Show $B=\bigcup \mathcal{A}$ is well-ordered and $A\leq B, \forall A\in \mathcal{A}$

Question

The relation $\leq$ here means, $A\leq B$ iff \begin{align*} &\bullet A\subseteq B\\ &\bullet \forall x\in A, \forall y\in B, y\leq x \Rightarrow y\in A \end{align*}

Let $X$ be a partially ordered set and $\mathcal{A}$ a set of well-ordered subsets of $X$ where for all $A_1, A_2\in \mathcal{A}, A_1\leq A_2$ or $A_2\leq A_1$. Show that $B=\bigcup \mathcal{A}$ is well-ordered and $\forall A\in \mathcal{A},A\leq B$.

My first doubt comes from the definition of the set $B$, which I suppose means $\bigcup_{A\in \mathcal{A}}A$. Using this $B$, is this argument enough:

Each $A_i$ is well ordered, since $A_i\leq A_j, \forall i,j$ and $i\neq j$ then $A_i\cap A_j\neq \emptyset$ for $i\neq j$. Now every element of $A_i$ is comparable pair-wise, because every set $A_i$ has a non-empty intersection, there exists a well-order over $\bigcap_{A\in \mathcal{A}} A$.

I haven't reached the result, what am I doing wrong? Do I have to use Zorn's Lemma or even the Axiom of Choice directly?

Answer 1

We have the lemma:

\begin{equation*}\text{For all elements $A$ of $\mathcal{A}$ and $x$ of $A$ and $y$ of $B$ such that $y < x$, $y \in A$.}\end{equation*}

For, since $y \in B = \bigcup \mathcal{A}$, for some element $A'$ of $\mathcal{A}$, $y \in A'$. Then, by hypothesis, either $A' \leq A$ or $A \leq A'$. But, if $A' \leq A$, then, by the first condition for "$\leq$", $y \in A$, and, if $A \leq A'$, then, by the second condition for "$\leq$", $y \in A$.

Now let $S$ be a non-empty subset of $B$. Since $S$ is non-empty, it contains some object $s$. Then, since $S \subset B$, $s \in B = \bigcup \mathcal{A}$, so, for some element $A$ of $\mathcal{A}$, $s \in A$, so, since $s \in S$, $s \in S \cap A$, so $S \cap A$ is non-empty, so, since $A$ is well ordered, $S \cap A$ has a least element $l$.

Now let $x$ be an element of $X$ such that $x < l$. Now either $x \in A$ or $x \in' A$. But, since $x < l$, $x \in' S \cap A$, so, if $x \in A$, then $x \in' S$; and, if $x \in' A$, then, since $x < l$ and $l \in A$, by the lemma, $x \in' B$, so $x \in' S$. So no element $x$ of $X$ is both less than $l$ and contained in $S$, so $l$ is a least element of $S$, so $S$ has a least element.

Since that holds for every non-empty subset $S$ of $B$, $B$ is well ordered.

Answer 2

Consider a subset $\emptyset\neq S\subseteq B$ and take $s\in S$. Select $A_i$ such that $s\in A_i$ (such $A_i$ exists of course, by definition of $B$). Then the set $$\{t\in S:t\leqslant s\}$$ is contained in $A_i$. This because if $t\in A_j\succeq A_i$ (I use $\preceq$ and $\succeq$ for the ordering you've defined on the subsets of $X$, so there is no confusion with the order on the elements of $X$), then $t\in A_i$ since $t\leqslant s\in A_i$ (here I apply the definition of $\preceq$). On the other hand, if $t\in A_j\preceq A_i$, then $t\in A_j\subseteq A_i$ (again, by definition of $\preceq$). These are the only cases because of the hypothesis on $\mathcal A$ (that either $C\preceq D$ or $D\preceq C$, for all $C,D\in\mathcal A$)

So, being $A_i$ well-ordered, there is a minimum $m\in A_i$ to $J=\{t\in S:t\leqslant s\}$. Can you prove that $m$ is the minimum of $S$?

Hint: $B$ is totally ordered, since every two $x,y\in B$ belong to some $A_i$, which is totally ordered (any well-ordered set is also totally ordered), so for every $q\in S$, either $s\leqslant q\rightarrow m\leqslant s\leqslant q$, or $q\leqslant s\rightarrow q\in J\rightarrow m\leqslant q$