For $a,b \in \mathbb{R}$ fixed and $S$ is a bounded set above and below I want to prove the following:

If $a \geq 0$ then $\inf aS+b = a \inf S +b$ and $\sup aS+b = a \sup S +b$

I am not having issue proving the property about infimum but I am having trouble proving the supremum one... Should the $\sup aS+b = a\sup S -b$ and not $+b$ ?

$aS+b$ means $\{as+b: s\in S\}$

Drawing out the picture makes me believe $+b$ but proving it I keep on getting $-b$ using the epsilon definition of sup and inf.

Sample proof:

Claim: $\sup aS+b =a \sup S +b$.

Proof: Let $\sup S = v$ this means (by epsilon definition) $\forall \epsilon'>0 \exists s\in S$ such that $v-\epsilon'<s$.

We want to show: $\forall \epsilon>0 \exists y\in S$ such that $(av+b)-\epsilon<y$.

Let $\epsilon>0$ be arbitrary.

Take $\epsilon'= (\epsilon +b) /a$ then we know there exists $s\in S$ such that $v-(\epsilon +b) /a<s$ which means $av-\epsilon -b<as$ which means $(av-b) -\epsilon <as$.

So I got $(av-b) -\epsilon <as$ but I want $(av+b)-\epsilon <as$ to finish the proof... So I am doubting the CLAIM that $\sup aS+b=a \sup S -b$ BUT WHEN I DRAW THE PICTURE THE CLAIM SEEMS RIGHT...

  • $\begingroup$ Using the definition, one arrives at $\epsilon$. So to see what you do wrong, you might want to show what you did. It's probably one of those sign errors we all do once in a while (my prof used to put it this way: "If there were twenty different signs, we would nver mix them up, but with two signs, we are bound to do it sometimes") $\endgroup$ May 11, 2019 at 21:12
  • $\begingroup$ I'll update my post right now. $\endgroup$
    – javacoder
    May 11, 2019 at 21:13
  • $\begingroup$ I updated it, please check $\endgroup$
    – javacoder
    May 11, 2019 at 21:19

1 Answer 1


$$\sup(aS+b)=-\inf(-(aS+b))=-\inf(a(-S)-b)=-(\inf(a(-S))-b)=-\inf(-(aS))+b=\sup aS+b $$


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