If $c \in \mathbb{R}, c<0, A \subseteq \mathbb{R}$ prove that $\sup(cA) = c \cdot \inf(A)$

Question

This is what I have done, but I am not sure if it is correct.

$s = \sup(cA) \Rightarrow s \ge -ca, \, \forall a \in A$ and $\nexists t \in \mathbb{R} : t \ge -ca, t < s, \forall a \in A$

Restated with the sign applied to the inequality:

$\dfrac{s}{c}\le a, \forall a \in A$ and $\nexists t \in \mathbb{R} : \dfrac{t}{c}\le a, \forall a \in A, t < s \Rightarrow \dfrac{s}{c} = \inf(A)$

$\dfrac{s}{c} = \inf(A) \Rightarrow s = \sup(cA) = c \cdot \inf(A)$

Is this correct?

Answer 1

Your attempt at a proof is, at the least, insufficient. I suggest you to use less symbols and to spell out what you need to prove.

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Let $r=\inf(A)$; you want to prove

1. $cr\ge x$, for all $x\in cA$
2. for all $\varepsilon>0$, there exists $y\in cA$ such that $y>cr-\varepsilon$.

First condition. Since $r=\inf(A)$, we know that $r\le a$, for all $a\in A$; if $x\in cA$, then $x=ca$, for some $a\in A$; from $r\le a$ and $c<0$, it follows $cr\ge ca=x$.

Second condition. There exists $b\in A$ such that $r-\dfrac{\varepsilon}{c}>b$, because $-\varepsilon/c>0$. Then $$ c\left(r-\frac{\varepsilon}{c}\right)<cb $$ which is the same as $$ cr-\varepsilon < cb $$ and we can take $y=cb\in cA$.