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It is extremely well-known that Zorn's lemma is a theorem of ZFC. My interest is in a certain finitely-axiomatisable fragment of ZFC, sometimes called RZC (restricted Zermelo with choice) or ZBQC. The axioms of RZC are the following:

  • Extensionality
  • Empty set
  • Pair set
  • Union
  • Power set
  • Infinity
  • Foundation
  • Choice (in the sense that every surjection has a right inverse)
  • Separation for $\Delta_0$-formulae

Adrian Mathias also defines an extension of RZC, called MAC (Mac Lane set theory), by adding the transitive containment axiom:

  • Every set is contained in a transitive set.

(Apparently RZC and MAC are equiconsistent.)

Question. Is Zorn's lemma a theorem of RZC? If not, is it a theorem of MAC?

It is reasonably clear that well-founded induction is valid in RZC, but without the axiom of replacement it is not at all obvious to me whether Hartogs numbers exist. ($V_{\omega + \omega}$ is a model of RZC, but the only von Neumann ordinals in $V_{\omega + \omega}$ are precisely those below $\omega + \omega$, even though it has uncountable well-ordered sets.) Once we know that there are sufficiently large well-ordered sets, it seems to me that the usual proof of Zorn's lemma will go through in RZC.

Motivation. One can build a model of RZC out of any model of ETCS (elementary theory of the category of sets) and ETCS can be interpreted in any model of RZC. What I really want to know is whether ETCS proves that, say, every vector space has a basis, and it seems like a good first step would be to establish the claim for RZC.

Zhen Lin
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  • That's a very good question. I suspect that this may depend on the variant of choice you are using as well. If one analyzes the proof of ZL then one can see that indeed we go through the set using a transfinite induction. Last I recall validity of general transfinite induction is equivalent to the replacement schema. Surely this means it is much stronger than separation for $\Delta_0$. I think that the use of $V_{\omega+\omega}$ can be misleading. All the elements in $V_{\omega+\omega}$ are also elements of $V$, ZL holds in $V$ so they there are maximals, which are in $V_{\omega+\omega}$. – Asaf Karagila Nov 28 '12 at 21:06
  • Good point. I've clarified what I mean by the axiom of choice here. – Zhen Lin Nov 28 '12 at 21:11
  • ... apparently I can't read. Mathias mentions in his paper about Mac Lane set theory that the well-ordering theorem is provable in RZC minus Foundation and Infinity. Still, an explicit proof would be nice. – Zhen Lin Nov 28 '12 at 21:17
  • Are you sure that the variant of choice you are using is the one used by Mathias/MacLane? – Asaf Karagila Nov 28 '12 at 21:51
  • Mathias uses the usual formulation, i.e. the existence of a choice-function for families of non-empty sets. But this is the same as asking for surjections to split in the presence of disjoint unions. – Zhen Lin Nov 28 '12 at 21:57
  • Have you looked at Zermelo's original proof (from his 1904 paper) about $\theta$-chains and whatnot? I think there are two available translations to English, both can be found on Google books (I was looking for this recently, so I remember finding that). – Asaf Karagila Nov 28 '12 at 23:14
  • @Asaf: I'm not sure where I can find that online. I may be able to find it in the library, if you have a more precise reference. – Zhen Lin Dec 03 '12 at 10:44
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    http://books.google.co.il/books?id=XB2nd2ovakIC&lpg=PR4&dq=Zermelo&pg=PA88#v=onepage&q&f=false (p. 88 in case the link points elsewhere) – Asaf Karagila Dec 03 '12 at 12:05

3 Answers3

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Actually, one does not need to use any ordinals to prove Zorn's lemma in RZC!

Theorem: Every non-empty partial order $(S,<)$ in which every non-empty chain has an upper bound has a maximal element.

Proof: We shall work under the assumption that the claim is false. First let $F$ be a choice-function that maps each chain $C$ in $(S,<)$ to a strict upper bound for $C$ in $S$. (This is the only use of AC in this proof, and also shows that it suffices for $S$ to be well-orderable in absence of AC.) We say that $T$ is a tower iff $T$ is a well-ordered chain in $(S,<)$ and $∀x{∈}T\ ( \ x = F(T_{<x}) \ )$. Note that any two towers $T,U$ agree, meaning that one is a subchain of the other, since otherwise $T{∖}U$ and $U{∖}T$ are both non-empty and disjoint and so letting $m = \min_<(T{∖}U)$ and $n = \min_<(U{∖}T)$ we would obtain $m = F(T_{<m}) = F(U_{<n}) = n$ and hence a contradiction. Now let $V$ be the union of all the towers, and observe by the above note that it is a tower too, but that $V∪{F(V)}$ is a taller tower, contradicting the definition of $V$.

