# For $a,b \in \mathbb{R}$ fixed supremum and infimum property

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'.

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

Let $$\epsilon>0$$ be arbitrary.

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

So I got $$(av-b) -\epsilon but I want $$(av+b)-\epsilon 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...

• 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") May 11, 2019 at 21:12
• I'll update my post right now. May 11, 2019 at 21:13
• I updated it, please check May 11, 2019 at 21:19

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