# Infinitely iterated power set

1. Jul 10, 2005

### Don Aman

A buddy and I were wondering if there is a way to define a sort of infinite power set in the following way:

You can make an inclusion map from X to P(X) if map each element of X to the singleton set containing it in P(X). Thus you have this chain of maps

$$X\subset \mathcal{P}(X) \subset \mathcal{P}(\mathcal{P}(X)) \subset \dotsb \subset \mathcal{P}^n(X)\subset \dotsb$$

is there a limit, in the sense of category theory? Can it simply be the union (of the images under inclusion) of all these sets?

I think probably that construction should exist. The category of sets is complete, so all small limits exist.

On the other hand, if you just think about what the final result of such a process will look like, it doesn't really look like a set. For example, if you start with the empty set, you should get the collection of all possible grammatical pairings of opening and closing braces. I think this collection is not well-founded, i.e. is an element of itself, and is therefore not a set.

So my question is: does this construction work? Is it a limit? And if so, is the result a set?

2. Jul 10, 2005

### matt grime

but you're taking a colimit and not a limit.

3. Jul 10, 2005

### Hurkyl

Staff Emeritus
But it's not an inclusion map. :tongue: (It is a monic, though)

Set theoretically, if you have an operator T satisfying S <= T(S), then you can inductively define Tα for any ordinal number α. The value for a limit ordinal is as you guess: simply taking the nested union. I.E.

$$T^0(S) = S$$
$$T^{\alpha+1}(S) = T(T^\alpha(S))$$
$$T^\beta(S) = \bigcup_{\alpha < \beta} T^\alpha(S)$$

(Where β denotes any nonzero limit ordinal)

But, since S is not a subset of P(S), this approach does not work.

Now, as for the colimit in Set, remember that they're only defined up to isomorphism. What you get is merely the least upper bound of the cardinals |X|, |P(X)|, |P2(X)|, ...

Now, there is a (very) useful alternative: use the operator:

$$Q(S) := S \cup \mathcal{P}(S)$$

Here, we have that S is a subset of Q(S), so the aforementioned inductive definition works for applying Q ω (= {0, 1, 2, ...}) times. The result is called the superstructure on S (or something like that), and is fairly useful for various things, such as nonstandard analysis.

Last edited: Jul 10, 2005
4. Jul 11, 2005

### Don Aman

well I had in mind to make X into a subset of P(X) by identifying each element of X with the singleton set containing it in P(X). Like you do for the direct limit in group theory. But I think your idea of using P(X) + X instead is prettier, and achieves pretty much the same thing.

So it seems pretty straightforward. Why do I think that the resulting object won't be a well-founded set? Hmm