Let $|X|$ denote the cardinality of any set $X$, let $\mathbb N$ be the set of nonnegative integers, let $a$ be a cardinal $> 2^{|\mathbb N|}$, let $C$ be the set of cardinals $\le a$, let $\Omega$ be the set of ordinals of cardinality $\le a$, let $\mathcal P(\mathbb N)$ be the set of subsets of $\mathbb N$ partially ordered by inclusion, let $\mathcal T$ be the set of totally ordered subsets of $\mathcal P(\mathbb N)$, let $\mathcal W$ be the set of well-ordered subsets of $\mathcal P(\mathbb N)$, let $t:\mathcal T\to C$ be the map $T\mapsto|T|$, let $w:\mathcal W\to\Omega$ be the map sending $W$ to its corresponding ordinal, define the cardinal $t_0$ by $$ t_0:=\sup\ \{t(T)\ |\ T\in\mathcal T\}, $$ and define the ordinal $w_0$ by $$ w_0:=\sup\ \{w(W)\ |\ W\in\mathcal W\}. $$ Can one compute $t_0$ and $w_0$?
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Totally ordered under inclusion? – Andrés E. Caicedo Jun 19 '17 at 13:56
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@AndrésE.Caicedo - Yes. I've edited. Don't hesitate to let me know if it's still unclear. Thanks! – Pierre-Yves Gaillard Jun 19 '17 at 14:04
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Ugh. Using $\omega$ for anything other than the first infinite ordinal is just irresponsible. And $c$ often denotes the cardinal of the continuum, although usually in fraktur font. – Asaf Karagila Jun 19 '17 at 14:54
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@AsafKaragila - Sorry for the confusion. I hope it's ok now. Thanks for your comment! – Pierre-Yves Gaillard Jun 19 '17 at 15:14
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1I will post an answer soon, when I get home. To a proper keyboard. – Asaf Karagila Jun 19 '17 at 15:30
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1It looks like you used $a$ only to ensure that the ranges of your functions were in fact sets. You don't need to do this - the axioms of replacement do that for you automatically. – Noah Schweber Jun 19 '17 at 16:20
2 Answers
Let me rephrase the question a bit, since I think as written it's a bit harder to parse than necessary:
As you've defined it, the map $t$ just sends a chain in $2^\mathbb{N}$ to its cardinality; there's no need to refer to $a$ at all. So $t_0$ is the supremum of the cardinalities of chains in $2^\mathbb{N}$. Similarly, $w_0$ is the supremum of the ordertypes of well-founded chains.
Let's think about $t_0$ first. Clearly $t_0$ is at most $\mathfrak{c}$ (= the cardinality of $\mathbb{R}$); in fact, this upper bound sharp - $t_0=\mathfrak{c}$ - and is in fact achieved, so is a maximum and not just a supremum.
To show this we just need to find a single chain of cardinality $\mathfrak{c}$. The idea here is to use Dedekind cuts. Fix a bijection $f: \mathbb{N}\rightarrow\mathbb{Q}$, and for a real $r$ let $S_r=\{f^{-1}(q): q\le r\}$. Then the collection $\{S_r: r\in\mathbb{R}\}$ is a chain of size continuum.
Alright, now let's turn to $w_0$. Suppose I have a well-ordered chain of ordertype $\theta$, $S=\{X_\eta: \eta<\theta\}$ with $X_\eta\subseteq X_\zeta$ iff $\eta\le\zeta$. For simplicity let's assume $\theta$ is a limit ordinal; this will be good enough. Then for each $\eta<\theta$, there is some $n\in\mathbb{N}$ in $X_{\eta+1}\setminus X_\eta$; let $n_\eta$ denote the least such natural number. Since there are only countably many natural numbers, this means that $\theta<\omega_1$.
This tells us $w_0\le\omega_1$. It also tells us that $\omega_1$ can't be achieved, so if we want to show $w_0=\omega_1$ we'll have to work a bit harder: fixing $\theta<\omega_1$ we'll need to build a well-ordered chain of ordertype $\theta$.
