Axiom of Choice Implies Zorn's Lemma

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Theorem

Acceptance of the Axiom of Choice implies the truth of Zorn's Lemma.


Statement of Zorn's Lemma

Let $\left({X, \preceq}\right), X \ne \varnothing$ be a non-empty ordered set such that every non-empty chain in $X$ has an upper bound in $X$.


Then $X$ has at least one maximal element.


Proof 1

For each $x \in X$, consider the lower closure $x^\preceq$:

$x^\preceq = \left\{{y \in X: y \preceq x}\right\}$

Let $\mathbb S \subseteq \mathcal P \left({X}\right)$ be the image of $\cdot^\preceq$ considered as a mapping from $X$ to $\mathcal P \left({X}\right)$, where $\mathcal P \left({X}\right)$ is the power set of $X$.

From Ordering Equivalent to Subset Relation:

$\forall x, y \in X: x^\preceq \subseteq y^\preceq \iff x \preceq y$

Thus the task of finding a maximal element of $X$ is equivalent to finding a maximal set in $\mathbb S$.


Thus the statement of the result is equivalent to a statement about chains in $\mathbb S$:

Let $\mathbb S$ be a non-empty subset of $\mathcal P \left({X}\right), X \ne \varnothing$ such that every non-empty chain in $\mathbb S$, ordered by $\subseteq$, has an upper bound in $\mathbb S$.
Then $\mathbb S$ has at least one maximal set.


Let $\mathbb X$ be the set of all chains in $\left({X, \preceq}\right)$.

Every element of $X$ is included in $s^\preceq \left({x}\right)$ for some $x \in X$.

$\mathbb X$ is a non-empty set of sets which are ordered (perhaps partially) by subset.

If $\mathcal C$ is a chain in $\mathbb X$, then:

$\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathbb X$

Since each set in $\mathbb X$ is dominated by some set in $\mathbb S$, going from $\mathbb S$ to $\mathbb X$ can not introduce any new maximal elements.


The main advantage of using $\mathbb X$ is that the chain hypothesis is in a slightly more specific form.

Instead of saying that each chain in $\mathcal C$ has an upper bound in $\mathbb S$, we can explicitly state that the union of the sets of $\mathcal C$ is an element of $\mathbb X$.

This union of the sets of $\mathcal C$ is clearly an upper bound of $\mathcal C$.

Another advantage of $\mathbb X$ is that, from Subset of Toset is Toset, it contains all the subsets of each of its sets.

Thus we can embiggen non-maximal sets in $\mathbb X$ one element at a time.


So, from now on, we need consider only this non-empty collection $\mathbb X$ of subsets of a non-empty set $X$.

$\mathbb X$ is subject to two conditions:

$(1): \quad$ Every subset of each set in $\mathbb X$ is in $\mathbb X$.
$(2): \quad$ The union of each chain of sets in $\mathbb X$ is in $\mathbb X$.

It follows from $(1)$ that $\varnothing \in \mathbb X$.

We need to show that there exists a maximal set in $\mathbb X$.


Let $f$ be a choice function for $X$:

$\forall A \in \mathcal P \left({X}\right) \setminus \varnothing: f \left({A}\right) \in A$

For each $A \in \mathbb X$, let $\hat A$ be defined as:

$\hat A := \left\{{x \in X: A \cup \left\{{x}\right\} \in \mathbb X}\right\}$

That is, $\hat A$ consists of all the elements of $X$ which, when added to $A$, make a set which is also in $\mathbb X$.


From its definition:

$\displaystyle \hat A = \bigcup_{x \mathop \in \hat A} \left({A \cup \left\{{x}\right\}}\right)$

where each of $A \cup \left\{{x}\right\}$ are chains in $X$ and so elements of $\mathbb X$.

Suppose there exists a maximal set $M$ in $\mathbb X$.

Then, by definition of maximal, there are no elements in $x \in X \setminus M$ which can be added to $M$ to make $M \cup \left\{{x}\right\}$ another element of $\mathbb X$.

