Talk:Hartogs number

Invalid (or at least unclear) "Proof"

The proof given appears to be invalid or at least so unclear as to be unconvincing. JRSpriggs 06:14, 26 June 2007 (UTC)

I'm working on it. You could have tagged it with {{expert}}.... — Arthur Rubin | (talk) 18:53, 1 December 2007 (UTC)

Can we assume that, in ZF, if X is a set, then X × X is a set. The proof I'm familiar with in is NBG, where that's one of the finite set of axioms encompassing comprehension. — Arthur Rubin | (talk) 19:03, 1 December 2007 (UTC)

See Kripke–Platek set theory#Proof that Cartesian products exist for a proof that Cartesian products exist. Or use the axiom of powerset#Consequences and the axiom schema of specification. JRSpriggs 22:11, 1 December 2007 (UTC)

And thanks for reworking the proof, it is much clearer now. JRSpriggs 22:27, 1 December 2007 (UTC)

Greater than vs. not less than or equal

Forgive my ignorance, but this sentence confuses me:

If X cannot be wellordered, then we can no longer say that this α is the least wellordered cardinal greater than the cardinality of X, but it remains the least wellordered cardinal not less than or equal to the cardinality of X.

What is the difference between "greater than" and "not less than or equal to"? Solemnavalanche (talk) 19:52, 23 November 2008 (UTC)

See the article on cardinality. If A is a set of cardinality κ and B is a set of cardinality μ, then $\kappa \nleqslant \mu \,$ means that there is no injection from A to B while $\kappa >\mu \,$ means that there is an injection from B to A in addition to the absence of an injection from A to B. Does that clarify it? JRSpriggs (talk) 21:29, 23 November 2008 (UTC)

Thanks for the prompt response. It does clarify it somewhat, but it raises a further question. Does this mean that, absent the axiom of choice, there could exist pairs of sets for which no injection exists in either direction? That is, we could have $\kappa \nleqslant \mu \land \mu \nleqslant \kappa \,$? That clashes with my intuition, so I want to be sure I'm understanding you (and these articles) correctly. I see that the Law of Trichotomy article appears to say the same thing; but I'm having a difficult time imagining what such sets A and B would look like. Is there an article that addresses that question? Solemnavalanche (talk) 04:04, 24 November 2008 (UTC)

Well, it's good that it conflicts with your intuition, because the axiom of choice is intuitively correct.

But anyway, a good example is this: Let one of your sets be R, the set of all real numbers, and the other one be the Hartogs number of R. Then by definition there's no injection from the second to the first, and if there were an injection from the first to the second, that would imply that the reals could be wellordered.

Actually this generalizes, and shows that trichotomy implies the axiom of choice. --Trovatore (talk) 04:41, 24 November 2008 (UTC)

I think I'm starting to understand a bit more. So there exists an injection from R to its Hartogs number iff there exists a choice function on R. And so this must be related to the fact that ZF + GCH -> AC, is that right? Solemnavalanche (talk) 06:04, 24 November 2008 (UTC)

GCH only constrains |ℝ|, not H(ℝ) as the Hartogs number for ℝ. Regardless of GCH, |ℝ| = $\beth$₁ = ω_r for some ordinal r > 0; with Choice, H(ℝ) = ω_r+1. With GCH (and hence Choice), r = 1 so H(ℝ) = ω₂, still in the same position relative to ℝ. Without either Choice or GCH, ω₁ ≤ H(ℝ) ≤ ω_r+1, with the lower bound on H(ℝ) resulting from all countable ordinals being embeddable in ℝ since ℚ is embedded in ℝ. If X is infinite but Dedekind-finite (only finite ordinals embeddable in X) then H(X) = ω (= ω₀), no matter what the cardinality of X is. If X is finite, so is H(X), namely |X| + 1. Vaughan Pratt (talk) 23:44, 28 November 2025 (UTC)

Please see my comments at Talk:Tarski's theorem about choice for why the axiom of choice is equivalent to the trichotomy of cardinality. JRSpriggs (talk) 01:12, 29 November 2025 (UTC)

Clarity issues

The definition "$\alpha =\{\beta \in {\textrm {Ord}}\mid \exists i:\beta \hookrightarrow X\}$" introduces several pieces of unexplained and unlinked notation.

