# nLab Zorn's lemma

foundations

## Foundational axioms

foundational axiom

# Zorn's Lemma

## Idea

Zorn's lemma is a proposition in set theory and order theory. It says that: A preorder in which every sub-total order has an upper bound has a maximal element.

Depending on the chosen perspective on foundations this is either an axiom or a consequence of the axiom of choice; or neither if neither the axiom of choice nor Zorn’s lemma is assumed as an axiom. Conversely, Zorn's lemma and the axiom of excluded middle together imply the axiom of choice. Although it doesn't imply excluded middle itself, Zorn's lemma is not generally accepted in constructive mathematics as an axiom.

(Say something about how one can systematically get around this in many situations ….)

In as far as it holds or is assumed to hold, Zorn’s lemma is a standard method for constructing (or at least, proving the existence of) objects that are ‘maximal’ in an appropriate sense. It is commonly used in algebra and named after the algebraist Max Zorn (although he himself protested the naming after him).

## Statement and proofs

###### Definition
• Given a preordered set $(S, \leq)$, an element $x$ of $S$ is maximal if $y \leq x$ whenever $x \leq y$.

• A chain in $S$ is a subset $A \hookrightarrow S$ such that $\leq$ is a total order on $A$.

• An upper bound of a subset $A$ of $S$ is an element $x$ of $S$ such that $y \leq x$ whenever $y \in A$.

###### Definition

The proposition Zorn's Lemma states that each preordered set $S$ has a maximal element if every chain in $S$ has an upper bound.

###### Theorem

Suppose the axiom of choice holds. Then Zorn's Lemma holds.

###### Proof

By contradiction. (We freely use the axiom of excluded middle, which in any case follows from AC, the axiom of choice. The one use made of AC is noted below. WLOG we prove the result for posets.)

Suppose $S$ has no maximal element. Then every chain $C \subseteq S$ has a strict upper bound: an element $x$ such that $c \lt x$ for all $c \in C$. (Take an upper bound $y$ of $C$, then $y$ is not maximal by supposition, so any $x$ with $y \lt x$ will serve.) For each $W \subseteq S$ that is well-ordered by $\leq$, use AC to choose a strict upper bound $f(W)$ of $W$. Given a such a function $f: Well(S) \to S$ from well-ordered subsets of $S$ to $S$, let us say $W \in Well(S)$ is $f$-inductive if for every $x \in W$, we have $x = f(\{y \in W: y \lt x\})$.

Let $Y, Z \in Well(S)$ be $f$-inductive sets; we will show one is an initial segment of the other. Let $I$ be the union of all subsets of $S$ that are initial segments of both $Y$ and $Z$; then $I$ is maximal among such initial segments of both. Suppose $I$ is properly contained in both; let $y$ be the least element of $Y \setminus I$, and $z$ the least element of $Z \setminus I$. Then $\{x \in Y: x \lt y\} = I = \{x' \in Z: x' \lt z\}$, so by $f$-inductivity of $Y, Z$, we have $y = f(I) = z$. Thus $I$ extends to an initial segment $I \cup \{y\} = I \cup \{z\}$, contradicting maximality of $I$. Therefore we must have $I = Y$ or $I = Z$, so one of $Y, Z$ is an initial segment of the other.

So the collection of $f$-inductive sets is totally ordered by inclusion of initial segments, making their union $U$ the maximal $f$-inductive set (Lemma 1 below). Appending its strict upper bound $f(U)$, the chain $U \cup \{f(U)\}$ is a larger $f$-inductive set, contradiction. The proof is complete.

###### Remark

The empty set (with its unique structure as preordered set) is not an exception to Zorn's lemma, as the chain $\emptyset$ does not have an upper bound.

