Zorn's Lemma And Axiom of Choice

Question

How can I prove Zorn's lemma is equivalent to Axiom of choice?

Answer 1

Assume the axiom of choice. Let $(P,\le)$ a partially ordered set that every chain has an upper bound. Let $f$ to be a choice function from all non-empty subsets of $P$, and let $P_a = \{x\in P\mid a<x\}$ then $P_a=\varnothing$ if and only if $a$ is a maximal element.

We define by using transfinite induction:

Let $a_0$ be an element of $P$, if it is maximal then we have finished. Otherwise $P_0 = \{x\in P\mid a_0<x\}$ is non-empty, let $a_1 = f(P_0)$.

Suppose $a_\alpha$ was defined, if it is maximal then we are done, otherwise $P_\alpha=\{x\in P\mid a_\alpha<x\}$ is non-empty and $a_{\alpha+1} = f(P_\alpha)$.

If $\alpha$ is a limit ordinal, and for all $\beta<\alpha$ we have chosen $a_\beta$, then $\{a_\beta\mid \beta<\alpha\}$ is a chain without a maximal element in $P$, and therefore bounded with an upper bound above all the elements of the chain, thus $\bigcap_{\beta<\alpha} P_\beta\neq\varnothing$, and let $a_\alpha=f(\bigcap_{\beta<\alpha} P_\beta)$.

We argue that this must stop at some point, otherwise we have an injection from the proper class of the ordinals into the set $P$. Why did the process stop? It can only stop if we have reached a maximal element at some $a_\gamma$, as wanted.

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Assume Zorn's lemma, and let $X$ be a non-empty collection of non-empty sets. Let $(C,\subseteq)$ be the set of all choice functions from some elements of $X$.

This is a non-empty set, since we can always choose from finitely many sets. Given a chain of choice functions, the union is indeed a function. So the condition of Zorn's lemma is satisfied. We have a maximal element $f$.

If $f$ is not a choice function from all the members of $X$ then we can extend it to one more element, which contradicts the maximality.

Answer 2

Even though this is a standard result, which can be found elsewhere (probably in most standard textbooks and lecture notes on set theory); I think some brief discussion and giving some pointers could be useful.

Perhaps the first important thing is to stress that when you want show that something is equivalent to Axiom of Choice, you have to work in an axiomatic systems where AC is not an axiom. I would say ZF is used most frequently.

The proof of ZL $\Rightarrow$ AC is similar to most applications of Zorn's lemma. We want to show that there exists a selector (choice function). This selector is obtained from ZL as maximal partial selector. So the only thing is to show that union of a chain (w.r.t. inclusion) of partial functions, which select one element from each set is again such a function.

The proof of AC $\Rightarrow$ ZL which I like best relies on transfinite induction. We assume that there is a set which fulfills the assumptions of ZL and has no maximal elements. In this way we inductively construct a chain with no upper bound. The use of AC is to select some element from the non-empty set of upper bounds of already selected elements in each step. This kind of proof assumes that we already know something about ordinals and transfinite induction; but I think that this technique is very useful in general.

However, for various reasons, some authors prefer to avoid using transfinite induction. As an example I can mention the papers of Weston and Lewin bellow. As I mentioned, the proof can be found in almost all standard textbooks. I've included Devlin and Halmos, to give at least one example of a book which uses ordinals in this proof and one example of a book which avoids them.

Some references:

• J. D. Weston: A short proof of Zorn's Lemma, Archiv der Mathematik, Volume 8, Number 4, 279, DOI: 10.1007/BF01898788. One page long proof of AC $\Rightarrow$ ZL without using transfinite induction.

• Lewin, J. (1991). A simple proof of Zorn's lemma. The American Mathematical Monthly, 98(4), pp. 353-354. jstor Another short proof of AC $\Rightarrow$ ZL avoiding the use of ordinals.

• K. Devlin: Joy of Sets. Theorem 2.7.5 on p.60 gives the proof of AC $\Rightarrow$ ZL using transfinite induction. The opposite implications is given as a series of implications in theorems following this one.

• P. Halmos: Naive Set Theory, p.62. Halmos chooses approach avoiding the transfinite induction.

Answer 3

$\newcommand{\P}{\mathscr{P}}\newcommand{\C}{\mathscr{C}}\newcommand{\A}{\mathcal{A}}$I am posting this just to flesh out some details in the proof in Bergman's handout. The motivation for this proof is to capture the concept of a transfinite sequence without doing any explicit work with transfinite stuff. I still don't really know about that, so this proof appealed to me, but for my own sake I felt it needed simplification: this posting is as much for my own benefit as I hope it might be for any future readers.

Note that I allow, for convenience, "chains" in a poset to be empty. Similarly I allow "well ordered" sets to be empty and I allow "initial segments" to be empty. Otherwise, these terms carry their standard meanings below.

