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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 $\operatorname{Ord}(X) \cong X$.

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

2 Answers 2

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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 $(\alpha,<)$.

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    $\begingroup$ +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. $\endgroup$ Apr 15, 2020 at 15:14
  • $\begingroup$ @AlbertoTakase: You don't care if $W$ is empty, though. $\endgroup$
    – Asaf Karagila
    Apr 15, 2020 at 15:17
  • $\begingroup$ @AsafKaragila in this context isn't $\min$ equivalent to $\bigcap$? $\endgroup$ Apr 15, 2020 at 15:18
  • $\begingroup$ @AlbertoTakase: Yes, that is correct. $\endgroup$
    – Asaf Karagila
    Apr 15, 2020 at 15:19
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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.

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  • $\begingroup$ So we don't need replacement axiom? $\endgroup$
    – user745578
    Apr 15, 2020 at 13:55
  • $\begingroup$ 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? $\endgroup$
    – user745578
    Apr 15, 2020 at 13:56
  • $\begingroup$ 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. $\endgroup$
    – drhab
    Apr 15, 2020 at 13:59
  • $\begingroup$ Alright. Thanks for the effort though! $\endgroup$
    – user745578
    Apr 15, 2020 at 14:00
  • $\begingroup$ On your second comment: every set can be well-ordered in ZFC right? $\endgroup$
    – drhab
    Apr 15, 2020 at 14:00

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