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Definitions

  • $(a,b) := \{\{a\},\{a,b\}\}$
  • $\Bbb N := \{n \in \infty \mid \forall A (0 \in A \land \forall n (n \in A \to n \cup \{n\} \in A) \to n \in A) \}$, where $\infty$ is the set guaranteed to exist by the axiom of infinity.
  • $\Bbb Z := \Bbb N \times \Bbb N / \sim$ where $(a,b) \sim (c,d) \iff a+d=b+c$
  • $\Bbb Q := \Bbb Z \times (\Bbb N \setminus \{0\}) / \sim$ where $(a,b) \sim (c,d) \iff ad = bc$
  • $\Bbb R := \{S \subseteq \Bbb Q \mid \forall x \forall y (x \in S \land y < x \to y \in S) \land \\ \exists m (m \in \Bbb Q \land \forall x (x \in S \to x < m)) \land \\ \forall m (m \in S \to \exists n (n \in S \land m < n)) \}$

So $\Bbb R$ is constructed using Dedekind cuts.

Examples

We have: $$\varnothing = 0_\Bbb N \in \{0_\Bbb N\} \in (0_\Bbb N,0_\Bbb N) \in 0_\Bbb Z \in \{0_\Bbb Z\} \in (0_\Bbb Z,1_\Bbb N) \in 0_\Bbb Q \in 1_\Bbb R \in \Bbb R$$

Also: $$\varnothing = 0_\Bbb N \in 1_\Bbb N \in \{1_\Bbb N\} \in (0_\Bbb Z,1_\Bbb N) \in 0_\Bbb Q \in 1_\Bbb R \in \Bbb R$$

This might faciliate the calculation.

Question

Where is $\Bbb R$ in the von Neumann hierarchy and the constructible hierarchy?

In other words, what is the least ordinal $\alpha$ and the least ordinal $\beta$ such that $\Bbb R \in L_\alpha$ and $\Bbb R \in V_\beta$?

Kenny Lau
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    $\mathbb R$ arrises in $V_{\omega+n}$ for some small natural $n$ whose precise value is sensitive to details of the construction of $\mathbb R$ that we don't usually care about (such as: Are elements of $\mathbb Z$ and $\mathbb Q$ equivalence classes of pairs, or canonical representativies?) – hmakholm left over Monica Dec 20 '17 at 02:45
  • @HenningMakholm as is in my example, elements of $\Bbb Z$ are sets of pairs. – Kenny Lau Dec 20 '17 at 02:46
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    Then, unless I have a fencepost error somewhere, each element of $\mathbb Z$ is in $V_{\omega+1}$, pairs of such elements are in $V_{\omega+3}$ (assuming that we use Kuratowski pairs); each element of $\mathbb Q$ is in $V_{\omega+4}$, each Dedekind cut is in $V_{\omega+5}$, and $\mathbb R$ itself is in $V_{\omega+6}$. – hmakholm left over Monica Dec 20 '17 at 02:50
  • @HenningMakholm I did define pairing on the first line. – Kenny Lau Dec 20 '17 at 02:51
  • Notice that I paired elements of $\Bbb Z$ with elements of $\Bbb N_{>0}$ – Kenny Lau Dec 20 '17 at 02:54
  • @HenningMakholm Intuitively say, does $L_{\omega_1^L}$ contain all the sets constructible by a finite number of formulas quantifying over any constructible sets constructed by countable recursion ? – reuns Dec 20 '17 at 03:21

1 Answers1

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The question of where $\mathbb{R}$ lies in the cumulative hierarchy is sensitive to the exact definitions we use; it will always be in some $V_{\omega+n}$ for $n$ finite and small, but the precise value of $n$ may vary between reasonable choices of definition. At a glance I think your definition gives $n=6$ (Henning's calculation looks right to me), but I could be missing something.

However, the situation with regards to the constructible hierarchy is much more canonical and interesting:

  • First of all, it is consistent that $\mathbb{R}\not\in L$: this will happen exactly when there is a non-constructible real.

  • So really you want to ask about $L$'s version of the real numbers, $\mathbb{R}^L$. It turns out that this set shows up first in $L_{\omega_1^L+1}$ (where "$\omega_1^L$" is the least ordinal which $L$ thinks is uncountable - we may of course have $\omega_1^L<\omega_1$, just as we may have $\mathbb{R}^L\subsetneq\mathbb{R}$). The proof goes roughly:

    • First, by the condensation lemma every real in $L$ is in $L_{\omega_1^L}$.

    • Since being a real is a definable property, this means that $\mathbb{R}^L$ is a definable subset of $L_{\omega_1^L}$ - that is, $\mathbb{R}^L\in L_{\omega_1^L+1}$.

    • Finally, it can be shown (see the discussion here) that for every $\alpha<\omega_1^L$ there is a real $r$ in $L$ but not in $L_\alpha$. This shows that $\mathbb{R}^L\not\in L_\alpha$ for any $\alpha<\omega_1^L$, and so $\mathbb{R}^L\not\in L_{\omega_1^L}$ either.

    • Combining the previous two bullet points, we have that $\mathbb{R}^L$ appears in the $L$-hierarchy exactly at stage $\omega_1^L+1$.

Noah Schweber
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