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From baby Rudin (exercise 5 in chapter 1):

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 $$\inf A=-\sup (-A).$$

I tried to prove this in the following way, but I'm not sure if it is an actual valid proof:

Proof

Let $\alpha=\sup(-A)$ (it exists since $A$ being bounded below implies $-A$ being bounded above. This means that for all $x\in A$, $\alpha\geqslant-x$ holds (since $-x\in(-A)$). Multiplying both sides by $-1$, we get $-\alpha\leqslant x$. This proves that $-\sup (-A)=\alpha$ is a lower bound of $A$.

Now let's say we have $\beta>\alpha$. Yet again multiplying both sides by $-1$, we aquire $-\beta<-\alpha$. Since $-\alpha=\sup(-A)$, there must exist $x\in A$ such that $-\beta<x$. Multiplying both sides by $-a$, we get $\beta>-x$. But $-x\in(-A)$, so $\beta$ is not a lower bound of $-A$. Therefore, $\alpha$ must be the greatest lower bound of $A$.

user163916
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2 Answers2

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As has already been pointed out, your use of $\alpha$ is inconsistent. Specifically, you change the sign of $\alpha$ in the last sentence of the first paragraph: "This proves that $-\sup (-A)=\alpha$ is a lower bound of $A$." Up to that point, you appear to be using your first definition of $\alpha,$ which was $\alpha=\sup(-A).$

It seems to me that you can reverse the sign of $\alpha$ in the last sentence of the first paragraph and all occurrences after that, or you can reverse the sign of $\alpha$ in all occurrences before that, and in either case you will resolve the inconsistency.

David K
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First of all, the first paragraph in your proof shows that $-\alpha$ is a lower bound of $A$, not $\alpha$.

Second:

There is a problem in your proof. You say that:

if $\beta > \alpha$, then (because $-\alpha = \sup(-A)$) there exists such an $x\in A$ that $-\beta <x$

This is not true because $-\alpha =\sup(-A)$ is not true. By your definition, $\alpha = \sup(-A)$.

5xum
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