Re: Kuratowski Ordered Pair

On Dec 13, 4:53 am, Hero <Hero.van.Jind...@xxxxxx> wrote:

( a, a )=|= ( a ),

Actually, without the axiom of regularity, the above is not a theorem
of Z set theory. With the axiom of regularity, of course <a a> =/= a.

but how can it be that a = a and that a is
existing twofold?

If you put "existing twofold" into the language of set theory, we'd
have a better chance of making sense of it as a set theoretical
concern.

MoeBlee
.

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