###### Remark

The proof above (does anyone know where it appears in the literature?) can be seen as an application of general recursion theory à la Paul Taylor’s book, which in turn is inspired in large part by Osius. In particular, initial segments are the same as $P$-coalgebra maps between well-founded coalgebras, and the maximal $U$ constructed is a maximal attempt with respect to a partial algebra structure $f: P(S) \rightharpoonup S$, following Taylor’s formulation of the recursion scheme. For a slightly different arrangement of the facts, one very commonly found in the literature, see the account of the Bourbaki-Witt theorem below.

###### Lemma

Let $P_\alpha$ be a collection of subsets of a set $S$, each equipped with a well-ordering $\leq_\alpha$, such that for any $\alpha, \beta$, one of $P_\alpha, P_\beta$ (each with their orderings) is an initial segment of the other. Let $P$ be the union $\cup_\alpha P_\alpha$, equipped with the ordering $x \leq y$ if $x \leq_\alpha y$ in some $P_\alpha$ containing them both. Then $P$ is well-ordered by $\leq$, with each $P_\alpha$ an initial segment of $P$.

###### Proof

Observe that $\leq$ is well-defined, and is a total order: if $x, y \in P$, then $x \in P_\alpha$ and $y \in P_\beta$ for some $\alpha, \beta$, where one of $P_\alpha, P_\beta$ contains the other, say $P_\alpha \subseteq P_\beta$, whence $x, y$ are comparable in $P_\beta$. If $T$ is an inhabited subset of $P$, then $T \cap P_\alpha$ is inhabited for some $\alpha$, so there is a minimal element $t \in T \cap P_\alpha$ with respect to $\leq_\alpha$. This $t$ is minimal in $T$ with respect to $\leq$, for if $s \in T$, $s \leq t$, then $s \in P_\alpha$ by the initial segment condition, and then $t \leq s$ by definition of $t$.

###### Theorem

If Zorn's Lemma and the principle of excluded middle hold, then so do the well-ordering principle and the axiom of choice.

###### Proof

We first show that Zorn’s lemma implies the classical well-ordering principle. The axiom of choice easily follows from the well-ordering principle.

1. Let $A$ be a set, and consider the poset whose elements are pairs $(X, R)$, where $X \in P A$ and $R$ is a (classical) well-ordering on $X$, ordered by $(X, R) \leq (Y, S)$ if $X$ as a well-ordered set is an initial segment of $Y$. If $C$ is a chain in the poset consisting of subsets $X_\alpha$, then their union $X$ is a well-order by Lemma 1, and so the hypothesis of Zorn’s lemma is satisfied.

2. By Zorn’s lemma, conclude that the poset above contains a maximal element $(X, R)$, where $X \subseteq A$. We claim $X = A$; suppose not (here is where we need excluded middle), and let $x$ be a member of the complement $\neg X$. Well-order $X \cup \{x\}$, extending $R$ by deeming $x$ to be the final element. This shows $(X, R)$ was not maximal, contradiction. Hence any set $A$ may be well-ordered.

3. The axiom of choice follows: suppose given a surjection $p: E \to B$, so that every fiber $p^{-1}(b)$ is inhabited. Consider any well-ordering of $E$, and define $s: B \to E$ by letting $s(b)$ be the least element in $p^{-1}(b)$ with respect to the well-ordering. This gives a section $s$ of $p$.

### Bourbaki-Witt theorem

Many accounts of the proof of Zorn’s lemma start by establishing first the so-called Bourbaki-Witt theorem, which does not require AC and is of interest in its own right. However, it too does not admit a constructive proof; see Bauer for a demonstration that it is not valid in the effective topos. That said, the issue is subtle enough that the Bourbaki-Witt theorem nonetheless holds in topos with a geometric morphism to $Set$, for instance any Grothendieck topos, assuming B-W holds in $Set$ (Bauer-Lumsdaine 2013).

###### Theorem

(Bourbaki-Witt) Let $P$ be an inhabited poset such that every chain has a least upper bound, and $s: P \to P$ a function that is inflationary: satisfies $x \leq s(x)$ for all $x$. Then $s$ has a fixed point: an $x$ such that $s(x) = x$.

