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This proof has appeared numerous times on this website, but I have just two questions on it that I don't believe have been addressed in the past. The full theorem from Rudin is:

Theorem. 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 $\inf A = - \sup\left(-A\right)$.

My questions are:

(a) Are we guaranteed that $\inf A$ or $\sup\left(-A\right)$ exists? We are only given that $A$ is bounded below, so $\exists \beta, \forall x \in A, x \geq \beta$. But this doesn't directly imply that there exists a greatest $\beta$, nor does it directly imply anything about the nature or boundedness of the set $-A$. In other words, must be prove, or asssume, existence to write this proof? Or do we posit axiomatically that such an infimum exists, and establish this proof based on its properties?

(b) My second question hinges a bit more on the method by which I sought to prove this theorem, but I think it can be easily generalized, as I will try to do here. Say that we have some statement to the effect of $\forall x \in \mathbb{R}, a \leq b \implies y \leq c$. From $a \leq b$, multiplying both sides of the inequality by $-1$ returns $-a \geq -b$. Are we allowed to conclude from this that $-a \geq -b \implies -y \geq -c$? In other words, can be multiply both sides of an implication through by $-1$ and retain its structure? This was my first approach to the proof, and the result seems to check out, but I question whether this is an allowable step.

Thanks.

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The answer to (a) is the completeness axiom: every nonempty set of real numbers with a lower bound has a greatest lower bound.

As for (b) you have the implications

$$-a \ge -b \implies a \le b \implies y \le c \implies -y \ge -c.$$

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  • $\begingroup$ This is very clear, thank you! $\endgroup$ – Matt.P Sep 2 '18 at 14:57

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