# What fragment of ZFC do we need to prove Zorn's lemma?

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.

• 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
• books.google.co.il/… (p. 88 in case the link points elsewhere) – Asaf Karagila Dec 3 '12 at 12:05

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 $Z$.

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. 　◼

• You need to show, though, that there is a surjection from $Y$ onto $X$. Note that even when the axiom of choice fails the Hartogs number is still a well-defined notion. – Asaf Karagila Dec 3 '12 at 12:07
• Hmmm, I don't think I need that feature of the Hartogs number here, though. – Zhen Lin Dec 3 '12 at 13:22
• Also by "well-ordered chain" do you mean that the chain with the restriction of the order is well-ordered, or do you mean "well-orderable"? – Asaf Karagila Dec 3 '12 at 14:32
• Chain with the induced order. – Zhen Lin Dec 3 '12 at 16:08