Injection of $\mathbb N$ into proper classes In ZFC, given a proper class, is it possible to prove there is injection of the natural numbers $\mathbb N$ into it?
If not, are there proper classes which do not have such injection?
 A: The answer is yes.
Recall the cumulative hierarchy: $V_0=\emptyset$, $V_\lambda=\bigcup_{\alpha<\lambda}V_\alpha$ for $\lambda\in ON$ a limit, and $V_{\alpha+1}=\mathcal{P}(V_\alpha)$. 
ZF alone proves (the crucial axioms are Foundation and Replacement) that $V=\bigcup_{\alpha\in ON} V_\alpha$, and that each $V_\alpha$ is a set. EDIT: the symbol "$V$" denotes the whole set-theoretic universe; so what I wrote above really means "ZF proves that every set $x$ is in some $V_\alpha$." So if $C$ is a proper class, we must have that for every $\alpha\in ON$ there is a $\beta>\alpha$ such that $C\cap (V_\beta\setminus V_\alpha)\not=\emptyset$, that is, any proper class has elements of arbitrarily high rank.
So - without choice! - we can define an $\omega$-sequence of ordinals $\alpha_0<\alpha_1<\alpha_2< . . . $ such that for each $i$, there is an element of $C$ in $V_{\alpha_i+1}\setminus V_{\alpha_i}$. This in turn gives us - again without choice! - an $\omega$-sequence of sets $X_i$ such that


*

*$X_i\subset C$ and is nonempty for every $i$, and

*$X_i\cap X_j=\emptyset$ for $i\not=j$.
Now, apply choice to this sequence.

Note that there's no special role played by $\omega$ here; more generally, ZF proves that for any proper class $C$ and any ordinal $\gamma$, there is a sequence $\langle X_\alpha: \alpha<\gamma\rangle$ such that 


*

*$X_\alpha\subset C$ and is nonempty for every $\alpha$, and

*$X_\alpha\cap X_\beta=\emptyset$ for $\alpha\not=\beta$.
Then, in the presence of choice we can extract an embedding from $\gamma$ into $C$.

And choice is, so far as I know, necessary. I suspect there is a model $W\models ZF$ and a definable-without-parameters proper class $C\subset W$ such that $W$ contains no injection from $\omega$ into $C$; put another way, for each set $x\in W$ the set $x\cap C$ is Dedekind-finite. However, I currently cannot show this.
Another observation is that in fact ZF alone proves that any proper class surjects onto the ordinals (exercise, use the argument above). However, the ordinals need not inject into every proper class, even with choice! There are models of ZFC in which there are proper classes into which the ordinals do not embed. Instead, global choice is needed here.
