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So I'm going to prefix this question by saying that I probably don't have a great understanding of what I'm asking.

We build the cumulative hierarchy as follows:

$V_0=\emptyset$

For every $\alpha$, $V_{\alpha+1}=\mathcal{P}(V_\alpha)$

If $Lim(\lambda)$, then $V_\lambda=\bigcup _{\alpha<\lambda} V_\alpha$

We then define the Von Neumann universe to be the class $V=\bigcup_{\alpha} V_{\alpha}$

Once we have done this we can prove various things about this such as:

Every set is in some $V_\alpha$

Intuitively we are supposed to picture this as the ordinal numbers being a vertical line starting at $\emptyset$ and going upwards. Then for each $\alpha$ the collection of sets of that rank are horizontal lines, so we get a sort of V shaped picture.

What I am unsure about (and maybe this is a ridiculous question) is where in this construction the real numbers are? We have that the natural numbers are $\omega$- the first transfinite ordinal and that $\omega_1$-the supremum of all countable ordinals but I am unsure where the real numbers come in?

Thanks for any help (sorry if the question is nonsense)

Martin Sleziak
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hmmmm
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3 Answers3

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The real numbers show up in $V_{\omega+n}$ for some small finite $n$ whose precise value is sensitive to the exact details of how you choose to construct the reals.

Even before you have to choose between Dedekind cuts and Cauchy sequences, the rationals are usually constructed as (infinite) equivalence classes of pairs of integers, and the integers themselves as (infinte) equivalence classes of pairs of naturals. Each of these constructions can only happen after $V_\omega$ and contribute a level to $n$, and then the Kuratowski pairs you use in the next construction take a few additional levels to show up.

However, if you tune your constructions specially for the reals to exist early in the Von Neumann hiearchy, you can use canonical representatives rather than equivalence classes to represent integers and rationals. Then every rational is represented by a hereditarily finite set, and then $\mathbb Q$ itself as well as all its subsets will be present already in $V_{\omega+1}$ and you can have $\mathbb R\in V_{\omega+2}$ by Dedekind cuts.

Note, however, that for many set theorists "the reals" tend to mean simply $\mathcal P(\omega)$ rather than $\mathbb R$, and $\mathcal P(\omega)$ certainly arises already in $V_{\omega+2}$.

In any case, you cannot get $\mathbb R$ earlier than $V_{\omega+2}$, because every member of $V_{\omega+1}$ is at most countable.

hmakholm left over Monica
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The real numbers are not an intrinsic object to the universe of set theory. We have a good way of constructing them from the natural numbers, but actually every set of size continuum can be made into the real numbers.

In particular we have that $V_{\omega+1}$ is of size continuum, so in $V_{\omega+2}$ you already have a set of size continuum which can function as the real numbers (e.g. $\mathcal P(\omega)$ ordered by $A\prec B\iff\min(A\Delta B)\in A$).

If you wish to compute another construction of the real numbers, then you can do that manually. Suppose we wish to think of the real numbers as Dedekind-cuts, i.e. subsets of $\mathbb Q$, so we need to find when $\mathbb Q$ enters the universe; but again we have the same problem. What is $\mathbb Q$? Well, we can think of it as a quotient of $\mathbb Z\times\mathbb Z$, so again... when does $\mathbb Z$ enters the universe? Well, $\mathbb Z$ is a quotient of $\omega\times2$.

Let us consider the following rules:

Suppose that $A$ has rank $\alpha$. We know that pairs from $A$, $\langle a,b\rangle=\{\{a\},\{a,b\}\}$, which means that $A\times A\subseteq\mathcal{P P}(A)$, so $A\times A$ has rank of $\underline{\alpha+3}$.

Quotients of $A\times A$ are subsets of $A$, though, so they have rank of $\alpha+1$. Now you need to sit and calculate, if $\omega$ has rank $\alpha$ how do we get to $\mathbb R$?

Furthermore, if you also want the real numbers alongside with additions and other operations, you need to go higher as well, because those operations are only generated at higher stages.

For further reading:

