Cardinalities in non-standard models of $\mathsf{ZF}$ are not generally reflected externally, but certain internal facts about cardinalities are always externally reflected (e.g., any internally infinite set is externally infinite; if $|A| \leq |B|$ internally, then the same holds externally; the internal powerset of a given set is no larger than the external powerset; etc.). (By an abuse of notation, I'll write $|A|$ for $|\{a \in M : M \models a \in A\}|$.)


In $\mathsf{ZFC}$, $\omega$ embeds into any infinite set, so for any $M \models \mathsf{ZFC}$, we always have that $|\omega^M| \leq |A|$ for any $A \in M$ such that $M \models \text{“}A\text{ is infinite"}$ (i.e., $A$ is internally infinite).


In $\mathsf{ZF}$ on the other hand, it can be the case that there are infinite sets admitting no injection from $\omega$. At most we know that each initial segment of $\omega$ injects into a given infinite set. Externally, this means that if $M \models \mathsf{ZF}$ and $M \models \text{“}A\text{ is infinite"}$, then $|A| \geq |n|$ for each $n \in \omega^M$. This leaves open the possibility that $A$ is actually strictly smaller than $\omega^M$, but not that much smaller. In particular, we must have that $|\omega^M| \leq |A|^+$.


My question is whether that bound is ever exact.



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> **Question:** Is there a model $M$ of $\mathsf{ZF}$ with an $A \in M$ such that $M\models\text{“}A\text{ is infinite"}$ and $|\omega^M| = |A|^+$?
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Certain basic model-theoretic facts about two-cardinal models and Vaughtian pairs tell us that this is equivalent to the existence of a Vaughtian pair of models of $\mathsf{ZF}$ in which $\omega^M$ grows but some internally infinite $A$ does not grow. ​Stated more formally:


**Fact:** The following are equivalent:


* The above question has a positive answer.
* There is a model $M\models \mathsf{ZF}$ with an internally infinite element $A$ such that $|A| = \aleph\_0$ and $|\omega^M| = \aleph\_1$.
* There is a model $N \models \mathsf{ZF}$ with an internally infinite element $B$ and an elementary extension $N' \succ N$ satisfying that there is an $n \in N' \setminus N$ such that $N' \models n \in \omega$ but for all $b \in N'$, if $N' \models b \in B$, then $b \in N$.


It's possible to get a more detailed equivalent with the omitting types theorem, but it's a bit awkward to state.



Can a non-standard model $M$ of $\mathsf{ZF}$ contain an internally infinite set that is externally smaller than $\aleph_0^M$?
The answer is yes. Let $M$ be a countable transitive model of $\mathsf{ZF}$ containing an amorphous set $A$ (i.e., every subset of $A$ is either finite or co-finite).


Let $T$ be the theory consisting of the elementary diagram of $M$ together with a fresh constant $c$ and the sentences $c \in \omega$ and $n < c$ for each standard natural $n$. Let $(a\_i)\_{i<\omega}$ be an (external) enumeration of $A$. By the omitting types theorem and the argument in the question, it is sufficient to show that the type $\Sigma(x) = \{x \in A\}\cup\{x \neq a\_i : i < \omega\}$ is not principal (i.e., for each consistent formula $\varphi(x)$, it is not the case that $\varphi(x) \vdash \Sigma(x)$). (If we can show this then we can build a model of $T$ in which $\Sigma(x)$ is omitted, which is precisely the Vaughtian pair condition we need.) Suppressing constants from $M$, we'll write these formulas as $\varphi(x,c)$ to emphasize the role of $c$.


By comprehension, we have that for any $M$-formula $\varphi(x,y)$, there is a set $Q \subseteq A \times \omega$ such that for any $a \in A$ and $n < \omega$, $\varphi(a,n)$ holds if and only if $\langle a,n \rangle \in Q$. This means that it is sufficient to consider formulas of the form $\langle x,c \rangle \in Q$ for $Q \subseteq A\times \omega$.


Given $Q \subseteq A \times \omega$ and $n < \omega$, we'll write $Q(n)$ for $\{a \in A : \langle a,n \rangle \in Q\}$.



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> **Lemma.** For any $Q \subseteq A \times \omega$, there is a finite sequence $B\_0,\dots,B\_{k-1}$ of subsets of $A$ such that for all $n < \omega$, $Q(n) = B\_i$ for some $i<k$.
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*Proof.* (This is probably standard somewhere.) Let $Q'$ be the set satisfying that for each $n < \omega$, $Q'(n) = Q(n)$ if $Q(n)$ is finite and $Q'(n)= A \setminus Q(n)$ if $Q(n)$ is infinite. Let $A\_n = Q'(n) \setminus \bigcup\_{m<n}Q'(m)$. It must be the case that $A\_n = \varnothing$ for all sufficiently large $n$ (otherwise we would have a surjection from a subset of $A$ onto $\omega$, which cannot happen). This implies that there is a finite $C \subset A$ such that $Q'(n) \subseteq C$ for all $n$. Let $B\_0,\dots,B\_{k-1}$ be an enumeration of all subsets of $C$ and all complements of subsets of $C$ (which is necessarily finite). $\square$


Now fix a set $Q \subseteq A \times \omega$. We just need to show that either $T \vdash \neg (\exists a \in A) \langle a,c \rangle \in Q$ or there is an $i<\omega$ such that $T \cup \{\langle a\_i,c \rangle \in Q\}$ is consistent. An immediate compactness argument shows that $T \vdash \neg (\exists a \in A) \langle a,c \rangle \in Q$ if and only if $Q(n) = \varnothing$ for all sufficiently large $n<\omega$, so assume that $Q(n)$ is not eventually empty. By the lemma, there must be some non-empty $B \subseteq A$ such that $Q(n) = B$ for infinitely many $n<\omega$. Fix some $a\_i \in B$. We now have by compactness that $T\cup \{\langle a\_i,c \rangle \in Q\}$ is consistent, whence the formula $\langle x,c \rangle \in Q$ does not isolate $\Sigma(x)$.


Since we can do this for any $Q \subseteq A \times \omega$, we have by the omitting types theorem that there is an elementary extension $N \succ M$ with the property that $A^N = A^M$ and $\omega^N \supset \omega^M$.


To finish we just apply the standard Vaughtian pair argument. Using the pair $(N,M)$, we can build a pair $(D\_1,D\_0) \equiv (N,M)$ with $D\_0$ and $D\_1$ isomorphic and $\omega$-homogeneous. Then we can build an elementary chain $(D\_\alpha)\_{\alpha < \omega\_1}$ starting with $D\_0$ and $D\_1$ satisfying that $(D\_{\alpha+1},D\_\alpha)\cong (D\_1,D\_0)$ for each $\alpha$. The union $D = \bigcup\_{\alpha < \omega\_1}$ then has the property that $|A^{D}|=\aleph\_0$ yet $|\omega^{D}| = \aleph\_1$, which is what we were after.


