Rudin's exercise 1.5 says:
Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $$\inf A = - \sup(-A).$$
My only question on the exercise is whether an argument that I am making which allows me to generalize for an arbitrary element of $A$ to an arbitrary element of $-A$ is sufficiently rigorous. I am trying to establish the existence of $\sup(-A)$.
Since $A$ is nonempty and bounded below, I can pick an $x \in A$, which has the property that $x \geq \alpha$ for some lower bound $\alpha$. This implies that $-x \leq - \alpha$. But $x$ was an arbitrary element of $A$, and since there exists a one-to-correspondence between $A$ and $-A$, given by $$f: A \to -A, \; \; x \mapsto -x,$$ we may generalize from a statement for an arbitrary element of $A$ to an arbitrary element of $-A$. Hence, we can state that $\forall -x \in -A, \; -x \leq -\alpha$. Hence, $-A$ is bounded above, hence $\sup(-A)$ exists.
Is this argument sound? Is it fully rigorous, or am I missing anything?
The idea is fine, but not properly expressed. You wrote that you “can pick an $x\in A$, which has the property that $x\geqslant\alpha$ for some lower bound $\alpha$”. It would be better to talk about $\alpha$ first. That is, you should state first “if $\alpha$ is some lower bound of $A$” or something like that. But the real problem is that you wrote about “an $x\in A$”. Take a look at the rest of the proof. What you actually need here is that the assertion $x\geqslant\alpha$ holds for every $x\in A$, not just for a certain $x$ that you have picked.