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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)? –  kahen Nov 26 '12 at 13:07
    
@kahen Oops, typo. Thanks for pointing it out! –  Rudy the Reindeer Nov 26 '12 at 13:09
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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. –  Henning Makholm Nov 26 '12 at 13:26
    
@HenningMakholm Did I just phrase it badly? How about if I replace ZFC with ZFC' := ZFC $\setminus \{$ Axiom of Infinity $\}$? –  Rudy the Reindeer Nov 26 '12 at 13:35
    
@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. –  Henning Makholm Nov 26 '12 at 13:38

1 Answer 1

up vote 4 down vote accepted

$V_\omega$ 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? –  Rudy the Reindeer Nov 26 '12 at 13:37
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@Matt: Yes. Except for purposes where the difference is important, of course. :-) –  Henning Makholm Nov 26 '12 at 13:41
    
Thank you very much : ) –  Rudy the Reindeer Nov 26 '12 at 13:42
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@Matt: You could also read about the enumeration in this old thread. ($R(\omega)$ is also used to denote the same set as $V_\omega$.) –  Asaf Karagila Nov 26 '12 at 16:58
    
@AsafKaragila Thank you for pointing this out! –  Rudy the Reindeer Nov 26 '12 at 20:44

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