I recently stumbled upon Zorn's lemma. At the first glance I did not quite understand it, so I searched for articles explaining it.
After reading more about this lemma I thought I understood it, but almost everywhere I read about it, it was called non-intuitive, or similarly.
I would like to check if I truly understood the lemma, in an intuitive manner.
I will attempt to write an informal explanation of the lemma, so that you can correct me if I am mistaken with hopes that I will ultimately understand this theorem.
Informal explanation
The theorem states (Taken from wikipedia):
Suppose a partially ordered set $P$ has the property that every chain in $P$ has an upper bound in $P$. Then the set $P$ contains at least one maximal element.
What I understand from this is that a set $P$ can be divided in multiple totally ordered subsets, those being chains, and at least one of the upper bounds of these totally ordered subsets is a maximal element.
Also a type of a set $P$ with more than one maximal elements would be one where there are two such totally ordered subsets of $P$, let them be $S_1, S_2$, with the added property that $\forall \space x \in S_1 \space$ and $\forall \space y \in S_2 \space$, $x$ is not comparable to $y$.
This means that the upper bound of $S_1$, respectively $S_2$, are, or could be, maximal elements of $P$, right?
P.S.
I would appreciate simple answers, if possible, because I do not have a strong mathematical background, being a high school student. Also by simple I mean as few technical terms as possible.
Best Answer
I don't think your intuition is correct. In particular, I don't see a natural/intuitive way to make sense of "$P$ can be divided into multiple totally ordered subsets" in such a way that you can be sure that an upper bound for one of these chains is maximal in $P$.
Instead, let me suggest a different intuition for Zorn's Lemma. This is really just the usual proof, which I find to be fairly intuitive.
Suppose you give me a nonempty poset $P$. I want to find a maximal element of $P$. I start by picking any element $a_0$. If it happens to be maximal, then hooray: I'm done. If not, then I want to "climb up" the poset to find a maximal element.
Well, since $a_0$ is not maximal, there is some element $a_1$ with $a_0 < a_1$. Now maybe $a_1$ is maximal, in which case hooray. If not, then there is some element $a_2$ with $a_0 < a_1 < a_2$. If I continue in this way, then either I reach a maximal element at some point, or I have built a chain $$a_0 < a_1 < a_2 < a_3 < \dots.$$ Now here's where the hypothesis of Zorn's Lemma comes in. In general, the chain $a_0<a_1<a_2<\dots$ might go "all the way up". Think of $\mathbb{N}$ with its usual order (which has no maximal element) and the chain $0<1<2<\dots$.
But based on our hypothesis, the chain I built has an upper bound, i.e. there is some element $a_\omega$ which is greater than every element of the chain. If $a_\omega$ is maximal in $P$, then hooray. Otherwise, I can continue, building a chain $$a_0 < a_1 < a_2 < \dots < a_\omega < a_{\omega+1} < a_{\omega+2} < \dots$$ and this chain also has an upper bound, etc. etc. At some point, I will have exhausted all the elements of $P$, so the chain-building process must terminate at a maximal element.
Now there are a few points where we need some set theory to make the argument precise. First of all, each time I pick an element $a_{n+1}$ which is greater than $a_n$, I'm picking it in an arbitrary way from among all the elements of $P$ which are greater than $a_n$ (of which there might be many), and I need the Axiom of Choice to make all these arbitrary choices. Personally, I don't find this part unintuitive at all.
Second, when I write "etc. etc." I'm indicating that I might have to keep building chains for a really long time. Note that already my first chain was infinite, and then I have to keep going, and build another chain, and then a chain of chains, ... To formalize this, we use transfinite recursion/induction over the ordinals.
And finally, the most hand-wavey part of what I wrote: "at some point, we will have exhausted all the elements of $P$". This is because $P$ is a set, while the ordinals form a proper class. Speaking informally, we can prove in set theory that for any set $P$, there is upper bound for how long an infinite process of selecting new elements of $P$ (like the one described above) can go on. (Here I'm referring to Hartogs's Theorem.)
Edit: I've just noticed this other question which contains an informal description of the proof of Zorn's Lemma very similar to the one I gave above. You may find the question and Asaf's answer helpful.