Let S be a non-empty bounded set, and let $a,b \in \mathbb{R}$. Define $T = \{a \cdot s + b | s \in S\}$. Find a formula for Sup($T$).
Claim: $Sup(T) = a \cdot sup(S) + b$
Proof Attempt:
In order to establish the claim I have to show two things:
i) that $a\cdot sup(S) + b$ is an upper bound of the set $T$
ii) $a \cdot sup(S) + b$ is the least upper bound by establishing that: $\forall \ \epsilon > 0 \ \exists \ t = a \cdot s + b$ s.t. $a\cdot sup(S) + b – \epsilon \ \leq t$
part i) By definition $sup(S) \geq s$ for all $s \in S$. Therefore:
$$ s \leq sup(S) \\a \cdot s \leq a \cdot sup(S) \\ a \cdot s + b \leq a \cdot sup(S) + b$$
Which implies that $T$ is bounded above.
part ii) This is where I'm having a slight issue, I'm torn between letting $t = a \cdot s + b – \frac{\epsilon}{2}$ and $t = a \cdot s + b – \frac{\epsilon}{2a}$.
I bring this up because if I follow the same sort of reasoning I did in part i) that would mean:
$$ sup(S) – \epsilon \ \leq s \\ \Rightarrow \ a\cdot sup(S) + b – a\epsilon \ \leq a \cdot s + b $$
Which means that in order for me to get $\epsilon$ I would have to let $\epsilon = \frac{\epsilon}{a}$ And I'm not sure I am able to let $\epsilon$ be a certain set of values.
Best Answer
First, the claim is true only if $a > 0$ (otherwise you can't go from $s \leq \sup (S)$ to $as \leq \sup (aS)$).
For part 2, you indeed need to "let $\varepsilon = \frac{\varepsilon}{a}$". Formally it will be (omitting $b$ for simplicity for now, assuming $x = \sup(S)$).
We need: $\forall \varepsilon > 0 \exists t\in aS\colon t > ax - \varepsilon$.
We know that $\forall \varepsilon' > 0 \exists s \in S\colon s > x - \varepsilon'$. Let $\varepsilon' = \frac{\varepsilon}{a}$. Then there is $s$ for such $\varepsilon'$ and we can take $t = as$. We will have $t = as > a(x - \varepsilon') = ax - \varepsilon$ - exactly what we needed.
It is often acceptable to use some function $f(\varepsilon)$ as upper bound instead of $\varepsilon$ as long as $\lim\limits_{x \to 0} f(x) = 0$. For example, we proved that $\forall \varepsilon > 0\exists x \in X\colon x < f(\varepsilon)$. We usually could get $\varepsilon$ in r.h.s. by using some tighter bounds in intermediate reasoning, but as his always implies that $\forall \varepsilon > 0 \exists x \in X \colon x < \varepsilon$ it would just make the reasoning a bit harder and looking strange, so little point in doing so.