# Why is $\{\alpha: \alpha \mathrm{\ ordinal \ and \ \alpha \ equipotent \ with \ X}\}$ a set?

Given a set $$X$$, we define

$$|X|:= \min\{\alpha: \alpha \mathrm{\ ordinal \ and \ \alpha \ equipotent \ with \ X}\}$$

Here equipotent means there is a bijection between both sets.

Why is $$\{\alpha: \alpha \mathrm{\ ordinal \ and \ \alpha \ equipotent \ with \ X}\}$$ a set in ZF(C)? What axioms does one use?

I tried to use replacement with well-orders on $$X$$ but I'm not sure if this work. I know one can associate to every well-order on $$X$$ a unique ordinal $$\operatorname{Ord}(X)$$ with $$\operatorname{Ord}(X) \cong X$$.

• Even if it's not a set there is no problem in defining the minimum as long as the collection is non-empty. Apr 15, 2020 at 14:51
• I see, but I'd love to know if it is a set anyway.
– user745578
Apr 15, 2020 at 14:54

This is indeed a set. Namely consider first the set $$W$$ of well-orders on $$X$$.

Why is $$W$$ a set?

A well-order on $$X$$ is a binary relation, thus it is a subset of $$X \times X$$, i.e. an element of $$P(X\times X)$$. By the comprehension axiom (sometimes called separation), $$W = \{ E \in P(X \times X) : E \text{ is a well-order on } X \}$$ is a set.

You know already that for each well-order $$E$$ on $$X$$, there is a unique ordinal $$\alpha$$ so that $$(X,E)$$ is isomorphic to $$(\alpha, <)$$. Thus by applying the replacement scheme (which is indeed needed), we have that $$\{\alpha : \exists E \in W ( (X,E) \text{ is isomorphic to } (\alpha,<) \}$$ is a set. But this is exactly the set that you are interested in since any bijection $$f$$ from $$X$$ to an ordinal $$\alpha$$, induces the well-order $$E_f$$ on $$X$$, where $$x E_f y$$ iff $$f(x) < f(y)$$. $$f$$ is then also an isomorphism from $$(X,E)$$ to $$(\alpha,<)$$.

• +1: (For OP) An additional axiom is required to be able to take the $\min$ ultimately defining $|X|$. The axiom of choice is used to ensure $W$ is nonempty. Apr 15, 2020 at 15:14
• @AlbertoTakase: You don't care if $W$ is empty, though. Apr 15, 2020 at 15:17
• @AsafKaragila in this context isn't $\min$ equivalent to $\bigcap$? Apr 15, 2020 at 15:18
• @AlbertoTakase: Yes, that is correct. Apr 15, 2020 at 15:19

Let $$\beta:=\mathsf{Ord}(\mathcal P(X))$$.

Then $$\{\alpha: \alpha \mathrm{\ ordinal \ and \ \alpha \ equipotent \ with \ X}\}$$ is a subset of $$\beta$$ hence is a set by axiom of separation.

• So we don't need replacement axiom?
– user745578
Apr 15, 2020 at 13:55
• I'm sorry but I don't see how this works. $\mathcal{P}(X)$ is not well-ordered (it is not a total order), so how can we associate an ordinal to it?
– user745578
Apr 15, 2020 at 13:56
• I don't dare to say that (am not a set-theorist). In that sense my answer is incomplete and only shows that in ZFC the collection must be a set. Just do not accept this anwer and let's wait what set-theorists have to say about it. Apr 15, 2020 at 13:59
• Alright. Thanks for the effort though!
– user745578
Apr 15, 2020 at 14:00
• On your second comment: every set can be well-ordered in ZFC right? Apr 15, 2020 at 14:00