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.


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.


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.

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

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