TITLE: What level of rigour is expected in Real Analysis?
QUESTION [8 upvotes]: 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.

REPLY [16 votes]: $$
\inf A = -\sup(-A)
$$
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.