Fruits of procrastination

The existence or inexistence of a maximal element

Have you ever wondered why the real number line does not have a maximal element?
Take \Bbb{R}. Define an element \alpha. Declare that \alpha is greater than any element in in \Bbb{R}. Can we do that? Surely! We’re defining it thus. In fact, \alpha does not even have to be a real number! It can just be some mysterious object that we declare to be greater than every real number. Note that \alpha is greater than real number, but it is not the maximal element of \Bbb{R}, as for that it will have to be a part of \Bbb{R}. Why can’t \alpha be a part of \Bbb{R}? We’ll see in the next paragraph. 

However, it is when we assert that \alpha has to be a real number that we begin to face problems. If \alpha is a real number, then so is \alpha+1. Thereby, we reach a contradiction, showing that no real number can exist which is greater than all other real numbers.

Another approach is to take the sum of all real numbers. Let that sum be \mathfrak{S}, which is greater than any one real number. However, as the sum of real numbers is a real number by axiom (\Bbb{R} is a field), \mathfrak{S} is also a real number, which is smaller than the real number \mathfrak{S}+1. If we did not have the axiom that the sum of real numbers should be a real number, then we’d be able to create a number greater than all real numbers. The same argument would work if we were to multiply all real numbers.


Now I’d like to draw your attention to the proof of the fact that every ring must have a maximal ideal, as given on pg. 4 of the book “Commutative Algebra” by Atiyah-Macdonald. The gist of the proof is: take every ideal which is a proper subset of the ring, and find its union. This union is the maximal ideal.

Why this proof works is that we wouldn’t know which element to add to make the ideal bigger. If we could construct a bigger ideal for any ideal we choose, we can prove that no maximal ideal exists. But whatever element we choose to add to the previous ideal, we have no reason to suspect that that element does not already exist in it.

Let us generalize this argument. Let us take a set of elements, define an order between the elements, and then declare the existence of a maximal element which is part of the set. If we cannot prove that a bigger element exists, then there is no contradiction, and hence that element is indeed a maximal element of the set. This argument works if we were to prove that every ring has a maximal ideal, and does not if we were to prove that \Bbb{R} has a maximal element.


Breaking down Zorn’s lemma

Today I’m going to talk about Zorn’s lemma. No. I’m not going to prove that it is equivalent ot the Axiom of Choice. All I’m going to do is talk about what it really is. Hopefully, I shal be able to create a visually rich picture so that you may be able to understand it well.

First, the statement.

“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.”

Imagine chains of elements. Like plants. These may be intersecting or not. Imagine a flat piece of land, and lots of plants growing out of it. These may grow straight, or may grow in a crooked fashion, intersecting. These plants are totally ordered chains of elements. Now as all such chains have a maximal elements, imagine being able to see the tops of each of these plants. Not three things: 1. Each tree may have multiple tops (or maximal elements). 2. There may be multiple points of intersection between any two trees. 3. Different plants may have the same maximal element.

Moreover, there may be small bits of such plants lying on the ground. These are elements that are not part of any chain. If any such bit exists on the ground, then we have a maximal element. Proof: If it could be compared to any other element, it would be on a chain. If it can’t be compared to any other element, it’s not smaller than any element.

Let us suppose no such bits of plants exist. Then a maximal element of any chain will be the maximal element of the whole set! Proof: It is not smaller than any element in its own chain. It can’t be compared with the chains which do not intersect with this chain. And as for chains that intersect with this chain, if the maximal element is the same, then we’re done. If the maximal elements are not the same, then too the two maximal elements can’t be compared. Hence, every distinct maximal element is a maximal element of the whole set.

Assuming that the set is non-empty, at least one plant bit or chain has to exist. Hence, every partially ordered set has at least one maximal element. The possible candidates are plant bits (elements not in any chain) and plant tops (maximal elements of chains).