  1. Formalising real numbers in set theory
  2. In set theory, how are real numbers represented as sets?
Community
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Asaf Karagila
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  • Your rank rules give upper bounds, but you can probably do a bit better if you're careful. For example, $\omega\times\omega$ appears already in $V_{\omega+1}$. – Miha Habič Oct 13 '12 at 18:15
  • @MihaHabič: Yes. True, for the case of $\omega$ this is indeed true, but what happens when you take sets like $\mathbb Q$ which are infinite sets of quotients of sets etc.? – Asaf Karagila Oct 13 '12 at 18:19
  • I agree completely. I just thought, if you're not satisfied with saying $\mathbb{R}\in V_{\omega+\omega}$ then you probably want to make sure every increase in rank is necessary. – Miha Habič Oct 13 '12 at 18:27
  • @AsafKaragila Thanks very much, that helped a lot, although now I am slightly confused by something else. If every ordinal is a well ordered set and we are taking power sets and unions how do we get to a set that is not well ordered (the reals), i thought taking the power/union would preserve the well ordering? Thanks for any help – hmmmm Oct 14 '12 at 23:13
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    @hmmmm: Neither power sets nor unions preserve well-orderability in ZF. The axiom of choice is in fact equivalent to the claim "if $A$ can be well-ordered, then $P(A)$ can be well-ordered"; and in fact even requiring that the union of any collection of well-orderable sets is well-orderable is not enough to prove the axiom of choice holds. – Asaf Karagila Oct 14 '12 at 23:17
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    @hmmmm: You can read about this here as well. – Asaf Karagila Oct 15 '12 at 00:40
  • @AsafKaragila Thanks very much for the help, you wrote at the start of the post that any set of size $2^{\aleph_0}$ can be made into the reals. I've been looking in a few textbooks and stuff and can see no mention of this, I was wondering if there was somewhere I could read more about this? Thanks for any help – hmmmm Oct 16 '12 at 16:10
  • @hmmmm: This is trivial. It is called transport of structure. Once you know there is one set of size $2^{\aleph_0}$ which has a certain structure imposed over it, any other set of the same size has a bijection with it, and you can define [essentially] the same structure on your new set. – Asaf Karagila Oct 16 '12 at 17:41
  • Why has $A\times A\subseteq \mathcal P\mathcal P A$ rank $\alpha +3$ and not $\alpha + 2$? I thought we'd increase the rank by 1 with each $\mathcal P$. Also why are quotients of $A\times A$ subsets of $A$? $\mathbb Q$ is a quotient of $\mathbb Z\times\mathbb Z$ but we don't have $\mathbb Q \subseteq \mathbb Z$? Sorry for bumping this old question. – akkarin Mar 07 '16 at 17:42
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    @akkarin: (1) It should always read "at most", but anyway each pair is an element of $\mathcal{PP}(A)$, so each pair has rank $\alpha+2$. The set of pairs, if so, has rank $\alpha+3$ (some sets are closed under pairing, in which case you get $\alpha$ or $\alpha+1$). (2) Don't get me started on $\Bbb{Q\subseteq Z}$, these are just abstract sets and we can realize them in many possible ways. The fact is that there is a subset of $\Bbb N$ which encodes $\Bbb Q$ with its structure and everything, just because both sets are countable. And that was my point, if I remember from 3.5 years ago. – Asaf Karagila Mar 07 '16 at 18:42
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    (My point at least with regards to $\Bbb R$ being realized as $V_{\omega+1}$.) – Asaf Karagila Mar 07 '16 at 18:43
  • @AsafKaragila Just to clarify, I take it you are using the definition of the rank of a set $A$ to be the least ordinal $\alpha$ s.t. $A \in V_{\alpha}$? In other words you take rank $\varnothing$ to be $1$? Otherwise your response to akkarin doesn't make sense to me. – gowrath Aug 07 '17 at 08:47
  • @gowrath: Which response? That $V_{\omega+1}$ can be thought of as $\Bbb R$? Or that taking two power sets adds $2$ to the rank? – Asaf Karagila Aug 07 '17 at 08:50
  • @AsafKaragila Apologies. I am referring to the discussion about adding 2 or 3 to the rank. As I understand, rank$(A)$ is the least ordinal $\alpha$ s.t. $A \subseteq V_\alpha$ in which case, $A \times A \subseteq \mathcal{PP}(V_{\alpha}) = V_{\alpha+2}$, so rank $(A \times A) = \alpha + 2$. Have I made a mistake? – gowrath Aug 07 '17 at 08:54
  • @gowrath: You're probably right. I was young and foolish once and considered the rank to be defined by $\in$ rather than $\subseteq$ (which does have some limited merits). But in case, this is entirely irrelevant since it doesn't change much whether or not the rank is $+2$ or $+3$. Whenever such issues would come up in our classroom, the professor would have usually said "I think that $+5$ should be enough, so let's take $+\omega$" or something of that sort which was an overkill in any conceivable way. :) – Asaf Karagila Aug 07 '17 at 08:56
  • @AsafKaragila You're absolutely right—it doesn't matter. However, I'm young and foolish right now and the foolish mind stresses over petty details :). Thanks for replying to such an old question! – gowrath Aug 07 '17 at 09:00
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The reals are certainly in $V_{\omega+2}$, no? $V_{\omega+1}$ is the first set to contain an infinite set, i.e $\omega$ itself, and since the cardinality of $\mathbb{R}$ is that of $\mathcal{P}(\omega)$, $\mathbb{R} \in V_{\omega+2}$.

Or are you asking about $V_{\gamma}$ where $\gamma$ is the first ordinal whose cardinality is the same as that of $\mathbb{R}$? In that case, there can be no precise answer - since the continuum hypothesis is undecidable in ZFC, you don't know how many cardinals lie between $\omega$ and $\gamma$.

fgp
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