Prove $\sup\{as+b\mid s\in S\} = a \sup(S) +b$

Question

Let $S$ be a set of real numbers and suppose that $a > 0$ and $b$ are two real numbers. Prove that $\sup\{as+b\mid s \in S\} = a \sup(S) +b $.

I have $$\sup(as+b) = \sup(as) + \sup(b) = \sup(a)\cdot\sup(s) + \sup(b).$$

How do I proceed?

Answer 1

Let $s :=\sup(S)$. Then for all $x \in S$ we have $x \le s$ and so (as $a>0$) $ax \le as$ and also $ax +b \le as+b$, so $as+b$ is an upperbound for $\{ax+b\mid x \in S\}$. If $m$ is any upper bound for this set, then for $x \in S$ we have $ax+b \le m$ and so $x \le \frac{m-b}{a}$ and the right hand side is an upperbound for $S$, and $s=\sup(S)$ is the smallest upperbound for $S$ so $s \le \frac{m-b}{a}$ from which it follows again that $as+b \le m$.

So $a\sup(S) + b$ is the smallest upperbound for $\{ax+b\mid x \in S\}$ and so $\sup \{ax+b\mid x \in S\} = a\sup(S)+b$.

More slickly: $t_b(x)=x+b$ and $m_a(x) = ax$ are order isomorphisms of $\Bbb R$ to itself and these preserve sups so $$\sup \{ax+b\mid x \in S\} = \sup t_b[m_a[S]] = t_b(m_a(\sup(S)) = a\sup(S)+b$$

Answer 2

The supremum of a set is the smallest upper bound of the set. Hence, proceed as follows:

1. Show that $a s + b \leq a \sup(S) + b$ for all $s \in S$. That shows that $a \sup(S) + b$ is a upper bound of the set $\lbrace a s + b : s \in S \rbrace$.

2. Show that for each other upperbound $c$ your have that $a \sup(S) + b \leq c$.