18
$\begingroup$

Let ${\cal P}(\omega)$ denote the power-set of $\omega$. We order it by set inclusion $\subseteq$ and say that ${\cal C}\subseteq {\cal P}(\omega)$ is a chain if for all $A, B\in {\cal C}$ we have $A\subseteq B$ or $B \subseteq A$.

Zorn's Lemma implies that every chain is contained in a maximal chain with respect to set inclusion.

Consider the following statement in ${\sf (ZFC)}$:

(S) If $\kappa$ is a cardinal with $\aleph_0 \leq \kappa \leq 2^{\aleph_0}$ there is a maximal chain ${\cal M}\subseteq {\cal P}(\omega)$ with $|{\cal M}| = \kappa$.

The chain $\big\{\emptyset, \{0\}, \{0,1\}, \{0,1,2\}, \ldots\big\}\cup \{\omega\} \;\big( = \omega \cup \{\omega\}\big)$ is easily seen to be a countable maximal chain in ${\cal P}(\omega)$. Interestingly, chains in ${\cal P}(\omega)$ can have length $2^{\aleph_0}$. So statement (S) is holds under the assumption of the Continuum Hypothesis ${\sf (CH)}$.

Question. Is (S) also true in ${\sf (ZFC)}$ if we do not assume ${\sf (CH)}$?

Improve this question
$\endgroup$

3 Answers 3

Reset to default
22
$\begingroup$

In fact, $\mathsf{ZFC}$-provably the only possible cardinalities of maximal chains in $\mathcal{P}(\omega)$ are $\aleph_0$ and $2^{\aleph_0}$.

Let $C$ be a maximal chain. Say that $C$ is scattered if there is no embedding of $\mathbb{Q}$ into $C$ (as linear orders). If $C$ is not scattered, then it must have size continuum: letting $f:\mathbb{Q}\hookrightarrow C$, if $\vert C\vert<2^{\aleph_0}$ then there must be some real $r$ such that the sets $C_{<r}=\{c\in C: \exists q\in\mathbb{Q}_{<r}(c\subseteq f(q))\}$ and $C_{>r}=\{c\in C: \exists q\in\mathbb{Q}_{<r}(c\supseteq f(q))\}$ form a gap in $C$, i.e. no element of $C$ is above everything in $C_{<r}$ and below everything in $C_{>r}$. But then the set $\bigcup C_{<r}$ can be added to $C$, so $C$ is not maximal.

So any maximal chain of size $<2^{\aleph_0}$ must be scattered. But any uncountable scattered linear order admits an embedding from either $\omega_1$ or $\omega_1^*$, and no chain in $\mathcal{P}(\omega)$ admits such an embedding. So we are done.

EDIT: I just realized that there's a much simpler argument after concluding that maximal chains of size $<2^{\aleph_0}$ are scattered. If $C$ is a scattered chain, then there are at least as many successor elements as there are limit elements, and each successor element of a chain contains a natural number not in any earlier element of the chain. So "scattered chains are countable" does not rely on the more difficult fact I mentioned above. (Also, note that this argument is related to YCor's topological one.)

FURTHER EDIT: I don't know a reference for the claim that every uncountable scattered linear order admits an embedding from either $\omega_1$ or $\omega_1^*$. However, here's one proof:

Given an countable linear order $L$, say that an interval $I$ in $L$ is huge iff there is some $a\in I$ such that both $I_{<a}$ and $I_{>a}$ are uncountable. If every uncountable interval is huge then by recursively applying the definition of hugeness, starting with $I=L$, we can build an injection of the dyadic rationals between $0$ and $1$ into $L$. But the former is isomorphic to $\mathbb{Q}$, so we're done.

What if not all uncountable intervals are huge? Well, let $S$ be an uncountable non-huge interval. If either the forward or backward cofinality of $S$ is $\ge\omega_1$, we're done, so we can WLOG find a cofinal-in-both-directions image of $\mathbb{Z}$ in $S$. By non-hugeness of $S$, at most (hence exactly) one of the resulting subintervals is uncountable. Repeating this process, we get a sequence of pairs $(a_i,b_i)$ with $j<i\implies a_j<a_i<b_i<b_j$ for all countable ordinals $i$: going from $(a_i,b_i)$ to $(a_{i+1},b_{i+1})$ is just what we did above, and to define $(a_\lambda, b_\lambda)$ for $\lambda$ limit we simply repeat the argument above on the open interval $\{s\in S: \forall \alpha<\lambda(a_\alpha<s<b_\alpha)\}$. If this process stops at some countable ordinal (i.e. the resulting interval is empty) we get $S$ as a countable union of countable subintervals, and if the process continues through $\omega_1$ we get an embedding of $\omega_1+\omega_1^*$ into $S$.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ @ Noah Schweber: Would be so kind to give me a reference for the statement "every uncountable scattered linear order admits an embedding from either $\omega_1$ or $\omega_1^*$. $\endgroup$ Commented Sep 17 at 22:07
  • $\begingroup$ @JoelAdler See my edit. It's probably in Rosenstein's book, but I don't have that book on hand. $\endgroup$ Commented Sep 18 at 0:34
  • $\begingroup$ @ NoahSchweber Thanks a lot. I checked Rosenstein's book but couldn't find anything. $\endgroup$ Commented Sep 18 at 8:59
25
$\begingroup$

Already answered, but here's an alternative argument:

endow $2^\omega$ with the product topology. Then the closure of every chain is a chain. Hence every maximal chain is closed, hence compact metrizable (and totally disconnected). Now every compact metrizable space is either countable or contains a Cantor set. Hence it has cardinality either countable or continuum. And clearly no finite chain is maximal, so the only possible cardinalities are $\aleph_0$ and continuum.