Thus it follows that $\hat M = M$.

From Set Difference with Self is Empty Set it then follows that if $A$ is maximal, then $\hat A \setminus A = \varnothing$.


The mapping $g: \mathbb X \to \mathbb X$ can now be defined as:

$\forall A \in \mathbb X: g \left({A}\right) = \begin{cases} A \cup \left\{{f \left({\hat A \setminus A}\right)}\right\} & : \hat A \setminus A \ne \varnothing \\ A & : \text{otherwise} \end{cases}$

Thus what we now have to prove is that:

$\exists A \in \mathbb X: g \left({A}\right) = A$


Note that from the definition of $g$:

$\forall A \in \mathbb X: A \subseteq g \left({A}\right)$

Also note that $f \left({\hat A \setminus A}\right)$ is a single element of $\hat A \setminus A$.

Thus we obtain the crucial fact that $g \left({A}\right)$ contains at most one more element than $A$.


We (temporarily) define a tower as being a subset $\mathcal T$ of $\mathbb X$ such that:

$(1): \quad \varnothing \in \mathcal T$
$(2): \quad A \in \mathcal T \implies g \left({A}\right) \in \mathcal T$
$(3): \quad $ If $\mathcal C$ is a chain in $\mathcal T$, then $\displaystyle \bigcup_{A \mathop \in \mathcal C} A \in \mathcal T$


There is of course at least one tower in $\mathbb X$, as $\mathbb X$ itself is one.

It follows from its definition that the intersection of a collection of towers is itself a tower.

It follows in particular that if $\mathcal T_0$ is the intersection of all towers in $\mathbb X$, then $\mathcal T_0$ is the smallest tower in $\mathbb X$.

Next we demonstrate that $\mathcal T_0$ is a chain.


We (temporarily) define a set $C \in \mathcal T_0$ as comparable if it is comparable with every element of $\mathcal T_0$.

That is, if $A \in \mathcal T_0$ then $C \subseteq A$ or $A \subseteq C$.

To say that $\mathcal T_0$ is a chain means that all sets of $\mathcal T_0$ are comparable.

There is at least one comparable set in $\mathcal T_0$, as $\varnothing$ is one of them.


So, suppose $C \in \mathcal T_0$ is comparable.

Let $A \in \mathcal T_0$ such that $A \subsetneq C$.

Consider $g \left({A}\right)$.

Because $C$ is comparable, either $C \subsetneq g \left({A}\right)$ or $g \left({A}\right) \subseteq C$.

In the former case $A$ is a proper subset of a proper subset of $g \left({A}\right)$.

This contradicts the fact that $g \left({A}\right) \setminus A$ can be no more than a singleton.

Thus if such an $A$ exists, we have that:

$(A): \quad g \left({A}\right) \subseteq C$.


Now let $\mathcal U$ be the set defined as:

$\mathcal U := \left\{{A \in \mathcal T_0: A \subseteq C \lor g \left({C}\right) \subseteq A}\right\}$

Let $\mathcal U'$ be the set defined as:

$\mathcal U' := \left\{{A \in \mathcal T_0: A \subseteq g \left({C}\right) \lor g \left({C}\right) \subseteq A}\right\}$

That is, $\mathcal U'$ is the set of all sets in $\mathcal T_0$ which are comparable with $g \left({C}\right)$.

If $A \in \mathcal U$, then as $C \subseteq g \left({C}\right)$, either $A \subseteq g \left({C}\right) \lor g \left({C}\right) \subseteq A$

So $\mathcal U \subseteq \mathcal U'$.


The aim now is to demonstrate that $\mathcal U$ is a tower.

From Empty Set is Subset of All Sets, $\varnothing \subseteq C$.

Hence condition $(1)$ is satisfied.


Now let $A \in \mathcal U$.

As $C$ is comparable, there are three possibilities:

$(1'): \quad A \subsetneq C$

Then from $(A)$ above, $g \left({A}\right) \subseteq C$.