user21820
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    Nice proof! I wish this were included in every set theory textbook (although the proof by transfinite recursion is equally nice). – YuiTo Cheng Oct 08 '20 at 08:37
  • @YuiToCheng: The proof I was taught was by transfinite recursion on ordinals as well. But what is not usually taught is that the crazy amount of setup needed for ordinals is actually not needed for normal mathematics, since both Zorn's lemma and the well-ordering theorem can be proven without any ordinals. In other words, this shows that normal mathematics (outside set theory and logic) do not need ordinals, contrary to some popular beliefs. Even the long line in topology needs only an uncountable well-ordering whose proper initial segments are countable, not the specific ordinal $ω_1$. =) – user21820 Oct 08 '20 at 08:44
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    I totally agree with what your comment. I think one of the reasons many books offers a transfinite recursion proof of Zorn's lemma (or well ordering theroem) is it provides a more structured approach so that it can be reproduced very quickly by students (at least for those who have learnt of ordinals). – YuiTo Cheng Oct 08 '20 at 08:52
  • @YuiToCheng: That's an interesting idea. Too bad it is based on a false premise because once you have the well-ordering theorem you can do transfinite recursion along that well-ordering. It's an equally structured approach, minus the fuss with ordinals. =) – user21820 Oct 08 '20 at 09:25
  • Of course, the only catch is that it doesn't work anymore if one wants to do transfinite recursion along all ordinals, but that's set theory land already. =) – user21820 Oct 08 '20 at 09:27
  • I believe this proof is inspired by the transfinite induction proof. An ordinal is just a particular (and arguably nice) representative of that well-ordering, which is guaranteed to exist under axiom of replacement. Of course you can substitute well-ordering in ordinal in most arguments. I think the point is without replacement, well-orderings that obviously "are there" are not guaranteed to exist. But once we agree on replacement we might as well replace a well-ordering by the corresponding ordinal. – Lxm Jun 09 '22 at 17:19
  • @Lxm: This proof was not inspired by the usual proof (that I was taught). Also, I do not think that ordinals are "obviously there", nor do I agree on Replacement. In particular, there is no non-circular philosophical justification of Replacement, and more concretely there is no non-circular conceptual ontology for $ω_1$. So I don't think there is good reason to use ordinals when unnecessary. Of course, if you simply accept ZFC then ordinals are nice canonical representatives of well-orderings, but that's a big if. =) – user21820 Jun 10 '22 at 10:08
  • @user21820 You said the proof you was taught was transfinite recursion: at successor step choose a larger element which exists by assumption, and at limit step take the union which is still a chain; isn't your "tower" basically the same, except the elements are ordered by "themselves" rather than ordinals? Also I first read this proof here where the author also says it has the flavor of transfinite induction. – Lxm Jun 10 '22 at 19:23
  • @user21820 Without replacement $\omega_1$ still makes sense: it's the set of isomorphism classes of well-orderings of $\omega$, equipped with a natural order. It exists by power set and comprehension axiom; it just not necessarily exists as a von-Neumann ordinal. Similarly one can define $\omega_2$, $\omega_3$, etc. However, without replacement one cannot form $\omega_\omega$; but to me this set obviously should exist, it's the union of a sequence of inductively defined sets! See also this – Lxm Jun 10 '22 at 19:23
  • @Lxm: when I said $ω_1$ I meant the von Neumann ordinal... But have you thought carefully about your philosophical justification (if any at all) of full Replacement? Bounded replacement does not justify full replacement. As for the proof I was taught, I cannot remember the details (it was long and complicated) but I know the proof I gave here has nothing to do with it. – user21820 Jun 11 '22 at 04:28
  • Let us continue the discussion in chat – Lxm Jun 11 '22 at 04:41
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The argument proposed by Zhen Lin works once one knows that $Y$ (as defined there) is well-ordered by the "obvious" relation $\prec$, namely that one equivalence class is smaller than another if some (equivalently every) element of the first class embeds as a proper initial segment in some (equivalently every) element of the second class. Zhen Lin leaves this issue unsettled when $\mathbb N$ doesn't exist, but even when $\mathbb N$ does exist, the proposed argument doesn't work. The reason is that it uses the equivalence between "well-ordered" and "has no decreasing $\omega$-sequence"; this equivalence is not provable in the absence of the axiom of choice (more precisely, it needs the principle of dependent choice (DC), which is one of the weak forms of AC). So here's a proof, without $\mathbb N$ and without DC, that $Y$ is well-ordered.