• Is Ord the category of preordered sets or the ordinal numbers? The latter makes little sense because the lead is talking exclusively about cardinal numbers.
• Is the hooked arrow an embedding or an inclusion map?
• And what's with this variable i? i is usually an index, right? It's never referred to anywhere else, so why can't it just be deleted? "$\alpha =\{\beta \in {\textrm {Ord}}\mid \beta \hookrightarrow X\}$". Oh, right, there's a colon. It's not a variable, it's a function named $i$. Why not call it $f$?
• Not to mention set-builder notation in the first place, although that's hopefully well-known enough.

It's not stated anywhere that $X\subseteq {\textrm {Ord}}$, so I'm assuming that's false. That implies that $\hookrightarrow$ is not the natural injection $\iota$, which opens the question of decidability: can it be determined if such a function exists? 71.41.210.146 (talk) 05:41, 21 January 2017 (UTC)

• "Ord" refers to the class of ordinal numbers. If the axiom of choice holds, then the cardinal numbers can be regarded as a subclass of the ordinal numbers, namely the initial ordinals.
• The hooked arrow means that "i" refers to an injective function.
• Yes, "X" refers to an arbitrary set. Indeed, there would be little point in defining this (Hartog's number), if X were a set of ordinals. JRSpriggs (talk) 18:22, 21 January 2017 (UTC)

@JRSpriggs: Thanks! Article clarified, I hope. Now to just untangle the fact that the lead talks about cardinals while the proof is about ordinals. Perhaps I should just import (paraphrase) the explanation at http://planetmath.org/HartogsNumber, which is nicely clear. 71.41.210.146 (talk) 20:26, 21 January 2017 (UTC)

Thank you for clarifying the article. I am glad that I could help you. JRSpriggs (talk) 18:52, 22 January 2017 (UTC)

Hartogs['][s] function

Re the late exchange between User:JRSpriggs and User:Cherkash: I decided to go looking on Google Scholar to see how the terminology is actually used. I agree with Cherkash that the general style on WP is to keep the final s. On the other hand, if the term Hartogs' function without the s is the common usage, then we should use it. Or, if attested, we might be able to cut the Gordian knot by writing Hartogs function, using Hartogs as a noun adjunct rather than a possessive.

As it turns out, though, I was not able to verify (with any of the search terms) that this terminology is used for this at all. It seems to be used instead for some quite different notion. To quote this paper, "[a]n upper semi-continuous function h in a Stein space X is called a Hartogs function in X if the Hartogs domain $\{(z,x)\in C\times X;|z|<e^{-h(x)}\}$ is Runge–Stein in C×X." I don't know what all of that means, exactly, but it does not appear to be about ordinal numbers.

So maybe we should just remove the contested text altogether? --Trovatore (talk) 06:27, 21 December 2017 (UTC)

Going by the grammar argument alone, I agree with the dichotomy presented: we are effectively left to decide whether to use the noun adjunct form (aka noun's attributive form), or a possessive (with " 's " added). Either one is a grammatically acceptable way to use here, with the choice towards one or the other usually guided by tradition and commonality of use. In the first case, the spelling is "Hartogs function", in the second it's "Hartogs's function". As for the actual subject, if there's any doubt about the statement presented, then please by all means go ahead and either tag it with the "citation needed" tag (pending reliable sources) – or just go ahead and remove it altogether. cherkash (talk) 06:50, 21 December 2017 (UTC)

Well, if it appears standardly in the literature as Hartogs' function, then that's what we should say; I don't care what the general WP standard is in that case. However I don't see any evidence so far that that's actually the case. --Trovatore (talk) 06:54, 21 December 2017 (UTC)

Absolute nonsense "Proof"

Please erase it completely and precis either the original paper or my translation, a link to which has recently been added to this page.

Paul Taylor, 25 November 2025. — Preceding unsigned comment added by ~2025-36371-09 (talk) 20:54, 25 November 2025 (UTC)

What do you see as wrong in Derek Goldrei's proof?

The article claims that the proof is constructive, but Replacement as used in step 4 isn't usually considered constructive. For a constructive version of Goldrei's proof one would use John Myhill's Axiom schema of collection (1965) as weakened by Peter Aczel in 1968. Vaughan Pratt (talk) 02:42, 26 November 2025 (UTC)

Incidentally there's a subtlety in the case when X is Dedekind-finite defined as infinite with no injection ω ↪ X, which can happen in ZF without Choice. In that case α = ω because Dedekind-finite sets (in the above sense) only admit injections from finite ordinals (and, with the above definition, do so for all finite ordinals). This is true even when X is uncountable; counterintuitively the Hartogs number of any Dedekind-finite set is ω.

