# Re: How is this collection called?

From: Paul <paul_at_test.com>

Date: Wed, 31 Mar 2004 22:40:35 +0100

Message-ID: <uTGac.26323$h44.3505549_at_stones.force9.net>

> I wonder how your example works. It reminds me Kuratowski's definition of

Date: Wed, 31 Mar 2004 22:40:35 +0100

Message-ID: <uTGac.26323$h44.3505549_at_stones.force9.net>

Mikito Harakiri wrote:

>>Define a*b as "a U {b}" (where U is set union) >>Then a*(b*c) = aU{bU{c}} != (a*b)*c = aU{b}U{c}.

*>*> I wonder how your example works. It reminds me Kuratowski's definition of

*> ordered pair**>**> (a,b) = {{a}, {a,b}}**>**> Formally,**>**> ((a,b),c)={ {{a}, {a,b}}, {{{a}, {a,b}} ,{c}} }**> (a,(b,c))={ {{b}, {b,c}}, {{{b}, {b,c}} ,{a}} }**>**> are different sets, but we still consider them isomorphic. There was a**> recent thread on sci.math "Question about associativity of cartesian**> product".*Well strictly speaking we don't consider them isomorphic. We just group them together as an "equivalence class" and show that certain operators on this are "well-defined" in the sense that you can apply it to any members of some given equivalence classes, and the class of the result will always be the same.

So (a,b,c) could really be seen as just syntactic shorthand for the equivalence class of (a,(b,c)) and ((a,b),c). I don't know if it is in practice defined this way but it certainly could be I think.

Paul. Received on Wed Mar 31 2004 - 23:40:35 CEST