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I fail to find a duplicate.

I am wondering what level of rigour is needed in a typical undergraduate course in Real Analysis. To clarify my question, I provide an exercise from Rudin and my proposed solution:

(Exercise 5, 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 $$\inf A = -\sup(-A)$$

My answer:

$A$ is bounded below. As such, $-A$ must be bounded above. Suppose $\alpha$ is the greatest lower bound of $A$. It follows that $-\alpha$ is the least upper bound of $-A$. As such, we arrive at the desired expression $$\inf A = -\sup(-A)$$

This, for instance, feels very short, but I also feel that there is not much more to be said here. While this task might possibly be a bad example, I dare to guess that the most common pitfall for young students entering higher mathematics is that they underestimate the rigour needed to solve seemingly trivial problems. As such, I ask for an elaboration on this. The provided example does not necessarily have to be used in your answer.

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I believe this question should be made CW. – Sanath K. Devalapurkar Jun 15 '14 at 15:51
I would expect an undergraduate student in first semester real analysis to state precisely why $-A$ is bounded above. – Mark McClure Jun 15 '14 at 15:53
You fail to mention about the least upper bound property of $\mathbb{R}$. The reason why every non-empty upper bounded set in $\mathbb{R}$ has a supremum in $\mathbb{R}$. – fermesomme Jun 15 '14 at 15:57
Related. – Git Gud Jun 15 '14 at 16:11
@AndrewThompson: I think that you must be always rigorous while you are proving theorems of Real Analysis. I always had the habit of thinking from the first principles when I was a beginner. You may not always write your proofs mentioning first principles but I think that as a beginner, it is good to think from the first principles. – William Hilbert Jun 16 '14 at 6:35
up vote 16 down vote accepted

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

Rudin's book gives definitions of the concepts involved, and I would stick close to what those definitions say.

$A$ is bounded below, i.e. it has a lower bound $x$. That means $\forall a\in A,\ x\le a$. Consequently $\forall a\in A,\ -x\ge -a$.

$\forall b \in -A\ \exists a\in A\ b = -a$, hence $\forall b\in -A,\ -x\ge b$. Thus $-x$ is an upper bound of $-A$.

Thus we have proved that for every lower bound $x$ of $A$, $-x$ is an upper bound of $-A$. In particular $-\inf A$ is an upper bound of $-A$. In order to show that $-\inf A$ is the smallest upper bound of $-A$, one must show that no number less than $-\inf A$ is an upper bound of $-A$. Suppose $c<-\inf A$. Then $-c>\inf A$. Since $-c$ is greater than the largest lower bound of $A$, $-c$ is not a lower bound of $A$. Hence for some $a\in A$, $a<-c$, and so $-a>c$. Since $-a\in-A$, we have a member of $-A$ that is greater than $c$, so $c$ is not an upper bound of $-A$. ${}\qquad\blacksquare$

I'd write something like that in an exercise in a section in which the concepts of upper and lower bounds and infs and sups were introduced.

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BTW, in any exercise whose purpose is to prove that non-empty sets with a lower bound have an infimum, one way to do it follows just about the path I set out above. But here's another: Look at the set of all lower bounds of a non-empty set, and think about its supremum. – Michael Hardy Jun 15 '14 at 16:26
I fully admit I don't know how to make this proposal rigorous, but isn't it easier than all this? Since unary negation is an order-reversing "isomorphism" $\mathbb{R} \rightarrow \mathbb{R}$, and since $\sup$ and $\inf$ are defined only using order-theoretic concepts and are dual to each other, it should follow fairly straightforwardly that $\mathrm{sup}(-A) = -\mathrm{inf}(A).$ We should not have to "get our hands dirty" so to speak. – goblin Jun 15 '14 at 16:35
@user18921 : I'm not sure there is a difference between what you suggest and what I wrote. – Michael Hardy Jun 15 '14 at 16:45
@user18921: that's kind of what the questioner already did by saying "it follows that...". This answer fills in why that follows in terms other than intuition. If you throw a morphism theorem at the problem then you have to clearly state and might have to prove the theorem. Proof risks being more hassle than proving one simple case like this, but might well be less hassle than proving two. – Steve Jessop Jun 15 '14 at 17:29
@user18921: depends on context. Sometimes you want the big picture that illustrates this result is part of a whole family of related results (related by the morphism you're using to prove them). I'd say that sometimes you want a short easy proof that doesn't introduce unnecessary concepts. Aside from anything else, homework questions are generally leading somewhere and it's natural to let yourself be led rather than trying to rush ahead and do something very clever on the first one. You can always come back later and write another proof if the course doesn't take you there :-) – Steve Jessop Jun 16 '14 at 16:30

That is the expected answer. My experience with real analysis was questions that took a page or two followed by questions that follow as a simple corollary.

It is another common pitfall to assume that proofs have to be built up from first principles every time they are done.

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I agree with you, hence a +1. I believe that this answer should be made CW. – Sanath K. Devalapurkar Jun 15 '14 at 15:52
I don't entirely agree. See my posted answer. – Michael Hardy Jun 15 '14 at 16:23

Here's a proof along the lines that I would expect from an undergraduate student in first semester real analysis.

Proof: Let $A$ be a non-empty set of real numbers that is bounded below. By definition of bounded below, we may choose $\alpha\in\mathbb R$ such that $\alpha\leq x$ for every $x\in A$. This implies that $-x\leq -\alpha$ for every $x\in A$ so that $-\alpha$ is an upper bound for $-A$. Thus, $-A$ is a non-empty set of real numbers that is bounded above and, therefore, has a supremum, say $\beta$, by the axiom of completeness.

We must show that $-\beta$ is the infimum of $A$. First, note $\beta$ is an upper bound for $-A$ (by definition of supremum) or $\beta \geq -x$ for every $x\in A$. Thus, $-\beta\leq x$ for every $x\in A$ and $-\beta$ is a lower bound for $A$. Next, we must show that $\beta$ is the greatest lower bound of $A$. Thus, assume that $\beta<\gamma$. Then, $-\gamma<-\beta$ so (since $\beta$ is the supremum of $-A$), there is some $x\in A$ with $-\gamma<x<-\beta$. Therefore, $\beta<-x<\gamma$ with $x\in A$ so that $\gamma$ cannot be a lower bound of $-A$.$\Box$

To understand why these particular details are written out in grotesque detail, I would consider the material that you likely just learned. If you are trying to show that an infimum can be defined in terms of a supremum, then you have likely just learned these concepts, as well as concepts like upper and lower bounds. So I think you've really got to refer quite explicitly to those definitions. By contrast, I used the order properties, like $x<y \implies -y<-x$ without specific reference since that's probably at least a little bit in the past.

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Typo in the third line of the proof, $-\alpha$ is an upper bound of $-A$. – Daniel Fischer Jun 15 '14 at 16:20
@DanielFischer Thanks! – Mark McClure Jun 15 '14 at 16:22
In the second paragraph of the proof, you mix up $\beta$ and $-\beta$ in a few places, "we must show that" it's $-\beta$ that shall be the greatest lower bound of $A$, and so you need $-\beta < \gamma$, and get $-\gamma < \beta$, whence there is an $x\in -A$ with $-\gamma < -x$. – Daniel Fischer Jun 15 '14 at 16:24

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