Exercise 11.5

Question

Show that Zorn's Lemma implies the following:

\emph{Lemma (Kuratowski).} Let $\mathcal{A}$ be a collection of sets.
Suppose that for every subcollection $\mathcal{B}$ of $\mathcal{A}$ that is simply ordered by proper inclusion, the union of the elements of $\mathcal{B}$ belongs to $\mathcal{A}$.
Then $\mathcal{A}$ has an element that is properly contained in no other element of $\mathcal{A}$.

Dan Whitman · Accepted Answer

\begin{proof}
First, we know that $\subsetneq$ is a strict partial order on $\mathcal{A}$, which is trivial to show.
So consider any simply ordered subset $\mathcal{B}$ of $\mathcal{A}$ and let $A = \bigcup \mathcal{B}$ so that we know that $A \in \mathcal{A}$.
Clearly for any set $B \in \mathcal{B}$ we have that $B \subset \bigcup \mathcal{B} = A$ so that, since $B$ was arbitrary, $A$ is an upper bound of $\mathcal{B}$ in the strict partial order $\subsetneq$.
Since $\mathcal{B}$ was arbitrary, this shows the hypothesis of Zorn's Lemma so that $\mathcal{A}$ has a maximal element $A$.
Then clearly $A$ is not properly contained in any other element of $\mathcal{A}$.
\end{proof}

Exercise 11.5

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Exercise 11.5

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