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I am trying to understand how the Axiom of Choice implies Zorn's Lemma. A handout that I found to be rather easy to follow is here. The bit that I am stuck at is the tower.

In Halmos's Naive set theory, and other textbooks and handouts, the definition of tower comes without a warning and I am left trying to figure out how it advances the proof.

My questions are:

  • what the motivation behind the definition, what is the backstory that bring about the tower? What is it that we are trying to accomplish?

  • Is it by taking intersection of towers, we obtain "one long chain" not unlike a rope that extend upward? and no "limbs" coming out from the rope that extend upwards as well?

I went through this link but still none the wiser in the end. Explanation of Proof of Zorn's lemma in Halmos's Book- II- Definition of towers and $\mathscr{X}$

and I don't quite follow the answer given in Proof of Zorn's Lemma using the Axiom of Choice. Why is $\mathscr U$ a tower?

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    $\begingroup$ That is the problem when you insist on avoiding ordinals. Some things just get complicated for no good reason. $\endgroup$ – Asaf Karagila Dec 2 '16 at 16:08
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We're trying to find a maximal chain under inclusion, as indicated earlier. We're going to try and build one: an extremely long rope extending upwards, without branching. Ropes aren't easily composable, so the metaphor we'll use is a tower that we build layer by layer.

The way to build one is to repeatedly extend an existing tower by building another floor on top of it; this motivates the function $g$.

[The following paragraph amends item 1 in the definition of a tower: using $\emptyset$ instead of $\{ x \}$. I find this more obvious, but you can just skip this paragraph if you want to start with $\{x\}$.]

Now, we want to assume as little as possible about our poset, so ideally we just want to start off with a tiny chain and build up from there. The smallest chain is $\emptyset$, and we need a way to grow from there. We don't know anything else about our poset, so we'll just fix $x$ and say that $g(\emptyset) = \{ x \}$. (All told, this motivates item 1.)

To grow this chain $\{x\}$, we're obviously going to apply $g$, since that's our chain-growing procedure. Say it produces $\{x, y\}$. Then again, we'll grow the chain, and so on. (This motivates item 3.)

But now we've got arbitrarily large finite nested chains; how can we keep going to infinity? There's one obvious way: the chains are nested (and indeed each pair of chains has "one is an initial segment of the other"), so we can take the union. This is item 2.


A much more intuitive perspective is granted by the ordinals. We're performing recursion over the ordinals; item 3 gives us the definition on successor ordinals, and item 2 gives us the definition on limit ordinals. By Hartogs's lemma, the construction procedure must stop at some ordinal. The ~fischer document seems to be trying very hard to expunge the idea of the ordinals from the proof, but that's ultimately what's going on: Zorn is an attempt to build a copy of the ordinals inside a poset, and this is forbidden by Hartogs's Lemma.


The reason we need choice, by the way, is to decide any branching of our tower. At each stage, we might have several ways to extend the tower upwards; Choice lets us pick precisely one, so we end up with a single continuous tower at the end.

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