# The Universe of Discourse

Fri, 15 Apr 2022

A great deal of attention has been given to the encoding of ordered pairs as sets. I lately discussed the usual Kuratowski definition:

$$\langle a, b \rangle = \{\{a\}, \{a, b\}\}$$

but also the advantages of the ealier Wiener definition:

$$\langle a, b \rangle = \{\{\{a\},\emptyset\}, \{ \{ b\}\}\}$$

One advantage of the Wiener construction is that the Kuratowski pair has an odd degenerate case: if !!a=b!! it is not really a pair at all, it's a singleton. The Wiener pair always has exactly two elements.

Unordered pairs don't get the same attention because the implementation is simple and obvious. The unordered pair !![a, b]!! can be defined to be !! \{a, b\}!! which has the desired property. The desired property is:

$$[a, b] = [c, d] \\ \text{if and only if} \\ a=c \land b=d\quad \text{or} \quad a=d\land b=c$$

But the implementation as !!\{a, b\}!! suffers from the same drawback as the Kuratowski pair: if !!a=b!!, it's not actually a pair!

So I wonder:

Is there a set !![a, b]!! with the following properties:

1. !![a, b] = [c, d]!! if and only if !!a=c \land b=d!! or !! a=d\land b=c!!

2. !![a, b]!! has exactly two elements for all !!a!! and !!b!!

Put that way, a solution is $$[a, b] = \{ \{ a, b \}, \emptyset \}\tag{\color{darkred}{\spadesuit}}$$ but that is very unsatisfying. There must be some further property I want the solution to have, which is not possessed by !!(\color{darkred}{\spadesuit})!!, but I don't know yet what it is. Is it that I want it to be possible to extract the two elements again? I am not sure what that means, but whatever it means, if !!\{ a, b\}!! does it, then so does !!(\color{darkred}{\spadesuit})!!.

But that does also suggest another property that neither of those enjoys:

There should be formulas !!F_1!! and !!F_2!! such that for all !![a, b]!!:

1. !!F_1([a, b]) = a!! or !!b!!
2. !!F_2([a, b]) = a!! or !!b!!
3. !!F_1([a, b]) = F_2([a, b])!! if and only if !!a=b!!

I think this can be abbreviated to simply:

!![F_1([a, b]), F_2([a, b])] = [a, b]!!

There may be some symmetry argument why there are no such formulas, but if so I can't think of it offhand. Perhaps further consideration of !![a, a]!! will show that what I want is incoherent.

Today is the birthday of Leonhard Euler. Happy 315th, Lenny!

[ Addendum 20220422: Several readers pointed out that the !!F_i!! formulas are effectively choice functions, so there can be no simple solution. Further details. ]