Re: Cantor's definition of set

On Oct 29, 12:55 pm, John Jones <jonescard...@xxxxxxx> wrote:
On Oct 29, 5:46?pm, MoeBlee <jazzm...@xxxxxxxxxxx> wrote:

On Oct 27, 4:26 pm, John Jones <jonescard...@xxxxxxx> wrote:

You said,

Every thing is unique by virtue of being, by virtue of being a thing.

Good stuff. Of course, all things exist by their own efforts. We only
have to mention them and they can be in a set. But is this true? Can
you give me a list of things that cannot be in a set.

In an ordinary set theory in which we prove that every object is a
set, there is no object that is not a member of a set. In an ordinary
theory in which there are objects that are not sets, there are objects
that are not members of any set. For example, in Bernays class theory,
V = {x | x is a set} is not a member of any set.

MoeBlee

Your last paragraph was not clear.

It's as clear as Bermuda waters.

You imply that set membership is
optional

I said nothing about "optional".

but that everything is a member of a set.

In Z set theories, every x is a member of some y.

You will have to say
what V is, because a formal interpretation only suggests objects, even
when it uses {x | x is a set} .

In certain class theories, we prove there is a unique x such that for
all y, y in x iff y is a set. Then we define V = the unique x such
that for all y, y in x iff y is a set.

And should it rather be said that 'there is no object that could not
be a member of a set'?

First, we don't need subjunctives like 'could' in this context.
Second, it depends on the particular theory whether every object is a
member of a set or whether there are objects that are not members of
any set.

This is all ordinary, pedestrian, introductory set theory. Why don't
you read any of the many textbooks?

MoeBlee

.

Relevant Pages

  • Re: Tarski finite problem
    ... x has no limit ordinals ... proofs, if 'finite' means 1-1 with a member of omega), but I'm trying ... every non-empty set y of subsets of x has a member that is not a proper ...
    (sci.logic)
  • Re: Tarski finite problem
    ... x has no limit ordinals ... proofs, if 'finite' means 1-1 with a member of omega), but I'm trying ... every non-empty set y of subsets of x has a member that is not a proper ...
    (sci.logic)
  • Re: Cantor and the binary tree
    ... > Jan de Vos wrote: ... A number is in the set of even numbers iff it is even, ... WM argues that if every member of a set is finite the set must be finite. ...
    (sci.math)
  • Re: Cantors definition of set
    ... there is no object that is not a member of a set. ... For example, in Bernays class theory, ... Namely in a set theory that has urelements (or ... But in a set theory that has proper classes C ...
    (sci.logic)