# The Universe of Disco

Tue, 02 Nov 2021

!!\def\kp#1#2{\{\{{#1}\}, \{{#1},{#2}\}\}}!!

!!\def\nwp#1#2{\{\{\{{#1}\}, \emptyset\}, \{\{{#2}\}\}\}}!!

In any case the Kuratowski thing is still simpler. I wonder why Wiener didn't think of it first. … Perhaps he didn't consider the details important.

But today I thought of a specific technical advantage that Wiener's definition has over Kuratowski's. The Kuratowski definition is

$$\langle a, b\rangle = \kp ab$$

and in the special case where !!a=b!!, this reduces to:

$$\kp aa = \{\{a\}, \{a\}\} = \{\{a\}\}$$

This can complicate the proofs. For example suppose you want to prove that if !!\langle p,q\rangle = \langle r,s\rangle!! then !!p=r!! and !!q=s!!. You might like to start by saying that each side represents a set of two elements, and then compare the elements. But you can't, because either might be a set with one element, so there's a special case. Or maybe you have to worry about what happens if !!b=\{a\}!!, is it still all right?

With the Wiener pair, it easy to see that nothing like this can happen:

$$\langle a, b\rangle = \nwp ab$$

Here, no matter what !!a!! and !!b!! might be, !!\langle a, b\rangle!! is a set with two elements. The set !!\{\{a\}, \emptyset\}!! must have exactly two elements; it's impossible that !!\{a\} = \emptyset!!. And since !!\{\{a\}, \emptyset\}!! has two elements, it's impossible for it to equal !!\{\{b\}\}!!, which has one element. So !!\langle a, b\rangle!! is always a two-element set !!\{P_a, P_b\}!!. When !!a=b!! there's no special case. You just get $$\nwp aa$$ which doesn't look any different.

This analysis explains another possibly puzzling feature of Wiener's definition: Why !!\nwp ab!! and not the apparently simpler !!{\{\{a, \emptyset\}, \{b\}\}}!!? But that proposal, like Kuratowski's, has annoying special cases. For example, the proposal collases to !!\{\{\emptyset\}\}!! when !!a=b=\emptyset!!.

The components of a Wiener pair are easy to extract again: the !!P_b!! component is a singleton, satisfying !!\bigcup P_b = b!!, and the !!P_a!! component is a two-element set satisfying !!\bigcup P_a = a!!. This is all considerably simpler than with the Kuratowski pair, where you can't get !!b!! alone, you can only get !!\{a, b\}!! and then you have to take the !!a!! back out, except that if !!a=b!! you musn't. To extract the second component from a Wiener pair is easy. The first thing I thought of was $$\pi_2(p) = \bigcup \{x\in p: |x| = 1\}$$ (“give me the contents of the singleton elements of !!p!!”) but one could also do $$\pi_2(p) = \{x: \{\{x\}\} \in p\},$$ and it's obvious that both of these are correct. [ Addendum 20211112: Obvious perhaps, but wrong; the first one is not correct! ]

The analogous formula given by Wikipedia for the Kuratowski pair is not at all obvious:

$$\pi_2(p) = \bigcup\left\{\left. x \in \bigcup p\,\right|\,\bigcup p \neq \bigcap p \implies x \notin \bigcap p \right\}$$

It's so non-obvious that I suspect that it's wrong. (Wikipedia.) But I don't want to put in the effort it would take to check it.

[ Addendum 20211110: Matthijs Blom and I have confirmed that the formula is correct. ]

[ Addendum 20211112: my suggested definition of !!\pi_2(p) = \bigcup \{x\in p: |x| = 1\}!! does not work, because at this point we have not yet defined what !!|x|!! means. And we can't, because it requires functions and relations, which we can't define without an adequate model for orderded pairs. ]