But it's also worth mentioning that the continuum ℝ is not Dedekind-finite because the rationals ℚ can be given the order type of any countable ordinal, whence α ≥ ℵ₁, with equality if and only if no uncountable subset of ℝ can be well-ordered. Vaughan Pratt (talk) 18:07, 26 November 2025 (UTC)

A subtlety in Hartogs' Choice-free construction is his Lemma 4 (Satz 4 in his paper). He states and proves that in any totally ordered set L, if every element is strictly preceded by a well-ordered set, then L itself is well-ordered. (How would you prove that without Choice and without looking at Hartogs proof?)

One might wonder why modern constructions of the Hartogs number of any set X manage to sidestep this subtlety. The reason is that Hartogs starts by considering all possible total orderings. (In those days, "ordered set" defaulted to totally ordered set, different from today's default of partially ordered set or sometimes even preordered set.) If instead you consider only ordinals that inject into X, you get Lemma 4 for free and arrive at the same H(X) as Hartogs does by bypassing the whole notion of total orderings that might not be well-orderable for whatever reason.

With Choice, H(X) is just the next cardinal after |X|, e.g. 8 if |X| = 7, ℵ₈ if |X| = ℵ₇, usw.

Without Choice, size comparisons must now be made in terms of ordinals rather than cardinals. The cardinality of X might be huge, e.g. X = P(P(P(P(N)))), but if the only ordinals you can inject into X are the members of ω₁, namely the countable ordinals, then H(X) will just be ω₁ itself. This is the "size" of all uncountable X that could be called "Dedekind-countable" by virtue of only embedding countable ordinals, namely their sup as an ordinal. Vaughan Pratt (talk) 18:50, 28 November 2025 (UTC)

You asked "How would you prove that [in any totally ordered set L, if every element is strictly preceded by a well-ordered set, then L itself is well-ordered] without Choice and without looking at Hartogs proof?". Suppose S is a non-empty subset of L. Let x be an element of S. Either x is already the minimal element of S or there are elements below it by the totality of L. If there are elements below x, then the set of such elements is a non-empty subset of a well-ordered set so it contains an element y which is minimal in that subset. If so, then y < x and thus is less than the other elements of S by the transitivity of the total ordering. JRSpriggs (talk) 01:31, 29 November 2025 (UTC)

Well done. (I didn't mean it to be a hard challenge for those familiar with reasoning about infinite ordinals.) Vaughan Pratt (talk) 05:07, 29 November 2025 (UTC)

I do not understand your sentence "With Choice, H(X) is just the next cardinal after |X|, e.g. 8 if |X| = 7, ℵ₈ if |X| = ℵ₇, usw.". Please explain what you mean. JRSpriggs (talk) 01:34, 29 November 2025 (UTC)

Wait, I thought you'd already understood how Hartogs' construction works with sets when Choice holds. Are you questioning one or both of H(X) = 8 and H(X) = ℵ₈ in the respective cases for X, or something else? Vaughan Pratt (talk) 05:15, 29 November 2025 (UTC)

OK. I see it now. I think I was thrown off by the reverse order of your if-then constructions and by the "usw" which I still do not understand. JRSpriggs (talk) 14:31, 29 November 2025 (UTC)

Sorry, und so weiter. (Reading Hartogs auf deutsch for too long.) Vaughan Pratt (talk) 17:47, 29 November 2025 (UTC)

With or without Replacement?

The lede ends with this sentence: The existence of the Hartogs number was proved by Friedrich Hartogs in 1915, using Zermelo set theory alone (that is, without using the axiom of choice or the later-introduced replacement schema of Zermelo–Fraenkel set theory).

However the line preceding the proof says: and then one can apply the axiom schema of replacement to obtain the set of all β.

Which one is correct? Does Hartog's proof need Replacement or not? Vaughan Pratt (talk) 19:36, 24 January 2026 (UTC)

One explanation could be that Fraenkel noticed that Hartogs' 1915 proof used more than Zermelo's axioms and tentatively proposed Replacement in 1922, subsequently sharpened by my great-great-grandadvisor Skolem.

