Re: Two results of set geometry

On Sep 28, 11:45 am, Tony Orlow <t...@xxxxxxxxxxxxx> wrote:
MoeBlee wrote:
On Sep 26, 9:23 am,TonyOrlow<t...@xxxxxxxxxxxxx> wrote:
'E' means "thereexists"

Let me comment now on TO's nonstandard use of the
existential quantifier.

What TO would like to do is say Ex to denote that the set
exists, and ~Ex to denote that it doesn't. For example, in
ZF, if x = omega, then Ex, but in ZF-Infinity+~Infinity, we
instead conclude ~Ex.

MoeBlee reprimands TO for this misuse of the quantifier, and
with good reason, too -- if an object doesn't exist, then one
has no right to substitute it as a value of a variable in any
formula whatsoever, not even a formula which (appears to)
state its nonexistence!

Yet there is a way to salvage this nonstandard use of E. We
now consider a class theory (in which there may exist proper
class variables, such as NBG). Then E can represent the
one-place predicate "is a set." In this case ~Ex would denote
"x is a proper class."

I've seen a thread in which NBG and "is a set" is mentioned,
and IIRC the symbol M was used to denote "is a set." This
use of M is not much different from TO's use of E. (But M is
probably better, so that E can be reserved as the existential
quantifier as usual.)

Notice that in metamath, "is a set" is notated differently. The
symbol V is used to denote the (proper) class of all sets, so
that "x is a set" becomes "x in V."

Oh. So how does one state Peano's first axiom? We say E 0?

We can use "0 in V" instead, or using only primitives:

exists x(forall y(not(y in x))) or Ex(Ay(~(yex)))

Or, E 0eN?

We can use "0 in N" or "0 in omega" instead. No E required.

E XeU ^ ~E YeU ^ X=Y = false

This could be interpreted as "if x is a set, and Y is a proper
class, then x is not equal to Y." In other words:

(x in V and y not in V) -> not (x = Y)

Here we use V instead of U for the class of all sets, and omit
any reference to the nonstandard use of E.

.

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