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May 14

Kuratowski style ordered triples

We all know the usual representation of ordered pairs in set theory by Kuratowski's definition:

(x, y):={{x},{x, y}}

This has several desirable properties:

The variables x and y are at the same 'depth' (in this case 2) in the expression. Specifically, (x, y) ⊊ ℙ({x, y}).
No constants are used in the definition, unlike Wiener's definition: (x, y):={{{x}, ∅}, {{y}}} which uses the constant ∅.
The depth is as small as possible and for the given depth the number of {} pairs used (3) is as small as possible.

From here, the usual way to define on ordered triple is by

(x, y, z):=((x, y), z)

This fails the first condition, though you could fix it up as

(x, y, z):=((x, y), {{z}}).

It seems to me though that you could do better in terms of depth and number of {} pairs. I haven't found a better combination yet, but I did find that my first guess:

(x, y, z):={{x},{x, y},{x, y, z}}

does not work since one gets (a, a, b) = (a, b, b). It seems likely that no expression of depth 2 will work but I haven't proved this. It also seems likely there is an expression of depth 3 which will work, but I haven't found one yet. Anyone able to find a depth 3 or less expression that satisfies conditions 1 and 2? --RDBury (talk) 04:05, 14 May 2017 (UTC)[reply]

It seems like

(x,y,z)=\{\{\{x,y,z\},\{x\}\},\{\{y\}\}\}

should work for depth 3 and satisfy both your conditions. --Deacon Vorbis (talk) 16:45, 15 May 2017 (UTC)[reply]

No, it doesn't. Scratch that. --Deacon Vorbis (talk) 16:48, 15 May 2017 (UTC)[reply]

I think that

(x,y,z)=\{(x,x),(x,y),(y,z)\}

works for depth 3 (where the tuples are defined using Kuratowski's definition) --84.229.149.53 (talk) 16:50, 15 May 2017 (UTC)[reply]

Not sure about the previous answer, but after playing with it some more, I think

(x,y,z)=\{\{\{x,y,z\},\{x,y\},\{x\}\},\{\{y,z\},\{y\}\},\{\{z\}\}\}

should do the trick. --Deacon Vorbis (talk) 17:18, 15 May 2017 (UTC)[reply]

This still has (a, a, b) = { { {a, b}, {a} }, { {b} } } = (a, b, b), doesn't it? --JBL (talk) 20:23, 15 May 2017 (UTC)[reply]

Oh hell, you're right. Somehow I'd convinced myself differently. --Deacon Vorbis (talk) 21:53, 15 May 2017 (UTC)[reply]

I've verified that

(x,y,z):=\{(x,x),(x,y),(y,z)\}

does work. A variation that also seems to work is

(x,y,z):=\{(x,y),(x,z),(y,z)\}

. But if you try something similar for quadruples,

(p,q,r,s):=\{(p,q),(p,r),(q,r),(p,s),(q,s),(r,s)\}

you get (a, b, a, a)=(a, a, b, a). Finding a combination that works is tricky because there are plenty that seem like they ought to work, but when it comes to proving that they do some case pops up where they fail. --RDBury (talk) 23:00, 15 May 2017 (UTC)[reply]

A comprehensive brute-force search suggests (assuming my crude coding works correctly, not a given) that there is no set of ordered pairs of four objects that can distinguish the 16 ordered four-tuples with entries in an underlying set of size two. --JBL (talk) 17:05, 17 May 2017 (UTC)[reply]

Not to distract from your question, which is interesting, but the most usual way of representing finite tuples is to take them to be functions whose domain is the length of the tuple, considered as a finite ordinal. So a 3-tuple is a function from 3 to something, meaning it assigns values to 0, 1, and 2. That makes

(a,b,c)=\{(0,a),(1,b),(2,c)\}

. Of course this makes the ordered pair different from the ordered 2-tuple, but life is hard all over. --Trovatore (talk) 23:23, 15 May 2017 (UTC)[reply]