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Re: Real Analysis!!!
Posted:
Aug 12, 2012 6:15 AM
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On Aug 12, 11:24 am, William Elliot <ma...@panix.com> wrote:
> On Sat, 11 Aug 2012, Zuhair wrote:
> > Language: First order logic + (=,e,<>,||)
>
> > e is the membership symbol.
> > = is the identity symbol
> > <> is the ordered pair symbol, a two place function symbol
> > | | is the cardinality symbol, a one place function symbol
>
> > Axioms: Identity theory axioms +
>
> > (1) Sets with the same elements are identical
>
> > For all x. For all y. ((For all z. z e x <-> z e y) -> x=y)
>
> > (2) A singleton is its element
>
> {x} = x?
>
> > For all x,z. ((Exist! y. y e x) & z e x) -> x=z)
>
> for all x, not (some z with some y with x = {y}, z in x, x /= z)
> for all x, not (some y with x = {y} & some z with z in x, x /= z)
> for all x, (x isn't a singleton or for all z in x, x = z)
> for all x, (x isn't a singleton or x = {x})
Good.
>
> > (3) An element is singleton
>
> For all x, some y with x = {y}.
No this is not a theorem, I don't know why you keep writing that false
consequence of yours.
An element means a set that is an element of a set, not every set here
is an element of a set, only a singleton set is permitted to be an
element of a set, so it is NOT true that for all x. some y. x={y},
this is only true of singleton sets.
>
> > For all x. ((Exist y. x e y) -> (Exist! z. z e x))
>
> For all x, (x is a set -> x is a singleton).
> Every set is a singleton.
What do you mean by "set" as you wrote, note that the word "set" as I
used it applies to "every" object in this theory, it doesn't mean at
all the meaning attached to it in NBG where in NBG the term "set"
means a class that is an element of a class, the situation is not so
here, here in this theory the term "set" applies to every object
whether it is singleton or not. The above axiom is saying that if a
set x is an element of some set y then x must be singleton. This
theory doesn't say that every set is an element of a set. In this
theory sets that are not singletons (i.e. the empty set or the
multiple-ton sets) are not permitted to be elements of sets, Only
singletons can be elements of sets.
>
> > (4) There exist a set of all singletons satisfying phi
>
> { {x} | phi({x}) } = { y | phi(y), some x with y = {x} } is a set.
Yes
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