However it seems to me that Replacement is much stronger than needed for Hartogs' proof. What I don't see right now is whether his theorem can be proved entirely within Zermelo's axiomatization. And if not, would something weaker than Replacement suffice?

If ZF were some day to be found inconsistent, Replacement would at that time need to be suitably weakened. Vaughan Pratt (talk) 20:55, 24 January 2026 (UTC)

I believe the catch is that replacement is needed exactly to state this result in terms of (von Neumann) ordinals. Hartog's original result just says that there exists some well-ordered set that does not inject into X. For this purpose, one can just use equivalence classes of well-orderings of subsets of X. Bbbbbbbbba (talk) 00:15, 25 January 2026 (UTC)

Sure, but my concern is with the size of the equivalence classes allowed by Replacement. Hartogs' theorem may only need far smaller equivalence classes than what ZF might drive Z into inconsistency. I find full ZF very scary. Vaughan Pratt (talk) 06:11, 25 January 2026 (UTC)

If you only want the equivalence classes, you do not need replacement. Say you already have the set W of all well-orderings of subsets of X. You take the power set of W, and apply separation with a sentence saying "There exist a well-ordering w in S such that all well-orderings in S are order isomorphic to w, and none of the well-orderings in W \ S is order isomorphic to w". What you get is exactly the set of equivalence classes. You might need to prove that every w belongs to some equivalence class by another instance of separation: separate by "w' is order isomorphic to w" on W. But no replacement needed. Bbbbbbbbba (talk) 07:11, 25 January 2026 (UTC)

How do you construct S in Z? Vaughan Pratt (talk) 17:41, 26 January 2026 (UTC)

What do you mean exactly? In the first application of separation I mentioned above, S is just a bound variable, and the axiom of separation just says that S is in the set we are constructing (the set of equivalence classes) if it exists and satisfies the condition. The second application constructs a concrete equivalence class S for every specific well-ordering w by separation on W. Bbbbbbbbba (talk) 06:11, 27 January 2026 (UTC)

What I'm asking is, how do you describe equivalence classes without using Replacement?

Replacement says that the image of a definable function is a set. If you don't have Replacement, how do you know that these equivalence classes are sets? Vaughan Pratt (talk) 21:32, 27 January 2026 (UTC)

As I have explained multiple times, equivalence classes are subsets of W (a set we have already constructed) with definable membership criteria. This requires only axiom schema of separation, not the full axiom schema of replacement. Bbbbbbbbba (talk) 23:44, 27 January 2026 (UTC)

Reading between the lines, my impression is that you're trying to say:

use separation on P(P(W)) to pick out the set of those nonempty subsets S of W such that

for all s in S [for all w in W, w is in W iff w ~ s].

For suitably small X, this would obviously be the set of all blocks of the partition induced by the order-isomorphism equivalence w ~ w'. If that's what you meant, we're on the same page.

Here's the problem. How do you show that what's in the square brackets defines a set, even though the condition is a first order property?

Ordinals in models of Z behave sensibly only up to a certain point. ZF extends that point. Choosing X above the lower point behaves sensibly for Hartogs' Theorem only in ZF. This, or something like it, is what Fraenkel may have realized, and prompted his addition of Replacement to Z. Vaughan Pratt (talk) 05:56, 28 January 2026 (UTC)

First of all, I think your language about separation is a bit different from mine because when constructing a subset of ⁠${\mathcal {P}}(W)$⁠, I say I apply separation on ⁠${\mathcal {P}}(W)$⁠, not ⁠${\mathcal {P}}({\mathcal {P}}(W))$⁠. Yes the result will be an element of ⁠${\mathcal {P}}({\mathcal {P}}(W))$⁠, but the axion schema of separation does not directly mention that power set.

Second, the statement of the axiom schema of separation does allow ⁠$\varphi$⁠ to be a first-order sentence with quantifiers. Otherwise it would be much weaker; I do not see even a way to construct ⁠$W$⁠ in that case, since the sentence "⁠$w$⁠ is a well-ordering" hides a few quantifiers.

Or maybe you are worried about the "inner" separation, where the condition is just "⁠$w\sim s$⁠", but with the quantifier ⁠$\forall s$⁠ "on the outside"? For this instance of separation, ⁠$s$⁠ is an "additional" free variable in ⁠$\varphi$⁠, also known as a parameter, which is also allowed. Disallowing parameters would also result in a much weaker version of separation, as discussed in Talk:Axiom schema of specification#Why are the w1, w2, ..., wn necessary.

