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We are working in $\mathsf{ZF}$. Let $X$ be a set. Let $A$ be the class of all injections $f: \alpha \to X$ for arbitrary ordinals $\alpha$.

I am quite sure that, in fact, $A$ is a set, since if not, there is an injection from every ordinal $\alpha$ into our set $X$. Hence, $X$ is too large to be a set and must be a class. Contradiction.

In $\mathsf{ZFC}$ it is quite clear, that $A$ is a set, since there cannot be an injection from $\alpha$ to $X$ if $|\alpha| > |X|$. Therefore, for all such $\alpha$ where there is an injection we have that $\alpha < |X|^+$, i.e. $A \subseteq \mathcal{P}(|X|^+ \times X)$ and therefore $A$ is a set by separation.

The second argument makes use of $\mathsf{AC}$ since we need this to make sure that every set has a cardinality.

First of all, please let me know if I am right. If so, I would like to know how I can prove this by making use only of the axioms of $\mathsf{ZF}$. I tried to get a contradiction using the Replacement Scheme, but this did not seem to work.

namsap
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No. You're absolutely wrong when you say that the second argument uses the fact that $A$ "has a cardinality"

The axiom of choice makes sure that every cardinality is an aleph number, but cardinality and its basic arithmetics have absolutely nothing to do with the axiom of choice. Every set has a cardinality, because cardinality of a set $A$ is just the class of all sets which have a bijection with $A$. If $A$ has a bijection with an ordinal, we can pick the least such ordinal as a representative to that cardinality, and for infinite ordinals we call these representatives $\aleph$ numbers.

But even if $A$ cannot be well-ordered, it still has a cardinality. Using the axiom of regularity we can even create sets which represent each cardinality, even if they are not well-ordered.

Now that we've cleared that issue, yes, you are right there is only set-many such injections.

Here we can use Hartogs' theorem, as mentioned by Brian. Or, we can just prove this directly.

Note that if $f\colon\alpha\to A$ is an injection, with $\alpha$ an ordinal, then the restrictions of $f$ to initial segments of $\alpha$ create a well-ordered chain of subsets of $A$, ordered by $\subseteq$ (from which we can also reconstruct $f$, since $f(\beta)$ is the unique element that lies in the difference of the $\beta+1$-th and the $\beta$-th set in the chain).

This means that we can ask how many chains of subsets of $A$ are well-ordered by $\subseteq$ with the difference between two successive sets is a singleton. Since the collection of these can be defined with a first-order formula in the language of set theory (albeit very long and tedious), the collection of all such chains is a subclass of $\mathcal{PPP}(A)$, so the collection of all injections from ordinals into $A$ is also a set.

(As a side note, we didn't gain much here, since this is around 83% of the proof of Hartogs theorem as it is...)

Asaf Karagila
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    In our lecture, cardinality was defined as "If $x$ is a set, let $|x|$ denote the least ordinal $\alpha$ such that there is a bijection $f:\alpha \to x$, if this exists." Later, we proved that $|x|$ is in fact defined for every set $x$ using $\mathsf{AC}$ (implying the wellordering principle implying that cardinality is defined). So I thought that in $\mathsf{ZF}$ there are sets without a cardinality? I think you are using a different definition of cardinality - or are these in fact the same? (But then I don't understand why my lecturer did not prove the existence without $\mathsf{AC}$) – namsap Jan 23 '15 at 10:17
  • Many people don't know how to define the cardinals internally without the axiom of choice; many people believe that it's not important enough to talk about non-choice situations; many people simply confuse the idea of cardinality being an equivalence relation on sets, and cardinals being "material" representatives for each cardinality. I don't know who is your teacher, and I can't tell you why they did what they did. The point is that you were taught wrong. Here is something I wrote on the topic. [...] – Asaf Karagila Jan 23 '15 at 12:14
  • [...] You can also find several questions and answers on this site on this very topic. One, two, three, four, and there are probably even more. – Asaf Karagila Jan 23 '15 at 12:16
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You want the Hartogs number of the set $X$: it’s the least well-ordered cardinal that can’t be injected into $X$, and its existence can be proved in $\mathsf{ZF}$.

Briefly, if $X$ is a set, so is

$$\mathscr{W}=\{\langle S,\preceq\rangle\in\wp(X)\times\wp(X\times X):S\subseteq X\text{ and }\preceq\text{ well-orders }X\}\;.$$

For each $\langle S,\preceq\rangle\in\mathscr{W}$ there is a unique ordinal $\alpha$ such that $\langle\alpha,\in\rangle$ is isomorphic to $\langle S,\preceq\rangle$, and the axiom schema of replacement ensures that the family of those ordinals is a set, $A$. It’s not hard to see that $A$ is in fact some ordinal $\alpha$, so it must be the least ordinal that can’t be injected into $X$, and it’s clear that it must in fact be a cardinal.

Brian M. Scott
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