I've been struggling a bit with this proof from Rudin. I've written two proofs, both of which seem to differ rather considerably from other proofs of this I've seen. I'm hoping someone can look these over. I'm going to assume as a lemma a theorem proved in Rudin which establishes the existence of the infimum.

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

Proof 1. By the least-upper-bound property, any nonempty subset of $\mathbb{R}$ that is bounded above has a least upper bound.

Lemma. (Proved in Rudin's text, and I'll take it as given for the purpose of this proof). Suppose $S$ is an ordered set with the least-upper-bound property, $B \subset S$, $B$ is not empty, and $B$ is bounded below. Let $L$ be the set of all lower bounds of $B$. Then $\alpha = \sup L$ exists in $S$, and $\alpha = \inf B$.

So, by the aformentioned lemma, any nonempty subset of $\mathbb{R}$ that is bounded below has a greatest lower bound. Since $A$ is bounded below and is nonempty, $\inf A$ exists, and has the property that \begin{align*} \forall x \in A, \forall y \in \mathbb{R}, \left(x \geq \inf A \wedge x \geq y \implies y \leq \inf A \right). \end{align*} Since $A$ is nonempty, there exists at least one $x \in A, x \geq \inf A$, a property that is true for all such $x$. This implies that $-x \leq - \inf A$, for all $-x \in -A$ and we conclude that, $- \inf A$ bounds $-A$ above. Assume that there exists some other upper bound, $-y$, of $-A$, so $-y \geq -x$ for all $-x \in -A$. Note that \begin{align*} -y \geq -x \implies y \leq x \implies y \leq \inf A \implies -y \leq - \inf A, \end{align*} So, $-y \leq - \inf A$, meaning that $- \inf A$ is also the greatest lower bound of $-A$, so $- \inf A = \sup\left(-A\right)$, which implies that $\inf A = - \sup\left(-A\right)$.

Remark. I worry about beginning with the infimum, rather than the supremum. Upon working through this proof more, it seems that the existence of the supremum follows directly from the least-upper-bound property, whereas the infimum does not. I worry I could be assuming the conclusion of the proof in positing its existence, which is something I don't do in this second proof. I would be very interested in any critiques and comments on whether what I did here is mathematically valid.

Proof 2. Since $A$ is nonempty, there exists at least one $x \in A \subset \mathbb{R}$. So, $-A \subset \mathbb{R} = \{-x: x \in A\}$. Since $A$ is bounded below, there exists some real number, $\beta$, such that $\forall x \in A, x \geq \beta$. This implies that $\forall - x \in -A, -x \leq - \beta$. Thus, $- \beta$ is an upper bound of $-A$, so by the least-upper-bound property, $\sup \left(-A\right)$ exists, and has the properties that \begin{align*} \forall y \in \mathbb{R}, \forall -x \in -A, \left(-x \leq \sup \left(-A\right) \wedge -x \leq y \implies y \geq \sup\left(-A\right) \right). \end{align*}

We now need to establish that $- \sup \left(-A\right)$ is the greatest lower bound of $A$. First, we show that it is a lower bound of $A$. Let $x \in A$. First, we have that for all $-x \in -A$, $-x \leq \sup\left(-A\right)$, which implies that $x \geq -\sup\left(-A\right)$, so $-\sup\left(-A\right)$ bounds $A$ below. Then, take some $-y \in \mathbb{R}$ that is also a lower bound of $A$, so $\forall x \in A, x \geq -y$. We then have that
\begin{align*} x \geq -y \implies -x \leq y \implies y \geq \sup \left(-A\right) \implies -y \leq - \sup\left(-A\right), \end{align*} which follows from the definition of the supremum of $-A$. So, any other lower bound, $-y$, of $A$ is no larger than $- \sup\left(-A\right)$, so $-\sup\left(-A\right)$ is both an upper bound of $A$ and the greatest such upper bound, so $\inf A = - \sup\left(-A\right)$.

Remark. This proof seems a bit smoother and draws on the same principles used earlier, but with the inequalities reversed. Hopefully the quantifiers are written correctly and I haven't overlooked anything.

So, my questions are:

(1) How does these proofs look? Have I made any mistakes or leaps of logic?

(2) Is one proof preferable to the other, e.g., am I wrong to draw on a lemma from the text to establish the greatest-lower-bound property, which is arguably what I'm trying to prove? Or, is this result from the text merely prove its existence, which is obviously needed to prove this result regardless?

Thanks in advance.


1 Answer 1



  • Your first proof is not entirely correct. I follow you up to this point: $$y \leq \inf A \implies - y \leq - \inf A.$$ This should read: $$y \leq \inf A \implies -y \geq -\inf A.$$ (Note the change in direction of the inequality.) You should then conclude that $-\inf A$ is the least upper bound of $-A$, not the greatest lower bound, as you do. (I believe that this may have been a small mistake; you make the correct conclusion in the following sentences.)

  • Your second proof looks correct to me.

After fixing the small mistakes, I have one point of advice: There is something to be said for brevity and clarity. For example, using English ("for all" versus $\forall$) over logical symbols is good style. It makes your proofs easier to parse for everyone.

As a concrete example, the way you state the properties of $\sup (-A)$ in your second proof is hard to digest. It would perhaps be easier to simply state that $\sup(-A)$ exists, and then quote its properties (with English!) as they arise. (Of course, when you are being extremely careful and making sure that your logic is correct, logical quantifiers are a wonderful thing to use! This is especially if a course requires it of you.)

Here is an example proof that is condensed into one paragraph, and uses hardly any symbols:

Since $\inf A$ is a lower-bound for $A$, it is clear that $- \inf A \geq -a$ for all $-a \in -A$. Thus $-\inf A$ is an upper bound of $-A$. If $x \in \mathbb{R}$ is such that $x < -\inf A$, then $-x > \inf A$. By definition of the greatest lower bound, this implies some $a \in A$ with $-x > a \geq \inf A$. But then $x < -a$, and $-a \in -A$. Thus $x$ is not an upper bound for $-A$, and so it follows that $\sup(-A) = -\inf A$.

  1. I don't think that either proof is "preferable" to the other. In particular, I don't think that your "big picture" reasoning is ever circular, so both proofs are valid. They are merely different ways to go about proving the theorem.

I happen to like the idea of the first one, but that's just because I think Rudin's lemma is cool.

  • 1
    $\begingroup$ +1 for the advice about using English. Some textbooks make the subject of analysis boring and difficult by using excessive symbolism. $\endgroup$
    – Paramanand Singh
    Sep 3, 2018 at 4:11

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