In Halmos's Naive Set Theory about well-ordering set, it states that if a collecton $\mathbb{C}$ of well-ordered set is a chain w.r.t continuation, then the union of these sets is a well-ordered set. However, I cannot see why it is well-ordered. That is, I cannot see why each of its subset has a smallest element.
[Math] The union of well-ordered sets is a well-ordered set
elementary-set-theory
Related Solutions
Suppose you are given an uncountable, well-ordered set $S$.
Isn't it possible to provide a bijection $f:\mathbb{N} \rightarrow S$
No. $\mathbb N$ is countable. You will end up with a set of elements $\{s_1,s_2,s_3,\dots\}$, but there will exist elements you have not covered. There is nothing in your definition that demands you covered all of the elements.
An example (not a well ordered set, I know, but may still illustrate my point) is if you look at the set $$S=\left\{\frac12, \frac23, \frac34, \dots, \frac{n}{n+1},\dots\right\}\cup[1,2]$$
The procedure you described works on $S$, although $S$ is not well ordered. It takes $\frac12 = s_1$ as it is the least element. Then it takes $s_2=\frac23$ and so on. It produces $s_i = \frac{i}{i+1}$ which is an injection from $\mathbb N$ to $S$, but it does not cover the whole $S$.
Edit: In fact, you can even take the set $$T=\left\{\frac12, \frac23, \frac34, \dots, \frac{n}{n+1},\dots\right\}\cup\{1\},$$ whic is well ordered and is even countable, but your procedure still does not produce a bijection from $\mathbb N$ to $T$.
I assume that $\Bbb N=\{1,2,\dots\}$, and thus does not include $0$.
Your example $P$ is essentially the disjoint union of two copies of $\Bbb N$. I believe it will be helpful to rename the sets $\{\phi_1,\dots,\phi_n\}$ so that you do not get lost in nested sets. We can name $a_n=\{\phi_1,\dots,\phi_n\}$ and $b_n=\{\psi_1,\dots,\psi_n\}$ for each $n\in\Bbb N$, then $P=\{a_n\mid n\in\Bbb N\}\cup \{b_n\mid n\in\Bbb N\}$. Let $\preceq$ be the partial order, i.e.
\begin{align} a_n\preceq a_m \quad\text{ iff }\quad n\leq m\quad\text{ iff }\quad\{\phi_1,\dots,\phi_n\}\subseteq\{\phi_1,\dots,\phi_m\},\end{align}
and similarly
\begin{align} b_n\preceq b_m \quad\text{ iff }\quad n\leq m\quad\text{ iff }\quad\{\psi_1,\dots,\psi_n\}\subseteq\{\psi_1,\dots,\psi_m\}, \end{align}
while we also have $a_n\mathbin{\not{\!\!\preceq}} b_m$ for any $n,m\in\Bbb N$.
What does the collection of all totally ordered subsets of $P$ mean? We'll denote this set with $\mathcal F$. Usually we don't have to write out $\mathcal F$ explicitly, but for your example this is easy: if $x,x'\in P$ and $x\mathbin{\not{\!\!\preceq}}x'$ and $x'\mathbin{\not{\!\!\preceq}}x$, then we know that $x=a_n$ and $x'=b_m$ for some $n,m$, or vice versa.
Hence, a set of elements of $P$ is totally ordered if and only if it is empty, exclusively contains elements of the form $a_n$ or exclusively contains elements of the form $b_n$. More formally, $Y\in \mathcal F$ if and only if there exists some $X\subseteq\Bbb N$ such that $Y=\{a_n\mid n\in X\}$ or $Y=\{b_n\mid n\in X\}$.
For example, $\{a_{12},a_{1},a_{2748},a_{52}\}\in\mathcal F$, since $a_1\preceq a_{12}\preceq a_{52}\preceq a_{2748}$, but $\{a_4,b_{26},a_5\}\notin \mathcal F$, since $a_4\mathbin{\not{\!\!\preceq}}b_{26}$ and $b_{26}\mathbin{\not{\!\!\preceq}}a_4$. Note that $\mathcal F$ may also contain infinite sets, such as $\{a_n\mid n\text{ is even}\}$.
In general, for any partially ordered $P$ we may simply define $\mathcal F$ without knowing what its elements are. A subset $X\subseteq P$ either is totally ordered, or it is not. If $X$ is totally ordered, then $X\in \mathcal F$, and if $X$ is not totally ordered, then $X\notin \mathcal F$.
Finally we have the claim that the union of a $\subseteq$-chain in $\mathcal F$ is totally ordered by $\preceq$. Let $(I,\leq)$ be some totally ordered set of indices, and let $\{X_{i}\mid i\in I\}\subseteq \mathcal F$ be a $\subseteq$-chain of members of $\mathcal F$. In other words, each $X_i\subset P$ is totally ordered by $\preceq$ and for any $i,j\in I$ we have $X_i\subseteq X_j$ iff $i\leq j$. Then we need to show $\bigcup_{i\in I}X_i$ is totally ordered by $\preceq$.
It's easy to see that $\preceq$ is a partial order on $\bigcup_{i\in I} X_i$, since $\bigcup_{i\in I} X_i\subseteq P$ and subsets of partial orders are partially ordered. Thus we only need to check that $\preceq$ is total.
Let $x,x'\in\bigcup_{i\in I}X_i$, then there are $i,j\in I$ such that $x\in X_i$ and $x'\in X_j$. Without loss of generality $i\leq j$, which implies $X_i\subseteq X_j$ and thus $x,x'\in X_j$. Now, since $X_j$ is totally ordered by $\preceq$, we have that $x\preceq x'$ or $x'\preceq x$. Therefore for any $x,x'\in\bigcup_{i\in I}X_i$ we have $x\preceq x'$ or $x'\preceq x$, which means $\preceq$ is indeed total on $\bigcup_{i\in I}X_i$.
Best Answer
You have to re-read all the passage from page 67-on :
The relevant fact is : the collection $\mathcal C$ of well-ordered set is a chain w.r.t continuation.
A collection $\mathcal C$ is a chain when, for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$.
If a collection $\mathcal C$ is a chain w.r.t continuation, it has "something more" : in addition to the property (common to all chains) that for all $A,B \in \mathcal C$ : $A \subseteq B$ or $B \subseteq A$, we have also that (supposing : $B \subseteq A$) $B$ is an initial segment of $A$, and the ordering of the elements in $B$ is the same as their ordering in $A$.
Thus, when we "merge" all the members of the collection $\mathcal C$ into the "mega-set" $U$ every subset "preserve" its "original" minimal-element.
I hope it may help ...
I'm not able to "manufacture" an example different from the "trivial" one built from $\mathbb N$.
If you consider a collection $\mathcal C = \{ X_n \}$ where all $X_i$ form a chain w.r.t continuation, we have that $X_n = \{ 0,1,2, \ldots n \} = n+1$.
Thus the union $U$ of the collection is $\omega$ itself.