But this is straightforward: it's a standard exercise that every countable linear order embeds into $\mathbb{Q}$ in an order-preserving way. Let $f$ be as above, fix an order-preserving $h: \theta\rightarrow \mathbb{Q}$, and let $X_\eta=\{f^{-1}(q): q\le h(\eta)\}$. Then $S=\{X_\eta: \eta<\theta\}$ is a well-ordered chain of ordertype $\theta$.

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I find your answer and Asaf's wonderful. I especially appreciate the self-containedness of yours. I wanted to accept one of the answers to close the question, and I decided to accept Asaf's for chronological reasons. - You write "The for each $\eta<\theta$, there is some $n\in\mathbb{N}$ in $X_{\eta+1}\setminus X_\eta$". I think "The" should be "Then", and also, as noticed by Asaf, we may have $\eta<\theta$ but not $\eta+1<\theta$. These two points are so minor that you may want to leave your post as it is, that's why I didn't want to try to edit it myself. – Pierre-Yves Gaillard Jun 19 '17 at 18:10
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First note that $t_0$ has an obvious upper bound, the cardinality of $\mathcal P(\Bbb N)$, which happens to be $2^{\aleph_0}$. But we also know that there is a chain in $\mathcal P(\Bbb N)$ which is isomorphic to the real numbers, so this upper bound is obtained, and therefore $t_0$ is exactly $2^{\aleph_0}$.
For $w_0$, note that every partially ordered set embeds into its power set (with inclusion), and two power sets are isomorphic (as ordered sets) if and only if the underlying sets have the same cardinality. This gives you a quick proof that every countable ordinal embeds into $\mathcal P(\Bbb N)$. So $w_0\geq\omega_1$.
But, if $W$ is a well-ordered chain, denoting by $A'$ the successor of $A$ inside $W$ gives us an injection $A\mapsto\min A'\setminus A$ from $W$ into the natural numbers (this might need minor adjustments for a maximal element of $W$, but that does not matter overall). So it means that no member of $\mathcal W$ is uncountable. Therefore $w_0=\omega_1$, the first uncountable ordinal.
Two remarks:
Your definitions are cumbersome. You can just talk about the cardinals or ordinals. No need to choose $a$ and $C$ and $\Omega$ and all that.
If you already know that every countable ordinal embeds into the real numbers, then by the argument from $t_0=2^{\aleph_0}$ you already have that every countable ordinal embeds into $\mathcal P(\Bbb N)$. There are other, similar, methods which may be easier depending on your prior knowledge.
Or perhaps a direct approach might be clearer to you. It's all the same, really.

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1And I wasn't even home for half an hour (plus I had to hang the laundry, which took another ten minutes). Dude, you're getting sloppy... ;) – Asaf Karagila Jun 19 '17 at 16:16
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Thanks!!! You write: "every partially ordered set embeds into its power set (with inclusion)". Is this fact standard? Is it obvious? (At least it's not obvious to me...) – Pierre-Yves Gaillard Jun 19 '17 at 18:09
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1@Pierre-Yves: Somewhat standard. Map an element to the cone below it. You can find it on several questions and answers on this site as well. – Asaf Karagila Jun 19 '17 at 18:11
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1@Pierre-YvesGaillard There is a small subtlety here: in general you need to use the inclusive cone ${b: b\le a}$,since we could have distinct $a_1, a_2$ with ${a: a< b_1}={a: a< b_2}$. That said, in this case that doesn't happen, so we can use whichever type of cone we prefer. – Noah Schweber Jun 19 '17 at 18:22
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Thank you Asaf, thank you @NoahSchweber! (I agree that this is obvious, but I hadn't realized it. Sigh...) – Pierre-Yves Gaillard Jun 19 '17 at 18:40
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Asaf and @NoahSchweber - This is just to tell you that I've just asked a related question... – Pierre-Yves Gaillard Jun 19 '17 at 19:58
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