###### Proof

Without loss of generality, assume $P$ has a bottom element $0$; otherwise just pick an element $a \in P$ and replace $P$ with the upward set $\uparrow a$.

Say that a subset $I \subseteq P$ is $s$-inductive if: $0 \in I$, $I$ is closed under $s$ ($s(I) \subseteq I$), and $I$ contains the sup of any chain in $I$. The intersection of any family of $s$-inductive sets is also $s$-inductive, so the intersection $M$ of all $s$-inductive sets is $s$-inductive.

The idea is that $M$ is totally ordered (is a chain) by the following intuition: the elements of $M$ are $0, s(0), s s(0), \ldots s^n(0), \ldots$, where we continue by transfinite induction: at limit ordinals $\alpha$ define $s^\alpha(0)$ to be the sup of $\{s^\beta(0): \beta \lt \alpha\}$, and at successors define $s^{\alpha + 1}(0) = s(s^\alpha(0))$. We cannot have a strictly increasing chain that goes on forever, by cardinality considerations, so at some point we hit an $\alpha$ such that $s^{\alpha + 1}(0) = s^\alpha(0)$, which provides a fixed point. (What could be nonconstructive about that? See this comment by Lumsdaine.) In any case, once we prove $M$ is totally ordered, the sup of $M$ is obviously a fixed point.

Without getting into ordinals, one may prove $M$ is totally ordered as follows. Call an element $c \in M$ a cap if $x \lt c$ for $x\in M$ implies $s(x) \leq c$. For each $c \in M$, put $M_c = \{x \in M: x \leq c \vee s(c) \leq x\}$. A routine verification shows $M_c$ is $s$-inductive if $c$ is a cap, so in fact $M_c = M$ by minimality of $M$. Then, show that the set of caps is $s$-inductive. This is also routine if we avail ourselves of the just-proven fact that $M_c = M$ for any cap (but consult Appendix 2 in Lang’s Algebra if you get stuck). So again by minimality of $M$, every element of $M$ is a cap. Finally, if $c, d \in M$, use the fact that $c$ is a cap to conclude either that $d \leq c$ or $s(c) \leq d$, whence $c \leq s(c) \leq d$; thus we see any two elements are comparable.

For a discussion of how to get from Bourbaki-Witt to Zorn, see either Lang or this $n$-Category Café post by Tom Leinster, especially the main post which gives three very closely related results bearing on Zorn’s lemma, and a follow-up comment here which brings in Bourbaki-Witt.

###### Remark

An alternative proof of Bourbaki-Witt would be along lines similar to those used to show AC implies Zorn’s lemma. Given the inflationary function $s$, consider the function $f: Well(S) \to S$ which takes a well-ordered subset $W$ to $s(\sup W)$. If $s$ has no fixed points, then $f(W)$ will always be a strict upper bound of $W$, and from there one derives a contradiction exactly as in the proof of Zorn’s lemma.

###### Remark

The Bourbaki-Witt theorem is an example of a fixed-point theorem. We should point out its kinship with the quite remarkable fixed-point theorem due to Pataraia, who observed that the conclusion of the Bourbaki-Witt theorem may be strengthened quite considerably, and proved constructively (!), if we change the hypothesis to say that $s$ is a monotone operator (preserves the order $\leq$) on an inhabited dcpo. See fixed point for a brief account, and this blog post for some appreciative commentary.

## Usage/applications

It is very common, when starting with a preordered set $S$, to apply Zorn's lemma not to $S$ itself but to an up-set (an under category) in $S$. That is, one starts with an element $x$ of $S$ and proves the existence of a maximal element comparable to $x$.

Zorn's lemma may be used to prove all of the following:

Some of these are equivalent to Zorn's lemma, while some are weaker; conversely, some additionally require excluded middle.

## References

Revised on April 25, 2017 09:10:48 by Urs Schreiber (131.220.184.222)