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Is there an easy way to prove this? I found a book that suggests the injection $h:V_\omega\to\omega$ defined by

$$h(\{x_1,x_2,\dots,x_n\})=2^{h(x_1)}+2^{h(x_2)}+\cdots+2^{h(x_n)},$$

but I hit some snags while proving that the function is well-defined. Are there any other clever ways to prove this fact? (Note: I am making a formal proof, so I need to be able to justify my steps.)

Edit: Let me see if I can make my difficulty more clear. Before I can start the proof by induction, I need a precise statement of the induction hypothesis, and this is what I am trying to hammer out. The hypothesis:

For each $n\in\omega$, there exists a unique bijection $h_n:V_n\to|V_n|$ such that $$h_n(x)=\sum_{m<|V_{n-1}|}\operatorname{if}(h_{n-1}^{-1}(m)\in x,2^m,0).$$ Moreover, the $h_n$ as defined satisfies $h_n\restriction |V_{n-1}|=h_{n-1}$.

Does this hypothesis contain enough to prove the claim? The reason for the odd way of stating the sum is that it avoids the issue in the linked question (we need to show that addition is a well-defined operation over sets), but as a consequence I have to prove that the $h_n$ is a bijection and not just an injection, and I think I also need the restriction part in order to prove that the union $h=\bigcup_{n\in\omega}h_n$ is the required bijection.

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Mario Carneiro
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  • As for the suggested injection: you can use the Basis Representation Theorem for base $2$. – Lord_Farin May 18 '13 at 08:38
  • @Lord_Farin Actually, that's the easy part (or at least, I have a good idea how to do it). The hard part is that the notated sum is in fact well defined (since sets don't have an ordering, it is tricky to even some up with a formula that gives the right hand side of the equality there). – Mario Carneiro May 18 '13 at 08:41
  • Also related, http://math.stackexchange.com/questions/3953/a-nice-enumeration-of-r-omega/ (note that $R(\omega)$ is another notation for $V_\omega$). – Asaf Karagila May 18 '13 at 12:33
  • Mario, the reference by Kaye and Wong in my answer here sketches the needed details, and the paper by Mycielski they reference does as well. Is there something specific in the argument that is giving you trouble? – Andrés E. Caicedo May 18 '13 at 21:18
  • @AndresCaicedo I've edited the question to reflect my concerns. In my formal proof, I can't make the function $h$ explicitly self-referential, except in certain well-defined ways (namely a function of the form $f(n)=g(f(n-1))$). I also will need to prove the basis representation theorem, since I don't have that available, but that's a tangential issue. – Mario Carneiro May 19 '13 at 01:15
  • @AndresCaicedo Would defining the relation $n,E,m$ in your linked answer help me prove this? I understand that with $h$ as defined in the question, we have $E=h,\circ!\in!\circ, h^{-1}$, but is there an easy way to use $E$ instead of $h$ to prove $|V_\omega|=\omega$? – Mario Carneiro May 19 '13 at 01:24
  • I think the way you are stating things is really how things are done all along. The typical proof of a recursion theorem such as the one you need here (to prove the existence of $h$, say) goes by establishing the existence of approximations (the $h_n$), and showing that these approximations are compatible and unique ($h_{n+1}$ extends $h_n$; $h_n$ is the only function with domain $V_n$ satisfying the requirements $h_n$ satisfies) and so their union is well defined (and gives us the required function $h$). – Andrés E. Caicedo May 19 '13 at 01:50
  • @AsafKaragila Your proof in the linked answer works well on the composite function $h$, but as you can see in my induction hypothesis, I need to prove that $h_n$ is bijective, and that has a finite range. How do you know that $h_n$ is onto, and how do you get the specific range $|V_n|$ (or $2\uparrow\uparrow n$, to be explicit)? – Mario Carneiro May 19 '13 at 20:31
  • Is there a reason you insist on this function of yours and not the one I suggest in the answer I linked? – Asaf Karagila May 19 '13 at 20:47
  • @AsafKaragila The function you use in the link is what I am calling $h$, but I can't define $h$ directly, since it is self-referential. Instead, I am defining $h$ using the approximations $h_n$ (do you have a better way to do this?). But since $h_n$ depends on $h_{n-1}^{-1}$, I need to prove in the induction hypothesis that $h_n$ is at least injective, and I probably need that it is surjective to make sure that $h_{n-1}^{-1}$ is defined everywhere I need it in the definition of $h_n$. – Mario Carneiro May 19 '13 at 21:10
  • @AsafKaragila I also can't simply prove that there exists a unique function $h:V_\omega\to\omega$ satisfying $h(x)=\sum_{y\in x}2^{h(y)}$, because for me, $h$ is potentially a class-function (because I am not assuming infinity), so I can't do anything that requires positing its existence. – Mario Carneiro May 19 '13 at 21:16
  • @Mario: You can define its restrictions to $V_n$'s and then show that it is equal the increasing [class-]union. – Asaf Karagila May 19 '13 at 21:20
  • @AsafKaragila Isn't that what I'm doing? The point I'm making is that in order to show the properties I need, I need more than just $h$ is bijective; I need that each $h_n$ is bijective, and your argument doesn't give that. – Mario Carneiro May 19 '13 at 21:27
  • No, you don't need that every $h_n$ is bijection. You need that every $k$ is eventually in the range. But do note that if $h(V_n)=k$, then every number smaller than $k$ appears in the range of $h$. So you just need to show that $h$ is strictly increasing on the $V_n$'s themselves. – Asaf Karagila May 19 '13 at 21:39

