Re: Re :The empty set

On Jan 22, 11:40 am, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:
On Jan 22, 11:01 am, "Ross A. Finlayson" <r...@xxxxxxxxxxxxxxx> wrote:

If there exists an element z in the empty set, then where each other
non-empty set, notice by definition, has elements, then there exists
for each z so prototyped, an element of it, because then no sets are
empty, in triviality. That reads: "z is an element of the empty set"
is interchangeable with "there is not an empty set".

Pointless at best.

Then with regards to the previous statement that for any z, that the
statement "z is an element of the void" is interchangeable with the
statement "z is universal", I agree that's not a theorem of ZFC, with
that theory's lack of universality. While it does have that there's
only one, or no, proper classes in a set theory, it also has non-sets
in that set theory.

No, given an appropriate definition of 'x is a set', it is easy to
show in Z set theories:

Ax x is a set.

Basically we're arguing about vacuity. The empty set is axiomatized
to have no elements.

I don't know what particular system you have in mind, but in certain
ordinary treatments, we PROVE from the axiom schema of separation:

ExAy ~yex

and then from the axiom of extensionality:

E!xAy ~yex.

Then, to make a statement that is biconditional
or interchangeable about "the empty set with any element" is as absurd
(in ZF) as a statement about "the universal set with all elements".

I don't know what "biconditional or interchangable about" is supposed
to mean.

Yet, if it were so that there were no "truly" empty set in ZF, that
there exists z element of 0,

Then ZF would be inconsistent.

then while that would be interchangeable
with the statement that for any set in ZF, where each object is only a
set, that there exists a set that is its element, given that breaking
of the rules of ZF, then it is direct to say that for any set in ZF
that it has an element, than to furthermore suppose that element to be
universal.

Whatever that is supposed to mean, it seems pointless. If in ZF, the
empty set had a member, then ZF would be inconsistent, and thus would
prove every sentence in the language.

In breaking the axiomatization of null,

I have no idea what that is supposed to mean.

the next step is
closer acknowledgment of that notion than the gamut of other would-be
statements about objects that don't exist in ZF, that can't be, as
their ackowledgment breaks ZF.

Why don't you just learn some set theory rather than to continue to
post nonsense?

Back to the union of sets and their set-hood or lack thereof,
according to the axioms of comprehension, for each x, there exists y,
x element of y, where y is a set.

In Z set theory,

AxEy(xey & y is a set)

is a theorem, given a suitable definition of 'is a set'.

Then, the union of the y's for each
x is a set,

No, wrong, because there is no SET of such y's.

because for each x, there exists a set y containing it,

For each x there are many such y's, but there is no set of ALL such
y's.

and the union of sets is a set. Yet, that would be a universal set,
which contains each set x, so that union of sets is not a set. It is
not so that for all elements of the cumulative hierarchy (in order),
that their union is a set, yet for each, it is, in "anti-transfer",
where the transfer principle is not upheld there, for that would lead
to an inconsistency in ZF.

Nevermind the transfer principle (as if you had a clue as to what that
is). There is no set of all y such that xey.

union of sets is a set <=/=> union of sets is not a set

No such theorem in Z set theories.

That there is no quantification over sets in a regular set theory
means that there are no universally true statements about sets in a
regular set theory. Then, there arguably aren't generally true
statements about them either.

Why don't you just learn some set theory rather than to continue to
post nonsense?

In a broader theory than ZF, which restricts comprehension via
regularity,

I have no idea what you mean by "restricts comprehension via
regularity". Also, the axiom of regularity is a ZF axiom too.

that collection of sets is irregular, so ZF e ZF, the
universe of ZF's sets contains itself, and that collection is defined
by its elements. ZF is inconsistent. That there are non-sets in your
set theory, I can imagine you might find unamusing.

You present not just as nut case, but as an especially irritating one
since it has been explained over and over and over (for years) to you
what is incorrect in the argument you try to make above.

Also from the recent discussion about well-ordering the reals, a
variety of transfinite recursion schema were presented.

Another instance of you wasting people's time with your confusions.

MoeBlee

You're saying that in a naive universe that the subcollection of well-
founded sets would not be akin to the Russell set of Russell's
paradox, illustrating the same dilemma. The domain of discourse of ZF
is the Russell set of naive set theory. It is shown in naive set
theory, a set theory without the restriction on comprehension imposed
by regularity, that the regular subdomain so considered has irregular
elements. Where that is so, with less axioms than ZF's in what
otherwise has more unrestricted comprehension, it is proven by less of
those axioms that there is an irregular set among the collection of
regular sets. Then in axiomatizing via regularity the restriction of
comprehension to regular sets, it is illustrated that the universe of
ZF can't not contain an irregular set.

Then, the universe of ZF contains an irregular set, itself.

It's like a theory of small fishes, with an axiom that fish only eat
smaller fishes, and another axiom that there are no larger fishes. In
the larger sea, they are consumed by larger fishes. In the fishbowl,
it may well be observed that the small fishes do not eat each other,
released into the wild they would soon be a meal. Try to consider:
outside the fishbowl, beyond where your happy fishes illustrate
completion of all knowledge about the fish food chain. In the ocean,
the largest fish eventually decompose to food for the smallest.

With regards to the proof of the existence of the empty set as you
claim, which is otherwise the assertion of the existence of the
constant/ur-element of the set theory, then the axioms of ZF wouldn't
be pairwise independent.

Mind the transfer principle. In ZF: union of each set: set; union of
all sets: not a set.

In that discussion about how the rationals and irrationals are each
dense in the reals, if it was wasting time, then some of it was about
retroactively wasting the time of a couple generations of logicians.

Ross

--
Finlayson Consulting
.

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