Therefore $g \left({A}\right) \in \mathcal U$.

$(2'): \quad A = C$

Then $g \left({A}\right) = g \left({C}\right)$ and so $g \left({C}\right) \subseteq g \left({A}\right)$.

Therefore $g \left({A}\right) \in \mathcal U$.

$(3'): \quad g \left({C}\right) \subseteq A$

Then $g \left({C}\right) \subseteq g \left({A}\right)$

Therefore $g \left({A}\right) \in \mathcal U$.

Hence condition $(2)$ is satisfied.


From the definition of $\mathcal U$, it follows immediately that the union of a chain in $\mathcal U$ is also in $\mathcal U$.

Hence condition $(3)$ is satisfied.


The conclusion is that $\mathcal U$ is a tower such that $\mathcal U \subseteq \mathcal T_0$.

But as $\mathcal T_0$ is the smallest tower, $\mathcal T_0 \subseteq \mathcal U$.

It follows that $\mathcal U = \mathcal T_0$.


Consider some comparable set $C$, then.

From that $C$ we can form $\mathcal U$, as above.

But as $\mathcal U = \mathcal T_0$:

$A \in \mathcal T_0 \implies \left({A \subseteq C \implies A \subseteq g \left({C}\right)}\right) \lor g \left({C}\right) \subseteq A$

and so $g \left({C}\right)$ is also comparable.


We now know that:

$\varnothing$ is comparable
the mapping $g$ maps comparable sets to comparable sets.

Since the union of a chain of comparable sets is itself comparable, it follows that the comparable sets all form a tower $\mathcal T_C$.

But by the nature of $\mathcal T_0$ it follows that $\mathcal T_0 \subseteq \mathcal T_C$.

So the elements of $\mathcal T_0$ must all be comparable.


Since $\mathcal T_0$ is a chain, the union $M$ of all the sets in $\mathcal T_0$ is itself a set in $\mathcal T_0$.

Since the union includes all the sets of $\mathcal T_0$, it follows that $g \left({M}\right) \subseteq M$.

Since it is always the case that $M \subseteq g \left({M}\right)$, it follows that $M = g \left({M}\right)$.

The result follows.

$\blacksquare$


Proof 2

Suppose for the sake of contradiction that for each $x \in X$ there is a $y \in X$ such that $x \prec y$.

By the Axiom of Choice, there is a mapping $f: X \to X$ such that:

$\forall x \in X: x \prec f \left({x}\right)$

Let $\mathcal C$ be the set of all chains in $X$.

By the premise, each element of $\mathcal C$ has an upper bound in $X$.

Thus by the Axiom of Choice, there is a mapping $g: \mathcal C \to X$ such that for each $C \in \mathcal C$, $g \left({C}\right)$ is an upper bound of $C$.

Let $p$ be an arbitrary element of $X$.

Define a mapping $h: \operatorname{Ord} \to X$ by transfinite recursion thus:

\(\displaystyle h \left({0}\right)\) \(=\) \(\displaystyle p\) $\quad$ $\quad$
\(\displaystyle h \left({\alpha^+}\right)\) \(=\) \(\displaystyle f \left({h \left({\alpha}\right)}\right)\) $\quad$ $\quad$
\(\displaystyle h \left({\lambda}\right)\) \(=\) \(\displaystyle f \left({g \left({f \left[{\lambda}\right]}\right)}\right)\) $\quad$ if $\lambda$ is a limit ordinal $\quad$

Then $h$ is strictly increasing, and thus injective.


Let $h'$ be the restriction of $h$ to $\operatorname{On} \times h \left({\operatorname{On} }\right)$.

Then ${h'}^{-1}$ is a surjection from $h \left({\operatorname{On} }\right) \subseteq X$ onto $\operatorname{On}$.

By the Axiom of Replacement, $\operatorname{On}$ is a set.

By Burali-Forti Paradox, this is a contradiction.

Thus we conclude that some element of $X$ has no strict successor, and is thus maximal.

$\blacksquare$