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Let $V_\omega$ denote the set of all hereditarily finite sets. A set $S$ is called hereditarily finite if and only if its transitive closure is finite, that is, $TC(S) = \bigcup \{ S, \bigcup S, \bigcup \bigcup S, \dots \}$ is finite. Let $P(S)$ denote the power set of $S$ and let $\omega$ denote the natural numbers.

I am trying to understand what $V_\omega$ looks like and to this end I thought I could work out the relationship between $V_\omega$ and $P(\omega)$:

Of course, since $V_\omega$ is a model of $ZFC$ without the axiom of infinity, neither $\omega$ nor $P(\omega)$ are elements of $V_\omega$. Hence $P(\omega) \nsubseteq V_\omega$.

On the other hand, $\{\{\{\varnothing\}\}\}$ is in $V_\omega$ but not in $P(\omega)$. Hence $V_\omega \nsubseteq P(\omega)$.

So this is not going to give me any information about $V_\omega$. Yet, since hereditary finiteness is a stronger condition than finiteness ($\{\omega\}$ is a finite set that is not hereditarily finite) I am tempted to think that perhaps $V_\omega$ might somehow be in bijection with a subset of $P(\omega)$.

Question: Is there such a bijection? If not: what's a good intuition to think about $V_\omega$? What does $V_\omega$ look like?

Thanks for your help.

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    @HenningMakholm Of course, thank you. I should've written what you wrote in your last sentence.2012-11-26

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$V_\omega$ consists of exactly the sets you can write down in finite space using only the symbols {, }, and ,.

It is in bijection with $\omega$, by the rule

$f:\omega\to V_\omega \qquad f(n) = \{f(a_1),f(a_2),\ldots,f(a_{k_n})\}$ where $n=2^{a_1}+2^{a_2}+\cdots+2^{a_{k_n}}$ and all the $a_i$s are different.

(This bijection provides the standard proof that Peano Arithmetic is equiconsistent with ZFC$-$Infinity).

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    @AsafKaragila Thank you for pointing this out!2012-11-26