2 Answers2

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To prove that $h$ is well-defined and injective, it is most natural to proceed by induction on the rank of sets. The rank $\mathsf{rk}(x)$ of a set $x$ is an ordinal, defined recursively as follows (see also MathWorld for a definition for all sets):

$$\mathsf{rk}(x):= \sup\{\mathsf{rk}(y)+1: y \in x\}$$

Lemma: The rank of a set is well-defined.

This lemma can be proved using the Axiom of Foundation, see also this question. Since all suprema involved in computing $\mathsf{rk}$ for $x \in V_\omega$ are finite, the results are also finite ordinals.

The only set of rank zero is of course $\varnothing$; we define/obtain $h(\varnothing) = 0$.

Now given a set $X = \{x_1, \ldots, x_m\}$ of rank $n+1$, we use the Axiom of Replacement for $f(x) = 2^{h(x)}$, well-defined and injective by induction hypothesis.

We now have the set $Y = \{2^{h(x_1)},\ldots, 2^{h(x_m)}\}$; using your earlier question this gives rise to a well-defined number $h(X) = \sum Y$, so $h$ is well-defined on $X$ as well.

To show that $h$ is injective, let $X'$ be such that $\mathsf{rk}(X') \le \mathsf{rk}(X)$. Then we can obtain the sets $\{h(x_1),\ldots, h(x_m)\}$ and $\{h(x'_1),\ldots, h(x'_{m'})\}$ by the Basis Representation Theorem. The Axiom of Replacement can be applied to these sets since $h^{-1}$ is a functional relation, and we recover $X$ and $X'$; injectivity of both the Basis Representation and $h^{-1}$ ensures that $h(X) = h(X')$ implies $X = X'$. Thus $h$ is injective and well-defined on sets of rank at most $n+1$.

In conclusion, $h$ is well-defined on $V_\omega$; its injectivity on $V_\omega$ is trivially verified.

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Lord_Farin
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    Actually, you don't need Foundation to prove this, since we only care about sets $x\in V_\omega$, for which we already know they have a rank, more or less by definition. – Mario Carneiro May 18 '13 at 09:41
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    @Mario: Your comment is simply an argument as for why the axiom of foundation holds in $V_\omega$. – Asaf Karagila May 18 '13 at 15:42
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    @Asaf I disagree---Mario's comment simply points out that the Axiom of Foundation is irrelevant to the question and its answer. The relativization of Foundation to $V_\omega$ also seems to me to be irrelevant. – Trevor Wilson May 18 '13 at 22:43
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If you just need to prove that $V_\omega$ is countable, and if you're allowed to use the axiom of choice, then I think you're working too hard. Prove, by induction on natural numbers $n$ that $V_n$ is finite (because the power set of a finite set is finite). Then $V_\omega=\bigcup_{n\in\omega}V_n$ is the union of countably many finite sets, hence countable.

Andreas Blass
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  • I hope you enjoy Eilat! – Asaf Karagila May 19 '13 at 19:29
  • You are correct that using AC, it is relatively trivial. Unfortunately, I have a bug in me that says "if it can be proven without AC, it should be", and I know that this proof doesn't need AC (or even infinity or foundation). – Mario Carneiro May 19 '13 at 20:08