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Zorn's lemma:

If a poset $(\P,\prec)$ has the property that all chains $C\subseteq\P$ have an upper bound, then $\P$ has a maximal element.

Note taking $C=\emptyset$ implies $\P$ is at least nonempty.

Proof:

For the sake of contradiction, suppose no maximal element exists.

Fix a choice function $\phi$ on $2^{\P}\setminus\{\emptyset\}$. For any chain $C\subseteq\P$, the set $U(C):=\{x\in\P:\forall y\in C,\,y\prec x\}$ is nonempty. For there is an upper bound $m$ to $C$, and since $m$ is not maximal there is $m'\in\P$ with $m\prec m'$. Then $m'\in U(C)$.

If $\C$ denotes the set of all chains in $\P$, we have a valid function $\upsilon:\C\to\P$ taking $C\mapsto\phi(U(C))$. This will do the magic picking of larger elements for us.

Say $A\subseteq\P$ is an "attempt" if it is well ordered and if for all proper subchains $C$ of $A$, $\min A\setminus C=\upsilon(C)$. The motivation for this is explained better in the linked handout, but think of these "attempts" as transfinite sequences in disguise which attempt to reach a maximal element.

Suppose there are two attempts $A,A'$ with neither contained in the other; then define $\zeta=\min A\setminus A'$, $\zeta’=\min A’\setminus A$. With no loss of generality say $\zeta’$ is not less than $\zeta$. Consider $C:=\{x\in A:x\prec\zeta\}$: it is contained in both $A$ and $A’$ so by hypothesis it cannot equal either of them. $C$ shall then a proper initial segment of both $A$ and $A’$, using the fact that no element of $A’\setminus A$ can be less than $\zeta$.

But then we have: $$A\setminus A’\ni\zeta=\min A\setminus C=\upsilon(C)=\min A'\setminus C\in A'$$Which is absurd. Therefore for any two attempts $A,A'$, it is true that $A\subseteq A'$ or visa versa. With the same notation, in the situation $A’\subseteq A$ we find $C=A’$ is forced: $C$ is an initial segment of both $A,A’$ so the event that it is properly contained in $A’$ is impossible by the same reasoning.

We get to conclude that for any pair of attempts, one will be an initial segment of the other.

With this important fact proven, we can now prove that the union of any set $\A$ of attempts is an attempt:

Let $S\subseteq\bigcup\A$ be nonempty. For some $A_0\in\A$, $S\cap A_0$ is nonempty and has a minimum $\varsigma$ in $A_0$. If $A\in\A\setminus\{A_0\}$ and $A\subseteq A_0$, then $\varsigma$ remains a minimum for $S\cap A$ trivially; else, $A_0$ is a proper initial segment of $A$, so there cannot exist $x\in A\setminus A_0$ with $x\prec\varsigma$ and $\varsigma$ is still a minimum for $S\cap A$. Hence $\min S=\varsigma$ exists and $\bigcup\A$ is well ordered.

If $C$ is a proper initial segment of $\bigcup\A$ then with $S:=\bigcup\A\setminus C$, we know $\min S=\min S\cap A_0$ for some (any) $A_0\in\A$ satisfying $S\cap A_0\neq\emptyset$. There is $y\in A_0\setminus C$, necessarily. Suppose that $C$ is not contained in $A_0$, that there is also $x\in C\setminus A_0$. Then $x\in A$ for some attempt $A\in\A$ which must contain $A_0$ and $C':=C\cap A$ and $A_0$ are both initial segments of $A$.

Because $x\in C'\setminus A_0$, $y\in A_0\setminus C'$, the initial segment property allows both cases $x\prec y$ and $y\prec x$ to be ruled out - but since $A$ is well ordered, one of these must occur. This contradiction implies that $C$ is contained in $A_0$, hence it is a proper initial segment of $A_0$. Then we have: $$\min\bigcup\A\setminus C=\min S\cap A_0=\min A_0\setminus C=\upsilon(C)$$As desired.

Finally, take $\A$ to be the set of all attempts in $\P$: by the above, $\alpha:=\bigcup\A$ is an attempt and indeed the largest possible attempt. In particular $\alpha$ is also a chain, so $\upsilon(\alpha)$ exists (this is the whole reason for the song-and-dance about “attempts”, is so that the hypothetical largest [thing] is actually a chain and we can derive a contradiction). Since $\upsilon(\alpha)$ is a strict upper bound for $\alpha$, it is clear $\alpha\cup\{\upsilon(\alpha)\}$ (this is where the usual transfinite "add another element to the sequence" principle appears) is a strictly larger attempt than $\alpha$ - a contradiction!

$\blacksquare$