Why do we need "canonical" well-orders?

Question

(I asked this question on MO, https://mathoverflow.net/questions/443117/why-do-we-need-canonical-well-orders)

Von-Neumann ordinals can be thought of "canonical" well-orders, Indeed every well-order $W$ has a unique ordinal that is its "order type".

This raises the question of why a canonical order is needed, it seems to me that every application of ordinals can be done by using a "large enough" well-ordered set instead that is guarenteed by Hartog's lemma$^{1}$, for example, instead of performing a transfinite process on an ordinal, we perform it on the "large enough" well ordered set $X$ whose existence is guaranteed by Hartog's lemma. Using this method we can prove the first basic applications of ordinals such as Zorn's Lemma(See for example here). This seems to beg the question, if there is a deeper reason for ordinals than mere convenience.

$\small{1}$: For the purposes of this question let Hartog's Lemma state: For every set $S$, there exists a well-ordered set $X$, such that there is no injection from $X\to S$.

Notes:

-- This is not an entirely useless question that does not "affect things" in any way, since ordinals $\ge \omega+\omega$ need not exist in $\mathsf{ZFC}-\mathsf{Replacement}$, and indeed the above method gives a proof of Zorn's lemma in $\mathsf{ZFC}-\mathsf{Replacement}$.

-- I suppose one can ask a similar question about cardinal numbers: Why do we need cardinal numbers, when we can reason about cardinalities using simply injections and bijections on sets?

-- I realise that this probably a very näive question for a set-theorist, but it is something whose answer seems to be hidden for the non set-theorist, which is why I ask this question.

—- Ordinals seem to give a "uniform definablity" but is that actually useful?

Edit: Probably I should clarify that I do realise that the above observations imply that ordinals can be avoided by the "working mathematician", but I am more interested in why they are so important to the working set-theorist/logician(given that they literally are a set-theorists "bread and butter").

And even if convenience is the answer why do we need a formal notion that takes hours to develop when an informal notion seems to suffice(formally)?

Answer 1

Convenience is important, but let's ignore this point.

First let's talk about how easy it is to get ordinals: Say we start with $\sf ZF-Replacement$ and we want to have transfinite recursion, then the resulting theory is the full $\sf ZF$. (I apologize, but I don't remember if this result is by B. Rin or JDH) which means we get "all" of the ordinals again!

Interestingly enough, $\sf ZF(C)-Replacement+Recursion\ on\ Ord$ is strictly weaker than $\sf ZF(C)$, but usually we do out recursion on general sets.

Some reasons to want the axiom of replacement can be found here, here, or here, I will assume replacement for the rest of the answer.

One big advantage of the ordinals, it is that they are a proper class that is well-ordered, in particular you can do stuff along all of the ordinals. Using Hartog's lemma will only give you set-size applications.

For example: a class-size cumulative hierarchy, e.g. defining $V_α$ in von Neumann universe, or $L_α$ in the Constructible universe.

Those constructions are very important.

Another very important construction is $\sf HOD$.

Apart from constructions, there are few important result concerning absoluteness, one example was given by Eric in the comments. Another (non-obvious) result is that: if $M,N$ are transitive models such that $M⊨\sf ZFC$ and $N\models\sf ZF$ such that $M,N$ has the same ordinals and the same sets of ordinals, then $M=N$. This result uses the following fact about ordinals: for every set $x$, there is a set of set of ordinals $A$ that encodes uniquely $\operatorname{trcl}(\{x\})$ that is definable only with $x$.

Some other surprising results that are in the same spirit as the previous paragraph, the following are equivalent in $\sf ZF$:

1. There exists a definable injection $F:V\to \mathcal P(Ord)$

2. There exists a definable function $F:V\to V$ such that if $|X|>1$ then $∅≠F(X)⊆X$

3. For every $x\ne y$ there exists an ordinal $α$ such that $(V_α,∈,x)\not\equiv(V_α,∈,y)$

Another application of ordinals (and the cumulative hierarchy) is that it gives as a way to measure height of stuff. The rank of a set $x$ is the first ordinal $α$ such that $x\in V_{α+1}$. An example it is useful is: if $M$ is a model of $\sf ZF$, and $N\prec M$ of height $β$, then $\{x\in M\mid M∃α∈N\;\models"α\text{ is an ordinal and the rank of $x$ is $<α$}"\}=V_β^M$ satisfy $N\preceq V_β^M\prec M$, such $β$ is a "correct cardinal", so an elementary submodel of the universe implies that there exists a transitive elementary submodel of the universe.

This is only touching the edge of the iceberg of the usefulness of a cumulative hierarchy.

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A different point you had in your post is, why not just look at equivalence classes? The problem of this approach is that each equivalent class (apart from $0$) is a proper class, not a set.

The way to deal with this problem is something called Scott's trick, which allow you to look at disjoint proper classes as a collection of disjoint sets.

In $\sf ZF$, without $\sf AC$, this is indeed how we define general cardinals. The problem is that we need for this trick some cumulative hierarchy that cover all of $V$, whose construction uses ordinals.

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Indeed in all of the above, if you find some proper class $C$ that is well-ordered and set-like (every proper-initial segment is a set), then you can replace the ordinals with initial segments of $C$ and get basically the same results, but the problem is to find such $C$ without already using the ordinals.