Zorn's Lemma implies Kuratowski's Lemma

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Theorem

Let Zorn's Lemma be accepted as true.

Then Kuratowski's Lemma holds.


Proof

Recall Zorn's Lemma:

Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set such that every non-empty chain in $S$ has an upper bound in $S$.

Then $S$ has at least one maximal element.

$\Box$


Recall Kuratowski's Lemma:

Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set.

Then every chain in $S$ is the subset of some maximal chain.

$\Box$


Let $\struct {S, \preceq}$ be a non-empty ordered set.

Let $C$ be a chain in $\struct {S, \preceq}$.

Let $T$ be the set of all chains in $\struct {S, \preceq}$ that are supersets of $C$.


Let $\CC$ be a chain in the power set $\powerset T$ of $T$ under the ordering that is the subset relation.

Define $C' = \bigcup \CC$ be the union of $\CC$.

We have that the elements of $T$ are chains on $\struct {S, \preceq}$.

We have that $\CC$ is a subset of $T$.

Hence the elements of $\CC$ are also chains in $\struct {S, \preceq}$.

Thus $\bigcup \CC$ contains elements in $\struct {S, \preceq}$.

Hence:

$C' \subseteq S$

We have that $C'$ is a chain in $\struct {S, \preceq}$.

Let $x, y \in C'$.

Then $x \in X$ and $y \in Y$ for some $X, Y \in \CC$.

As $\CC$ is a chain in $\powerset T$:

either $X \subseteq Y$ or $Y \subseteq X$.

So $x$ and $y$ belong to the same chain in $\struct {S, \preceq}$.

Thus either $x \preceq y$ or $y \preceq x$.

Thus $C'$ is a chain on $\struct {S, \preceq}$.

Now let $x \in C$.

Then:

$\forall A \in T: x \in A$

Then because $\CC \subseteq T$:

$\forall A \in \CC: x \in A$

So:

$x \in \bigcup \CC$

and so:;

$C \subseteq C'$

Thus:

$C' \in T$


We now show that $C'$ is an upper bound on $\CC$.

Indeed, consider $x \in D \in \CC$.

This means:

$x \in \bigcup \CC = C'$

so:

$D\subseteq C'$

Thus $C'$ is an upper bound on $\CC$.


The chain in $T$ was arbitrary.

Hence every chain in $T$ has an upper bound.

Thus, by Zorn's Lemma, $T$ has a maximal element.

This must be a maximal chain containing $C$.

$\blacksquare$


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