The set of All sets
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From: zuhair on 19 Nov 2005 07:06 Robert Low wrote: > zuhair wrote: > > Similar condition happens with > > > > P(x) = Singlton x = {x} > > Assuming that this made sense, > > > Now P(P(x))= Singlton {x} = {{x}} which is also a singlton > > This would still be wrong. P({x})={ {}, {x} } > which has two elements. > > In general, if a set A has n(>=0) elements, P(A) has 2^n > elements. You are confusing P(x) for power set of x , which is not what I am refering to , here P means propositional function and not Power. Zuhair
From: Robert Low on 19 Nov 2005 07:19 Robert J. Kolker wrote: > Robert Low wrote: >> zuhair wrote: >> [stuff] >> In general, if a set A has n(>=0) elements, P(A) has 2^n >> elements. > Don't confuse zuhair with facts. In my (admittedly little) experience with zuhair, that's unlikely to happen. The word 'impervious' does spring to mind.
From: Robert Low on 19 Nov 2005 07:21 zuhair wrote: > Robert Low wrote: >>zuhair wrote: >>>Similar condition happens with >>>P(x) = Singlton x = {x} >>Assuming that this made sense, >>>Now P(P(x))= Singlton {x} = {{x}} which is also a singlton >>This would still be wrong. P({x})={ {}, {x} } >>which has two elements. >>In general, if a set A has n(>=0) elements, P(A) has 2^n >>elements. > You are confusing P(x) for power set of x , which is not what I am > refering to , here P means propositional function and not Power. Oh, I'm sorry. I thought you were writing something intelligible but incorrect. I see I was wrong about the first part of that.
From: zuhair on 19 Nov 2005 12:24 Robert Low wrote: > zuhair wrote: > > Robert Low wrote: > >>zuhair wrote: > >>>Similar condition happens with > >>>P(x) = Singlton x = {x} > >>Assuming that this made sense, > >>>Now P(P(x))= Singlton {x} = {{x}} which is also a singlton > >>This would still be wrong. P({x})={ {}, {x} } > >>which has two elements. > >>In general, if a set A has n(>=0) elements, P(A) has 2^n > >>elements. > > You are confusing P(x) for power set of x , which is not what I am > > refering to , here P means propositional function and not Power. > > Oh, I'm sorry. I thought you were writing something intelligible > but incorrect. I see I was wrong about the first part of > that. So propositional functions are something not intelligible?huh! How ignorant you are . Zuhair
From: G. Frege on 19 Nov 2005 12:58
On 19 Nov 2005 01:20:11 -0800, "zuhair" <zaljohar(a)yahoo.com> wrote: > > Define U to be the set of all x were x fulfills the propositional > function > > P(x) = x. > I guess, what you m e a n is: P(x) =df x = x > > U = Set of All sets, because everything is identical to itself. > This would indeed be the case, if such a set would exist (in the set theory we are working in/with). Actually, in the set theory MK (a descendant of NBG) you can define the /class/ of all sets: U =df {x : x = x}. Right, here P(x) is x = x, as you suggested above. Now in MK o n l y sets can be elements of sets, hence we would have: x e U <-> set x & x = x. for every x. It can be shown that U is n o t a set (but a so called /proper class/), hence U !e U, though (of course) U = U too. > > If U can be identical to itself, and so [an element] of itself. > Not in MK. > > so U = {U, ...} > Not in MK. But, surprise, in Quine's set theory NF there exists an universal s e t U (where U =df {x : x = x}) containing all sets, hence t h e r e we actually have U e U, hence U = {U, ...}. F. |