Let $Z$ be a nonempty subset of $Y$; we need to show that $Z$ has a smallest element. Consider any element $[R]$ of $Z$. By the notation $[R]$ I mean the isomorphism class of a well-ordering $R\in\mathcal F$ (still using the notation of Zhen Lin's answer). If $[R]$ is the smallest element of $Z$, we're done, so assume it is not. This means that $Z$ has other elements $[S]$ whose elements $S$ are embeddable as initial segments of $R$. Each such initial segment is, since $R$ is a well-order, of the form $\{x:xRa\}$ for some $a$ in the field of $R$. Let $Q$ be the set of those elements $a$ that arise in this way, i.e., the set of those $a$ in the field of $R$ such that the initial segment consisting of the $R$-predecessors of $a$ (ordered by the restriction of $R$) is in one of the isomorphism classes in $Z$. Because $R$ is well-ordered, $Q$ has a smallest element $b$. It is then easy to check that the isomorphism class of $\{x:xRb\}$ (again ordered by the restriction of $R$) is the smallest element of $Z$.

Andreas Blass
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This reformulation of the construction Hartogs number seems to work:

Lemma (Hartogs). If $X$ is a set, then there is a well-ordered set $Y$ for which there is no injection $Y \to X$.

Proof. Let $\overline{X}$ be any set containing $X$ as a proper subset. Let $\mathcal{F}$ be the set of all well-orderings of subsets of $\overline{X}$; this can be constructed using power sets, cartesian products, and $\Delta_0$-separation, and let $\mathcal{F}_0$ be the subset of $\mathcal{F}$ consisting of well-orderings of $X$. Let $Y$ be the set of isomorphism classes in $\mathcal{F}$; this can be constructed using power sets, function sets, and $\Delta_0$-separation. $Y$ has an obvious linear order (using induction over well-ordered sets), and in fact it is well-ordered.

  • If $\mathbb{N}$ (in the sense of a set for which mathematical induction is valid) exists, we proceed by supposing $Y$ is not well-ordered: then by mathematical induction we could produce an $\mathbb{N}$-indexed strictly descending sequence $$y_0 > y_1 > y_2 > \cdots$$ in $Y$, but the sequence $y_1 > y_2 > \cdots$ can then be embedded in a representative of the class $y_0$, implying $y_0$ is not an isomorphism class of well-orderings – which is a contradiction. [Does this need choice? Embeddings of one well-ordered set as an initial segment of another are unique if they exist, so I'm inclined to believe this is choice-free...]

  • If $\mathbb{N}$ does not exist, then [...?]

Once we have shown $Y$ is well-ordered, it is easy to prove that every member of $\mathcal{F}_0$ embeds as a proper initial segment of $Y$. A standard argument then shows there is no injection $Y \to X$: indeed, if there were, then $Y$ would embed as a proper initial segment of itself – an absurdity. 　◼

The standard proof of Zorn's lemma now goes through without problems:

Lemma (Zorn). If $X$ is a partially-ordered set such that every well-ordered chain in $X$ has an upper bound in $X$, then $X$ has a maximal element.

Proof. Let $\mathcal{G}$ be the set of well-ordered chains in $X$; this can be constructed using power sets and $\Delta_0$-separation. Using the axiom of choice, we may choose once and for all an upper bound for each well-ordered chain in $X$, so that we obtain a function $u : \mathcal{G} \to X$ such that $z < u(Z)$ for all $Z$ in $\mathcal{G}$ and all $z$ in $Z$. [I think the axiom of choice is only needed to construct $u$ and nowhere else.]

Let $Y$ be a well-ordered set so large that there is no injection $Y \to X$, and suppose for a contradiction that $X$ has no maximal elements. We define an injective monotone function $f : Y \to X$ by recursion:

  • If $f (y)$ has been defined for all $y < z$ in $Y$, then define $f (z) = u(\{ f(y) : y < z \})$ (where $\{ f(y) : y < z \}$ is constructed using $\Delta_0$-separation instead of replacement).

This produces the desired contradiction. 　◼

Zhen Lin
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