0
$\begingroup$

I was going through multiple proofs of Zorn's Lemma (Halmos', in Naive Set Theory, being amongst them), but I get stuck on the definition of towers, particularly the "union requirement" (requirement III in Halmos' proof and Proof Wiki).

Say $(X,\leq)$ is the poset under consideration and $\mathcal{X}$ is the set of all chains in $X$, partially ordered by inclusion. Let $\mathcal{N}$ be a subset of $\mathcal{X}$, and a candidate for being a tower. Now, the requirement I'm talking about is, if $\mathcal{K}\subseteq\mathcal{N}$ is a chain, then

$$\left(\bigcup_{K\in\mathcal{K}}K\right)\in\mathcal{N}.$$

What I don't understand is why this does not follow from the previous premises.

$\mathcal{N}$ is partially ordered under inclusion. Thus any chain $\mathcal{K}\subseteq\mathcal{N}$ must be totally ordered under inclusion. Doesn't that mean that there is a chain in $\mathcal{K}$ that already equals the union of all chains in $\mathcal{K}$ (so that said union is also in $\mathcal{N}$)?

For example, if $\mathcal{N} = \wp(\{1,2,3\})$ (the powerset of $\{1,2,3\}$), and say $\mathcal{K}=\{\{2\},\{2,3\},\{1,2,3\}\}$, then the union of all chains in $\mathcal{K}$ is $\{1,2,3\}\in\mathcal{K}$.

So, when would such union not be in $\mathcal{K}$? I searched for similar questions here and for other proofs online, but none seem to address this, which makes me think it's probably very obvious to everyone. Maybe I'm constantly misinterpreting something. Regardless, an explanatory example or just an explanation would be much appreciated.

$\endgroup$

1 Answer 1

0
$\begingroup$

Let $X=\mathcal{P}(\mathbb{N})$ and $\leqslant$ be set inclusion. Now, take$$\mathcal{K}=\bigl\{\{1\},\{1,2\},\{1,2,3\},\ldots\bigr\}$$and let $\mathcal N$ be the set of all finite subsets of $\mathbb N$. Then $\mathcal K$ is totaly ordered, right?! Besides, $\mathcal{K}\subset\mathcal N$. However, the union of all elements of $\mathcal K$ doesn't belong to $\mathcal N$.

$\endgroup$
6
  • $\begingroup$ Is $\mathcal{K}$ here finite or infinite? If it is finite, then the union of all elements of $K$ is $\{1,2,3,\ldots\}$ and does belong to $\mathcal{N}$, since it is a finite subset of $\mathbb{N}$. If $\mathcal{K}$ is infinite, then $\mathcal{K}$ is not a subset of $\mathcal{N}$. Where is my reasoning wrong? $\endgroup$
    – steve
    Oct 6, 2018 at 11:47
  • $\begingroup$ The set $\mathcal K$ is infinite. It's the set of all sets of the form $\{1,2,\ldots,n\}$, for some natural $n$. $\endgroup$ Oct 6, 2018 at 11:50
  • $\begingroup$ I get it now. Thank you so much, this is exactly what I was looking for! By the way, I assume there is no example that doesn't play around with infinities? $\endgroup$
    – steve
    Oct 6, 2018 at 12:20
  • $\begingroup$ I'm glad I could help. And, no, there is no example that doesn't play around with infinities. $\endgroup$ Oct 6, 2018 at 12:30
  • $\begingroup$ How could the union of $\cal K$ be $\Bbb N$ if $0$ is not an element of the union? :) $\endgroup$
    – Asaf Karagila
    Oct 6, 2018 at 13:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .