The assumption of a Cauchy sequence in proving completeness of a set.

Question

In a lot of proofs in proving that a set is complete we have to go about taking a Cauchy sequence in the set. I've always questioned why are we allowed to "just" do this? As in how are we certain that Cauchy sequences live in our set? I guess this question could be stated for taking any sorts of sequences. How do we know for certain that sequences exist in our set? Is it because we can "make" a sequence out of anything and then customize it for the necessary situation?

Answer 1

The definition of completeness is that all Cauchy sequences converge. A space without Cauchy sequences (necessarily empty) is still complete, because all of its $0$ Cauchy sequences converge.

If this is the case, then the space is still complete. This is a little feature of the "for all" quantifier $\forall$. We still define statements of the form $\forall x \in \emptyset, P(x)$ to be true. So, even if we consider every Cauchy sequence in a space without Cauchy sequences, the statement would be considered true. This particular truth is often dubbed "vacuous" truth. (In such circumstances, it would be equally, and vacuously, true to say all Cauchy sequences diverge!)

Basically, it's not your problem to worry about whether Cauchy sequences exist or not! Just work with an arbitrary Cauchy sequence, and find a limit. If Cauchy sequences exist, then your argument provides a recipe to find the limit, and proof that it is the limit. If Cauchy sequences do not exist, then your argument was not necessary after all, and completeness is vacuously true.

Answer 2

It is because, unless our set $S$ is empty, then we can find an element $a$ of $S$ and define $(\forall n\in\mathbb N):a_n=a$ (which, by the way, will be a Cauchy sequence). And also because, by definition, $S$ is complete if every Cauchy sequence of elements of $S$ converges to an element of $S$.