Zorn's Lemma implies Kuratowski's Lemma

Theorem
Let Zorn's Lemma be accepted as true.

Then Kuratowski's Lemma holds.

Proof
Recall Zorn's Lemma:

Recall Kuratowski's Lemma:

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 $C$.