Improve this answer
$\endgroup$
3
  • 4
    $\begingroup$ That's a very nice argument from an unexpected corner - thanks YCor! $\endgroup$ Commented Sep 8 at 8:25
  • 1
    $\begingroup$ And the proof of the key lemma here (CH for compact metrizable spaces) is at a glance easier than the proof of the key lemma in mine (uncountable scattered orders embed $\omega_1$ or $\omega_1^*$). $\endgroup$ Commented Sep 8 at 22:59
  • 1
    $\begingroup$ (Actually per my edit that trickier lemma is not needed in my answer, although a bit of analysis of linear orderings is still required.) $\endgroup$ Commented Sep 9 at 19:22
5
$\begingroup$

This is a comment on Noah Schweber's answer but won't fit into a comment box. It's a reference for the fact that every uncountable linearly ordered set embeds either $\omega_1$ or $\omega_1^*$ or $\eta$. Quoting from p. 446 of P. Erdős and R. Rado, A partition calculus in set theory, Bull. Amer. Math. Soc. 62 (1956), 427–489 (pdf):

Lemma 1. Let $S$ be an ordered set, and $|S|^\prime=\aleph_n$; $\omega_n,\omega_n^*\not\leqq\overline S$. Then, corresponding to every rational number $t$, there is $S_t\subset S$ such that $|S_t|=|S|$; $S_t\subset L(S_u)$ for $t\lt u$.

Sierpiński, in a letter to one of us, had already noted the weaker result that, if $|S|=\aleph_1$; $\omega_1,\omega_1^*\not\leqq\overline S$, then $\eta\leqq\overline S$.

Here $n$ is an ordinal number (finite or infinite); $|S|^\prime$ is the cofinality cardinal of the cardinal $|S|$; $\overline S$ is the order type of the linearly ordered set $S$; $S_t\subset L(S_u)$ means that every element of $S_t$ is less than every element of $S_u$; and $\eta=\overline{\mathbb Q}$.

I will paraphrase Lemma 1 and sketch a proof. If $A$ and $B$ are subsets of an ordered set, the notation $A\lt B$ means that $a\lt b$ for all $a\in A$ and $b\in B$.

Lemma. Let $S$ be a linearly ordered set of cardinality $|S|=\kappa$ where $\operatorname{cf}\kappa=\omega_\alpha$. If $S$ embeds neither $\omega_\alpha$ nor $\omega_\alpha^*$, then there are sets $S_t\subset S$ ($t\in\mathbb Q$) with $|S_t|=\kappa$ and $S_t\lt S_u$ whenever $t\lt u$.

Proof. Let $\kappa$ be a fixed cardinal with $\operatorname{cf}\kappa=\omega_\alpha$.

Sublemma 1. If $S$ is a linearly ordered set, $|S|=\kappa$, and if $|\{x\in S:x\lt a\}|\lt\kappa$ for all $a\in S$, then $S$ embeds $\omega_\alpha$.

Sublemma 2. If $S$ is a linearly ordered set, $|S|=\kappa$, and if $|\{x\in S:x\gt a\}|\lt\kappa$ for all $a\in S$, then $S$ embeds $\omega_\alpha^*$.

To save typing, a "blob" is a linearly ordered set of cardinality $\kappa$ embedding neither $\omega_\alpha$ nor $\omega_\alpha^*$.

Sublemma 3. If $S$ is a blob, then for some $a\in S $ we have $|\{x\in S:x\lt a\}|=|\{x\in S:x\gt a\}|=\kappa$; that is, $\{x\in S:x\lt a\}$ and $\{x\in S:x\gt a\}$ are both blobs.

Sublemma 4. If $S$ is a blob, then there are blobs $X,Y,Z\subset S$ with $X\lt Y\lt Z$.

Let $S$ be a blob. We construct blobs $S_t\subset S$ ($t\in\mathbb Q$) by applying Sublemma 4 repeatedly as follows.

Step 1. Find blobs $A_1,S_0,A_2\subset S$ with $$A_1\lt S_0\lt A_2.$$

Step 2. Find blobs $A_3,S_{-1},A_4\subset A_1$ and $A_5,S_1,A_6\subset A_2$ with $$A_3\lt S_{-1}\lt A_4\lt S_0\lt A_5\lt S_1\lt A_6.$$

Step 3. Find blobs $A_7,S_{-2},A_8\subset A_3$ and $A_9,S_{-\frac12},A_{10}\subset A_4$ and $A_{11},S_{\frac12},A_{12}\subset A_5$ and $A_{13},S_2,A_{14}\subset A_6$ with $$A_7\lt S_{-2}\lt A_8\lt S_{-1}\lt A_9\lt S_{-\frac12}\lt A_{10}\lt S_0\lt A_{11}\lt S_{\frac12}\lt A_{12}\lt S_1\lt A_{13}\lt S_2\lt A_{14}$$.

And so on.

Improve this answer
$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.