TITLE: Set of maximal chains in $\langle\mathcal{P}(\mathbb{N}),\subseteq\rangle$
QUESTION [7 upvotes]: Let $S$ be the set of all maximal chains in the poset $\langle$$\mathcal{P}$$($$\mathbb{N}$$),$$\subseteq$$\rangle$ partitioned into equivalence classes by their order types. 
How many different order types are there? Can you give me some insight into the structure of $S$ when quasi-ordered by embedding of order types? What is the supremum of ordinals that can be embedded into this quasi-order?

REPLY [4 votes]: Somehow nobody dares to answer this question. (Edit: Thomas' answer didn't show up on my computer in time.  We have the same result. I am talking about jumps, he talks about successors.)
Let $L$ be a linear order.  Two points $x,y\in L$ form a jump
if $x<y$ and the open interval $(x,y)$ is empty.
Let $M\subset\mathcal P(\mathbb N)$ be a maximal chain.
Then $M$ is a infinite complete linear order with endpoints 
and at most countably many jumps.
By the maximality of $M$, $\emptyset$ and $\mathbb N$ have to be in $M$.
Hence $M$ has endpoints.
Similarly, if for all nonempty $S\subseteq M$, $\bigcup S\in M$.
$\bigcup S$ is just the supremum of $S$ in $M$.
Similarly, every nonempty subset $S$ of $M$ has an infimum in $M$, namely 
$\bigcap S$.
It follows that $M$ is complete.
Now, if $x<y$ is a jump in $M$, then there is $n_{x,y}\in\mathbb N$ such that $n_{x,y}\in y$ and $n_{x,y}\not\in x$.
For two different jump $x,y$ and $x',y'$, $n_{x',y'}\not=n_{x,y}$.
Since there are only countably many elements of $\mathbb N$, 
there are only countably many jumps.
Finally, why is $M$ separable?  For each $n\in\mathbb N$ let 
$a_n=\bigcup\{a\in M:n\not\in a\}$.
Then $a_n\in M$.
Let $a,b\in M$ be such that $a\subsetneq b$.
For some $n\in\mathbb N$, $n\in b$ but $n\not\in a$.
Now $a\subseteq a_n\subsetneq b$.
This shows that every interval of the form $[a,b)$ with $a\subsetneq b$
has an element of the form $a_n$.
Now if $a<b$ and $a,b\in M$, then either $a,b$ form a jump in $M$ and the open interval $(a,b)$ is empty or there is $c\in M$ such that $a<c<b$.
In this case there is $n$ such that $a_n\in[c,b)\subseteq(a,b)$.
It follows that 
$\{a_n:n\in\mathbb N\}$ is actually dense in $M$.
This proof indicates another structure of $M$:
For $n\in\mathbb N$ let $a^n=\bigcap\{a\in M:n\in M\}$.
Clearly, $a_n<a^n$ and $a_n,a^n$ form a jump of $M$.
So, by the previous argument, left partners of jumps are dense in $M$.
It follows that there is no maximal chain whose ordertype is that of $\mathbb R$!
You do have chains of ordertype $\mathbb R$, but these are not maximal.
Similarly, you have chains of every separable ordertype with at most countably many jumps.
If $L$ is an infinite complete linear order with countably many jumps and the jumps are dense in $L$, let $(d_n)_{n\in\mathbb N}$ be an enumeration of all
left partners of jumps in $L$.
Now the map 
$$e:L\to\mathcal P(\mathbb N);x\mapsto\{n:d_n<x\}$$
is order preserving, so $e[L]$ is a chain.
The first element of $L$ gets mapped to $\emptyset$ and the last element of $L$
gets mapped to $\mathbb N$.
To see that $e[L]$ is maximal, let $a\subseteq\mathbb N$ be such that $e[L]$
is a chain.
If $\{d_n:n\in a\}$ has a largest element, it is of the form $d_n$ and hence the left partner of a jump with right partner $x$.
Now $e(x)=a$.
If $\{d_n:n\in a\}$ has no last element, then let $x=\sup\{d_n:n\in a\}$.
Now also $e(x)=a$.
This shows that $a\in e[L]$.
Hence $e[L]$ is indeed a maximal chain.