None of these has anything to do with "ordinals" per se, only well-ordered sets. I think in general the concept of "ordinal" is a meta-concept in Z that does not correspond to any in-universe object. Bbbbbbbbba (talk) 09:52, 28 January 2026 (UTC)

Thank you for all that. Unlike what you are replying to, S remains undefined in your account. Sorting it out is above my paygrade. Vaughan Pratt (talk) 12:59, 28 January 2026 (UTC)

That is because variables that can be reasonably named ⁠$S$⁠ appear many times in the argument, sometimes universally quantified and sometimes existentially quantified. Let me try to clarify the situation by formalizing the two instances of separation we are using after we get ⁠$W$⁠.

What we want to prove is:

⁠$\exists {\text{ well-ordered set }}\langle A,\leq \rangle ,\;\forall w_{0}\in W,\;\exists S_{0}\in A,\;\langle A,\leq \rangle _{<S_{0}}\sim w_{0}$⁠.

(⁠$\langle A,\leq \rangle _{<S_{0}}\equiv \langle A_{<S_{0}},\leq _{<S_{0}}\rangle$⁠ denotes a proper initial segment of ⁠$\langle A,\leq \rangle$⁠.)

To this end, we first construct an ⁠$A$⁠ with this instance of separation:

⁠$\exists A,\;\forall S,\;S\in A\Leftrightarrow (S\in {\mathcal {P}}(W)\land (\exists s\in W,\;\forall w,\;w\in S\Leftrightarrow (w\in W\land w\sim s)))$⁠.

(⁠$\exists s\in W$⁠ can be replaced by ⁠$\exists s\in S$⁠, but the former form is probably slightly more convenient. Replacing it with ⁠$\forall s\in S$⁠ would allow ⁠$S=\emptyset$⁠ which would be undesirable.)

Then we define the relation ⁠$\leq$⁠ as:

⁠$\{(S,S')\in A\times A\mid \exists s\in S,\;\exists s'\in S',\;s{\text{ is an intital segment of }}s'\}$⁠.

(OK, this is yet another application of separation that I have not explicitly mentioned before. I am not writing out the axiom instance explicitly here because I imagine it would be hard to read and not that important.)

Now, to prove the original proposition, we need to find a ⁠$S_{0}\in A$⁠ for every ⁠$w_{0}\in W$⁠. To this end we use the second instance of separation:

⁠$\forall w_{0},\;\exists S_{0},\;\forall w,\;w\in S_{0}\Leftrightarrow (w\in W\land w\sim w_{0})$⁠.

This gives us an ⁠$S_{0}$⁠, but does not yet assert that ⁠$S_{0}\in A$⁠. For that purpose we have to refer to the definition of ⁠$A$⁠ and prove ⁠$S_{0}\in {\mathcal {P}}(W)$⁠ and ⁠$\exists s\in W,\;\forall w,\;w\in S_{0}\Leftrightarrow (w\in W\land w\sim s)$⁠ separately, but the proofs are hopefully both obvious.

Finally we need to show that ⁠$\langle A,\leq \rangle _{<S_{0}}\sim w_{0}$⁠ by constructing a order isomorphism between those two well-orders, i.e., a bijection between their domains that is order-preserving. Let ⁠$w_{0}=\langle Y,\preceq \rangle$⁠. Then the bijection is:

⁠$\{(S,x)\in A_{<S_{0}}\times Y\mid \langle Y,\preceq \rangle _{\prec x}\in S\}$⁠.

(Another instance of separation...)

There remain some cumbersome details. We need to show:

• every ⁠$S\in A_{<S_{0}}$⁠ has an image ⁠$x\in Y$⁠, and
• that image is unique, and
• every ⁠$x\in Y$⁠ has a preimage ⁠$S\in A_{<S_{0}}$⁠, and
• that preimage is unique, and
• for all pairs ⁠$(S,x)$⁠ and ⁠$(S',x')$⁠ in the bijection, ⁠$S\leq S'\Leftrightarrow x\preceq x'$⁠, and
• ⁠$\langle A,\leq \rangle$⁠ is indeed a well-ordered set (a point I had neglected when I began writing this reply).

