# Union of Sets in Simply-Ordered Collection

This question about Kuratowki's Lemma appears in Munkres' Topology, section 11:

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 $\mathcal{A}$.

I'm racking my brain to figure out when the hypothesis is not true. If you have such a simply-ordered subcollection $\mathcal{B}$, wouldn't the union of its elements necessarily belong to $\mathcal{B}$ and therefore $\mathcal{A}$? My basic thought process is if $B_0$, $B_1 \in \mathcal{A}$, then $B_0 \prec B_1$ or $B_1 \prec B_0$, so every element of $\mathcal{B}$ is either a subset or superset of every other element in $\mathcal{B}$. As a result, there is some $B_m \in \mathcal{B}$ where $B_m = \cup_{X \in \mathcal{B}} X$. I feel like I'm missing something. An example of where this proposition isn't true would be helpful.

Many thanks!

• If $A$ is a non-empty collection of sets. – DanielWainfleet Apr 15 '17 at 13:27

Let $\mathcal A$ be the collection of all finite subsets of $\mathbb N$ and consider the subcollection $\mathcal B=\{B_1,B_2,B_3\dots\}$ where $B_n=\{1,2,3,\dots,n\}.$ Then $\mathcal B$ is simply ordered by inclusion, and the union of the elements of $\mathcal B$ is the infinite set $\{1,2,3,\dots\}$ which does not belong to $\mathcal A.$

• You mean $\cup B\not\in A.$ The OP did ask for an example where the hypothesis is not true, and this works for that. – DanielWainfleet Apr 15 '17 at 14:05
• Ah, I see, a perfect example. Thanks a bunch! – Doug Apr 15 '17 at 19:57
• @user254665 Thanks for catching the typo. – bof Apr 15 '17 at 20:25

Zorn's Lemma: If $R$ is an irreflexive transitive binary relation on $A$ and every $R$-chain has an upper bound then for any $s\in A$ there is an $R$-maximal $t\in A$ with $s=t$ or $sRt.$

An $R$-chain is a set $T\subset A$ such that when $s,t\in T$ and $s\ne t$ then $sRt$ or $tRs.$ An $R$-maximal $t\in S$ satisfies $\neg (tRs)$ for all $s\in A.$

Zorn's Lemma is equivalent to the Axiom of Choice (AC).

Let $A\ne \phi$ be as in Kuratowski's Lemma. For $s,t \in A$ let $sRt\iff s\subsetneqq t.$

If $V$ is a vector space and $A$ is the set of linearly independent subets of $V$, Kuratowski's Lemma implies that $V$ has a maximal linearly independent subset, which would be a vector space basis (Hamel basis) for $V$. It has been shown that without AC it is consistent that some vector space doesn't have a basis. So we do need Zorn in your Q.