What level of rigour is expected in Real Analysis?

Question

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.

Answer 1

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

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 $-A$. ${}\qquad\blacksquare$

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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.

Answer 2

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.

Answer 3

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.