Proof that $\inf A = -\sup(-A)$

Question

Let $A$ be a nonempty subset of real numbers which is bounded below. Let $-A$ be the set of of all numbers $-x$, where $x$ is in $A$. Prove that $\inf A = -\sup(-A)$

So far this is what I have

Let $\alpha=\inf(A)$, which allows us to say that $\alpha \leq x$ for all $x \in A$. Therefore, we know that $-\alpha \geq -x$ for all $x \in -A$. Therefore we know that $-\alpha$ is an upper bound of $-A$. $\ \ \ \ $

Now let $b$ be the upper bound of $-A$. There exists $b \geq-x \implies-b \leq x$ for all $x \in A$. Hence, \begin{align} -b & \leq \alpha\\ -\alpha & \leq b\\ -\alpha & = - \inf(A) = \sup (-A) \end{align}

By multiplying $-1$ on both sides, we get that $\inf(A) = -\sup (-A)$

Is my proof correct?

Answer 1

Aside from dropping a $-$ sign in your last line, that's good work. (Also, I'd say "let $b$ be an upper bound," instead, but it's clear that's what is meant.)