The proofs might be hairy and involve a bunch of more separation instances, but there is no obvious obstacle to doing them all in Z. Bbbbbbbbba (talk) 16:01, 28 January 2026 (UTC)

Thanks for spelling that out. I believe that all your uses of separation are sound, so that convinces me that the proof of Hartogs' number doesn't need Replacement. I'm happy to leave it to others to raise objections to your argument, which looks fine to me. (But what do I know? I just work in geophysics these days.).

Meanwhile I'm going to modify the line before the proof to reflect our present agreement. Vaughan Pratt (talk) 04:12, 29 January 2026 (UTC)

Mulling this over overnight, it occurs to me to ask whether Hartogs' proof can be completed in Z. After forming the set β of all ordinals embeddable in X as sets, we then want to show that the Hartogs number of X is the least ordinal not in β. This is not an ordinal representable by any well-ordering of X. Can we do this in Z?

Since we only need ordinals up to the next cardinal after $|X|$, one approach might be to represent ordinals as equivalence classes of all well-orderings of $P(X)$, which would still keep them as sets. We should be able to order these ordinals not by inclusion but by order embeddings between the members of these classes, which should accomplish an equivalent ordering.

On the one hand, it is surely far easier just to appeal to Replacement, which might have been how Fraenkel was looking at Hartogs' proof. On the other, something along the above lines, if it could be made to work, would at least disprove the claim that the Hartogs number of X can't be shown to exist without appealing to Fraenkel's Axiom Schema of Replacement, which might only spring into action for functional relations far beyond the power set function. Vaughan Pratt (talk) 03:44, 30 January 2026 (UTC)

It is less a matter of whether the proof can be completed in Z, and more a matter of the theorem itself should be formulated in Z. If we keep the statement in the same form as it is usually stated in ZF, i.e., with von Neumann ordinals, then it is simply unprovable since ⁠$V_{\omega +\omega }$⁠ is a model of Z that does not contain any uncountable von Neumann ordinal. So you are right in seeking an alternative representation.

However, "equivalence classes of all well-orderings of ⁠$P(X)$⁠" cannot be a canonical representation of ordinals because it is not unique. Different values of X give rise to different representations of the same ordinal. The problem is that, as far as I know, there exists no canonical representations of ordinals in Z. This is why Hartog's original paper does not actually talk about ordinals, except for a side note that "we could use ordinal numbers to make some simplifications".

Cantor viewed cardinals and ordinals as "abstractions", but it seems to have been a vague concept without a clear definition. I am wondering if what this really means is that one should translate every instance of "there exists an ordinal/for all ordinals" into "there exists a well-ordered set/for all well-ordered sets", and "equal to" into "order isomorphic to" when it refers to ordinals, etc. It should be a meta-theorem that translating a formula this way does not affect their truth value (although some formulas may not translate cleanly, for example when a set could contain both ordinals and other objects; in a sense we are treating ordinals as another sort of objects). If so, then the statement that Hartog did prove (using Z only) is exactly the translation of Hartog's theorem. Bbbbbbbbba (talk) 05:16, 30 January 2026 (UTC)

What I meant was that the whole proof should be done for ordinals represented as equivalence classes of well-orderings of subsets of $P(X)$. That keeps the representation uniform across the whole proof.

A cute side effect is that, within the proof for X, there is a largest ordinal, namely the cardinal $|P(X)|$. This would be the Hartogs number for X iff it is the next cardinal after $|X|$, i.e. CH for X. Vaughan Pratt (talk) 17:19, 30 January 2026 (UTC)

You could do that, I guess, but the result is not really any more interesting than one stated simply in terms of well-ordered sets. For that matter, ⁠$\vert P(X)\vert$⁠ is not the "largest ordinal" representable in this representation, just a ordinal with the largest cardinality. You can always construct the successor of an infinite ordinal represented as a well-ordered set by moving the first element in the well-order to the end. Bbbbbbbbba (talk) 23:06, 30 January 2026 (UTC)

I was wrong. By Hartogs' Theorem, the equivalence classes of the well-orderings of $P(X)$ are all the ordinals strictly less than the next cardinal after $P(X)$.

The irony is that in order to have the Hartogs number for X as one of the ordinals in the proof, you end up representing uncountably more ordinals than needed, at least when X is infinite.

What's nice about the ordinals, as opposed to well-ordered sets, is that they're linearly ordered. Vaughan Pratt (talk) 05:55, 31 January 2026 (UTC)