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The following is the proof of the Hausdorff Maximality theorem in Rudin's real and complex analysis:

Hausdorff's Maximality Theorem Every nonempty partially ordered set $P$ contains a maximal totally ordered subset.

PROOF Let $\mathcal{F}$ be the collection of all totally ordered subsets of $P$. Since every subset of $P$ which consists of a single element is totally ordered, $\mathcal{F}$ is not empty. Note that if the union of any chain of totally ordered sets is totally ordered.
Let $f$ be a choice function for $P$. If $A \in \mathcal{F}$, let $A^*$ be the set of all $x$ in the complement of $A$ such that $A \cup \{x\} \in \mathcal{F}$. If $A^* \neq \emptyset$, put $$g(A) = A \cup \{ f (A^*) \}.$$ If $A^*=\emptyset$, put $g(A) = A$.

I'm having some trouble understanding what the collection of all totally ordered subsets of $P$ means, and what the union of chain of totally ordered sets means. More specifically, as an example, let $$P = \{ \{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}, \{\phi_1, \phi_2, \phi_3, \dots\}, \dots, \{\psi_1, \psi_1\}, \{\psi_1, \psi_2, \psi_3\}, \{\psi_1, \psi_2, \psi_3, \dots\}, \dots \}$$ where $P$ contains $(\phi_n)_n$ and $(\psi_n)_n$ for all $n\in \mathbb{N}$. Clearly, $P$ is a partially ordered set by set inclusion. Then if $\mathcal{F}$ is the collection of all totally ordered subsets of $P$, does it mean that $\mathcal{F}$ is $$\mathcal{F} = \Big\{ \{ \{\phi_1, \phi_2\}\}, \{\{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}\}, \{\{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}, ...\}, \dots, \{\{\psi_1, \psi_2\} \}, \{\{\psi_1, \psi_2\}, \{\psi_1, \psi_2, \psi_3\} \}, \{\{\psi_1, \psi_2\}, \{\psi_1, \psi_2, \psi_3\}, \dots \}, \dots \Big\} ?$$ Also what does it mean when it says the union of any chain of totally ordered sets is totally ordered? Does it mean a union of two chains of totally ordered sets (e.g. union of $\{ \{\phi_1, \phi_2\}\}, \{\{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}\}$ and $\{\{\psi_1, \psi_2\} \}, \{\{\psi_1, \psi_2\}, \{\psi_1, \psi_2, \psi_3\} \}$)? Or is it a union of all the members of the chain of totally ordered sets? I took it as the latter since the former doesn't makes sense.

However, I don't understand why it is necessary to define it in such a way. For instance, $\{\{\phi_1, \phi_2, \phi_3, \dots\}\}$ is totally ordered since the union of the chain $\{ \{\phi_1, \phi_2\}\}, \{\{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}\}, \{\{\phi_1, \phi_2\}, \{\phi_1, \phi_2, \phi_3\}, ...\}$ is $\{\{\phi_1, \phi_2, \phi_3, \dots\}\}$. However, isn't $\{\{\phi_1, \phi_2, \phi_3, \dots\}\}$ always totally ordered since it is a subset of a single element?

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  • $\begingroup$ No, a chain of subsets of P is not a totally ordered subset of P. $\endgroup$ Jun 29 '21 at 2:43
  • $\begingroup$ isn't chain, by definition, totally ordered? or is that a subchain? $\endgroup$
    – MoneyBall
    Jun 29 '21 at 3:11
  • $\begingroup$ They are synonyms. $\endgroup$ Jun 29 '21 at 8:55
  • $\begingroup$ some small comments. (1) Your $P$ is missing $\{\phi_1\}$ and $\{\psi_1\}$. (2) Your $\mathcal F$ is missing a lot of elements, see my answer below (in fact $\mathcal F$ is uncountable, and thus cannot be written down with "..."). (3) You don't take the union of two chains, but the union of a single chain that consists of totally ordered sets. (4) In your last paragraph, the union is $\{\{\phi_1,\phi_2\},\{\phi_1,\phi_2,\phi_3\},\dots\}$, not $\{\{\phi_1,\phi_2,\dots\}\}$ (the latter would be like taking the union of the union, and then putting the result in a singleton set). $\endgroup$
    – Vsotvep
    Jun 29 '21 at 13:11
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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 $\bigcup_{i\in I}X_i$.

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  • $\begingroup$ Thank you for the detailed answer, took me a while but now I think I understand it! $\endgroup$
    – MoneyBall
    Jul 1 '21 at 4:45

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