### 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.