Elementary embedding from hereditarily finite to hereditarily countable sets

Question

Let $(H(\omega),\in\vert_{H(\omega)})$ be the structure of hereditarily finite sets and $(H(\omega_1),\in\vert_{H(\omega_1)})$ be the structure of hereditarily countable sets. Does there exist an elementary embedding from $(H(\omega),\in\vert_{H(\omega)})$ into $(H(\omega_1),\in\vert_{H(\omega_1)})$ with respect to the language of set-theory, I.e. that the first order language with one binary predicate $\in$.

Answer 1

They aren't even elementary equivalent (which is a stronger fact than the mere non-existence of an elementary embedding). HINT: only $H(\omega_1)$ has elements which are infinite.

In more detail:

$H(\omega_1)$ satisfies "There is an $x$ with a non-surjective injection to itself" - that is, "There is a Dedekind-infinite set." For example, take $x=\omega$. Keep in mind that every function between two hereditarily countable sets is itself a hereditarily countable set, so $H(\omega_1)$ sees all the basic combinatorics of hereditarily countable sets that $V$ does. By contrast of course $H(\omega)$ does not satisfy that sentence.

---

A more interesting problem is to show that $H(\omega_1)\not\equiv H(\omega_2)$. Here we need to think about cardinalities:

Only $H(\omega_2)$ satisfies "There are two Dedekind-infinite sets of different cardinalities."

And this extends to showing that $H(\omega_m)\not\equiv H(\omega_n)$ for all distinct finite $m,n$. In fact, there's a stronger phenomenon occurring here. Say that an ordinal $\alpha$ is $H$-isolated iff there is no $\beta\not=\alpha$ with $H(\omega_\alpha)\equiv H(\omega_\beta)$. The arguments above really show that every finite ordinal is isolated, and with a bit of thought we can push this much further (think about counting limit ordinals, counting limits-of-limit ordinals, etc.). Indeed, it's not hard to show that the smallest non-isolated ordinal must be rather large (of course by a counting argument it is guaranteed to be $<(2^\omega)^+$).