The strange difference between “divergent sequences” in real analysis and abstract algebra

by ayushkhaitan3437

I have been working on Commutative Algebra. A lot of the initial proofs that I’ve come across use Zorn’s lemma. The statement of Zorn’s lemma is simple enough (which I have blogged about before):

Suppose a partially ordered set P has the property that every chain (i.e. totally ordered subset) has an upper bound in P. Then the set P contains at least one maximal element.

I came across the proof of the fact that every field has an algebraic closure, which also uses Zorn’s lemma. The argument was: continue adding field extensions infinitely. So the sequence that we have generated is something like F_1\leq F_2\leq \dots (ad infinitum). Now \bigcup_{i=1}^{\infty} F_i is also a field (or rather can be made into a field). Also, \bigcup_{i=1}^{\infty} F_i is the limit of the sequence. We then argue that as \bigcup_{i=1}^{\infty} F_i is the maximal element in the chain of nested fields, it is the algebraic closure (use Zorn’s lemma here).

An analogy in real analysis would be: take a divergent sequence. Can we say anything about its limit except for the fact that it is \infty (or -\infty)? There’s really not that much to say. Say we have two divergent sequences \{n\} and \{n^2\} for n\in\Bbb{N}. All we can say is \{n^2\} approaches \infty faster than \{n\}. Nothing else.

But here, we’ve taken something like a divergent sequence, and said something intelligent about it. This has to do with the fact that the limit of the divergent sequence is still a field, and satisfies all the axioms of a field, while the limit of a divergent sequence in real algebra does not act like a real number by any stretch of imagination. This is a rather strange fact, and should be noted for a full appreciation of the argument.

Also, we did not know right away that the limit of the sequence of field extensions would be a field. We had to make a minor argument that the limit is exactly the union of all fields in the sequence.

So ya.