C}{b}$

Hence $\forall b \in B^*,b \leq \frac{\sup C}{\sup A}$

Hence $\sup B \leq \frac{\sup C}{\sup A}$

Hence $\sup A \times \sup B \leq \sup C$

  • Hence $$\sup A \times \sup B = \sup C$$

  • The other assertion follows the same proof:

If $a\in A$ and $b\in B$ then $ab \geq \inf A \inf B$ so $\inf C\geq \inf A\inf B$.

  • If $a\in A^*$ and $b\in B^*$ (WLOG),then $ab \geq \inf C$

Hence $a\geq \frac{\inf C}{b}$

Hence $ \forall b \in B^*, \inf A \geq \frac{\inf C}{b}$

Hence $\forall b \in B^*,b \geq \frac{\inf C}{\inf A}$

Hence $\inf B \geq \frac{\inf C}{\inf A}$

Hence $\inf A \times \inf B \geq \inf C$

  • Hence $$\inf A \times \inf B = \inf C$$
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  • $\begingroup$ What is $A^*$? The closure of $A$? I have never seen this notation. $\endgroup$
    – Taladris
    Commented Feb 20, 2014 at 13:27
  • $\begingroup$ @Taladris The non zero elements of $A$. More generally the invertible elements of a ring. $\endgroup$ Commented Feb 20, 2014 at 13:28
  • $\begingroup$ What does WLOG mean? $\endgroup$
    – ayv2
    Commented Feb 20, 2014 at 13:55
  • $\begingroup$ @ayv2 without loss of generality $\endgroup$ Commented Feb 20, 2014 at 13:55
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If $a\in A$ and $b\in B$ then $ab \le \sup A \sup B$ so $\sup C\le \sup A\sup B$.

If $\langle a_n\rangle$ be a sequence on $A$ (that is, $a_n\in A$ for all $n$) that converges to $\sup A$, and $\langle b_n\rangle$ be a sequence on $B$ that converges to $\sup B$, then $a_n b_n\le \sup C$ for all $n$. Take $n\to\infty$ then we get $\sup A\sup B \le \sup C$.

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  • $\begingroup$ Do I follow the same steps for the inf(C) = inf(A)inf(B) $\endgroup$
    – ayv2
    Commented Feb 20, 2014 at 11:55
  • $\begingroup$ @tretori the statement $ab \leq \sup A \sup B$ is false. Consider the case of $A = B = \{-2,0\}$ $\endgroup$ Commented Feb 20, 2014 at 11:59
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    $\begingroup$ @Omnomnomnom However, all elements of $A$ and $B$ are positive. $\endgroup$
    – Hanul Jeon
    Commented Feb 20, 2014 at 12:00
  • $\begingroup$ @ayv2 Proof of $\inf C=\inf A\inf B$ also have same steps. At first, you can prove that $\inf A\inf B \le \inf C$. Second, take a sequence $\langle a_n\rangle$ and $\langle b_n\rangle$ satisfy that $a_n\to \inf A$ and $b_n\to \inf B$ as $n\to\infty$. By definition of $\inf C$ you get $a_nb_n\ge \inf C$. Finally, take a limit for $n$ to intinity. $\endgroup$
    – Hanul Jeon
    Commented Feb 20, 2014 at 12:02
  • $\begingroup$ @tetori ah, my mistake $\endgroup$ Commented Feb 20, 2014 at 12:05
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If either $A=\{0\}$ or $B=\{0\}$, then the statement is trivial, so assume that $A\neq\{0\}$ and $B\neq\{0\}$.

On one hand, we must have $\sup{C} \le \sup{A}\sup{B}$, since for any $x\in{C}$ we have $x=ab$, where $a\in{A}$ and $b\in{B}$, and thus $x=ab\le\sup{A}\sup{B}$.

On the other hand, if we assume that $\sup{C} \lt \sup{A}\sup{B}$ then we can write: $$\frac{\sup{C}}{\sup{A}}\lt\sup{B}$$ therefore there exists some non-zero element $b\in{B}$, such that $\frac{\sup{C}}{\sup{A}}\lt b$. We can rewrite this as $\frac{\sup{C}}{b}\lt \sup{A}$, so there is an element $a\in{A}$, such that $\frac{\sup{C}}{b}\lt a$. Thus, we get: $$\sup{C}\lt ab\in{C}$$ which is a contradiction, and the required equality ensues.

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