Real Analysis – Proof of Supremum of Scaled Set

Question

Let $A\subseteq\mathbb{R}$ and $c\in\mathbb{R}^+$. Define the set $cA=\lbrace ca | a\in A\rbrace$.
Prove that $\sup(cA)=c\sup(A)$.

Here is my attempt:

Let $s_1=\sup(cA)$ and $s_2=\sup(A)$. Then, if $b_1$ is any upper bound to the set $cA$, $b_1\geq s_1\geq ca$ for any $a\in A$. If $b_2$ is any upper bound to the set $A$, then $b_2\geq s_2\geq a$ for any $a\in A$.

I'm not too sure what to do after this. What I would like to do is something like $\frac{s_1}{c}\geq a$ for any $a\in A$, so $\frac{s_1}{c}=s_2$ so $\sup(cA)=c\sup(A)$, but I don't think it's this simple.

Edit
Here is my second attempt:

Let $s=\sup(A)$. Then, if $b$ is any upper bound of the set $A$, $b\geq s\geq a$ for any $a\in A$. It follows that $cb\geq cs\geq ca$ for any $a\in A$. As $b\geq a$ for any $a\in A$, $cb$ is an upper bound for $cA$, and because $b$ is any upper bound of $A$, $cs$ is the least upper bound. Therefore, $cs=\sup(cA)$, which completes the proof.

Answer 1

Firstly, what $\sup$ is: (with a little category theory flavour)

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$\sup(A)$ if it exists, is:

1. an upper bound of $A$
2. so that for every $b$ an upper bound of $A$, we have $\sup(A) \le b$

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Now we will prove that $c\sup(A)$ is $\sup(cA)$.

1. Let $cx \in cA$ where $x \in A$. Then, $x \le \sup(A)$ by the first property of $\sup$. Therefore, $c \sup(A)$ is an upper bound of $cA$.
2. Let $b$ be another upper bound of $cA$. If $c \ne 0$, then $b/c$ is an upper bound of $A$, so we have $\sup(A) \le b/c$ by the second property of $\sup$, whence $c \sup(A) \le b$. If $c = 0$, then in fact $cA = \{0\}$, and $c \sup(A) = 0$, so if we have $0 \le b$ ($b$ is an upper bound) then it follows that $0 \le b$ ($b$ is greater than or equal to $c \sup(A)$).