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More formally, we can state the Transfinite Recursion Theorem as follows. Given a class function $G\colon V\to V$, there exists a unique transfinite sequence $F\colon\mathrm{Ord}\to V$ (where $\mathrm{Ord}$ is the class of all ordinals) such that $F(\alpha) = G(F\upharpoonright\alpha)$ for all ordinals $\alpha$. (Wikipedia, transfinite induction)

First question is, what does $\upharpoonright$ mean? Also, what exactly is $F$ in this usage? $F$ seems to be some form of function, but it says its transfinite sequence...

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As Brian mentions, the $\upharpoonright$ symbol denotes the restriction of the function (on the left) to the set (on the right).

In the theorem, $F$ is a function whose domain is the class $\mathbf{On}$ of all ordinals. As $\mathbf{On}$ is ordered by $<$ it is nice to think about $F$ instead as a "sequence" indexed by all ordinals: $F = \langle x_\alpha : \alpha \in \mathbf{On} \rangle$ (so that $F(\alpha) = x_\alpha$). Note that given any ordinal $\alpha$ as $\mathbf{On} \cap \alpha = \alpha$ it follows that the restriction $F \upharpoonright \alpha = \left( \langle x_\alpha : \alpha \in \mathbf{On} \rangle \right) \upharpoonright \alpha$ is just the $\alpha$-sequence $\langle x_\xi : \xi < \alpha \rangle$. The theorem then tells us what the $\alpha$th coordinate of $F$ is: $$x_\alpha = G ( \langle x_\xi : \xi < \alpha \rangle ).$$ Note that since $\alpha$ is a set it follows that the restriction $F \upharpoonright \alpha = \langle x_\xi : \xi < \alpha \rangle$ is also set, and is thus a legitimate argument for the function $F$.


Depending on the axiom system used, the exact meaning of this theorem may differ.

  • If you are using some set theory with classes as objects, then $F$ is just some class which is a function defined on all ordinals. As classes really exist in such theories, we don't have to look any deeper than this.
  • More likely, you are instead looking at ZF(C) where the only objects are sets. Then this theorem — theorem schema, actually — says something quite different. Here classes are just notational placeholders for formulae, so $G$ is really some formula $\theta ( x , y , \ldots )$ (with possible parameters), and $G(x) = y$ is an abbreviation for $\theta ( x , y , \ldots )$. The theorem then says that given any appropriate formula $\theta ( x , y , \ldots )$ (one defining a function on all of $\mathbf{V}$) there is another formula $\phi (x , y , \ldots )$ (defining a function on $\mathbf{On}$) such that for any ordinal $\alpha$ we have that $$\phi ( \alpha , y , \ldots ) \Leftrightarrow \theta ( \{ \langle \xi , z \rangle : \xi < \alpha , \phi ( \xi , z , \ldots ) \} , y , \ldots ).$$ The important point here is that such a formula can be constructed, and so we can ignore the fine details of the construction and just think of $F$ as a function in the normal sense; this is, in my opinion, the whole point of introducing classes as meta-linguistic objects in the first place.
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Just got confused on another matter: why would we need more than two variables or parameter (I am referring to the predicate or formula) when describing $G(x) = y$? –  Transfinite Lover Sep 16 '12 at 9:56
    
@TransfiniteLover: We want to have parameters available to us because not everything is definable without parameters! For example, we can use Transfinite Recursion to get $\langle A^{(\alpha)} : \alpha\in\mathbf{On} \rangle$ where $A^{(\alpha)}$ is the $\alpha$th Cantor-Bendixson derivative of $A$ in some topological space $X$. Now, if all of $X$, its topology, and $A$ are definable without parameters, one could instead apply Transfinite Recursion to unparametrized formulae in which these definitions take part. (cont...) –  Arthur Fischer Sep 16 '12 at 10:30
    
@TransfiniteLover: (...inued) However, any of these objects is undefinable without parameters (say the topology on $X$ involves a specific nonprincipal ultrafilter on $\omega$) then we couldn't get away with unparametrized formulae, but we might still want to talk about this sequence of sets. Of course, with ordered pairs we would never need more than one parameter. –  Arthur Fischer Sep 16 '12 at 10:30
    
The point "there is another formula $\phi(x,y,\ldots)$" is subtle. Of course we can give that formula explicitly, in prose: $x$ is an ordinal and there exists a (set!) function $f\colon x+1\to V$ such that $f$ obeys the recursion. The fact that we can quantify over (set) functions makes transfinite recursion in fact easier than common recursion on $\mathbb N$ in Peano arithmetic. There, in order to see that "there is a formula" without referencing sets, one needs Gödel-like tricks to encode finite sequences in numbers (e.g. with the Chinese Remainder Theorem). –  Hagen von Eitzen Sep 16 '12 at 11:55

$F\upharpoonright\alpha$ is the restriction of the function $F$ to the set $\alpha$. You may be more accustomed to $F|\alpha$ or $F|_\alpha$ for this. Every sequence, transfinite or otherwise, is a function. For instance, a sequence $\langle x_n:n\in\omega\rangle$ of real numbers is just a function $x:\omega\to\Bbb R:n\mapsto x_n$. A sequence $\langle k_0,k_1,\dots,k_{n-1}\rangle$ of natural numbers is just a function $k$ from $n$, thought of as $\{0,1,\dots,n-1\}$, to $\Bbb N$: $k(i)=k_i$.

Here $F$ is a proper class sequence/function defined on the ordinals and taking sets as values. Note that even though $F$ is a proper class, the axiom schema of replacement (together with some of the other axioms) ensures that $F\upharpoonright\alpha$ is a set for every ordinal $\alpha$, so that it is meaningful to talk about $G(F\upharpoonright\alpha)$: $G$ requires sets as input.

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