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Suppose that $S\neq\emptyset$ is a bounded set of numbers and that $a$ is a number. Define $aS=\{ax\mid x\in S\}$.

Prove that sup $aS$ = $a$ sup $S$ if $a \geq 0$

I can intuitively see why this is true by just trying out some cases for $a$ and $S$ but I can't seem to prove why this is true.

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Hint: Prove that

  1. $\sup (aS)\leq a\sup (S)$
  2. $\sup(aS)\ge a\sup(S)$

Hint to prove 1: Show that for every $y\in aS$ we have $y\leq a\sup S$. If you prove this, then you will have proven that $a\sup (S)$ is an upper bound for $aS$. Can you conclude?

Hint to prove 2: Similar to 1, but first note that $\sup(aS)\ge ay$, for all $y\in S$ and therefore, if $a>0,$ $\displaystyle \frac{\sup(aS)}{a}\ge y$, for all $y\in S$. The case $a=0$ is trivial. Conclude.

Edit: Before going about proving the above mentioned inequalities it is necessary to prove that $\sup (aS)$ does exist, i.e., it's necessary to show that there exists the least upper bound os $aS$, but that's a consequece of the Least-upper-bound property. That was probably given to you as an axiom, so there's nothing to prove. It simply is true.

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So we try by applying the definition and see how it goes.

So we deal with 2 cases, either $\sup S=\infty$ or $\sup S=M<\infty$.

If it is the second case, this means that for all $x\in S$, $x\leq M$, so naturally $ax\leq aM$. By definition of the least upper bound, we have $\sup aS\leq aM=a\sup S$.

For the reverse, let $\epsilon>0$ be given. If $a>0$, then $\frac{\epsilon}{a}>0$. By the definition of sup, there exists $x\in S$ such that $M-\frac{\epsilon}{a}\leq x\leq M$. Hence $aM-\epsilon\leq x\leq aM$. Since this $\epsilon$ is arbitrary, so $\sup S=M$.

If $a=0$ then we have nothing to say.

And as for the infinity case, it should be quite evident, by choosing a sequence in $S$ such that it goes to infinity.

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  • $\begingroup$ $S$ is bounded. $\endgroup$ – Git Gud Jan 22 '13 at 20:18

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