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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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    I think you used the wrong "not subset" symbols. Is it supposed to be $\nsubseteq$ (`\nsubseteq`) instead of $\subsetneq$ (`\subsetneq`)?2012-11-26
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    @kahen Oops, typo. Thanks for pointing it out!2012-11-26
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    "Of course, since $V_\omega$ is a model of $ZFC$ without the axiom of infinity, neither $\omega$ nor $P(\omega)$ are elements of $V_\omega$." That's not really valid. _Every_ model of ZFC is also, in particular, a model of ZFC$-$Infinity, but we cannot conclude that $\omega$ or $\mathcal P(\omega)$ cannot be elements of any model of ZFC.2012-11-26
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    @HenningMakholm Did I just phrase it badly? How about if I replace ZFC with ZFC' := ZFC $\setminus \{$ Axiom of Infinity $\}$?2012-11-26
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    @Matt: That makes no difference -- _removing_ axioms is never going to _stop_ anything from being a model. I suspect what you're aiming for is ZFC$-$Infinity$+(\neg$Infinity$)$, but it would be easier just to note that $\omega$ and $\mathcal P(\omega)$ are both infinite, and therefore in particular not _hereditarily_ finite.2012-11-26
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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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    Thank you! So one can think of $V_\omega$ as $\omega$ if one is talking about models?2012-11-26
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    @Matt: Yes. Except for purposes where the difference is important, of course. :-)2012-11-26
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    @Matt: You could also read about the enumeration in [this old thread](http://math.stackexchange.com/questions/3953/a-nice-enumeration-of-r-omega/). ($R(\omega)$ is also used to denote the same set as $V_\omega$.)2012-11-26
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    @AsafKaragila Thank you for pointing this out!2012-11-26