Discussion of Class Notation

June 23, 2007

Classes

Dennis:

zc is z with carrot

e is class membership sign

*20.01 f {zc (Gz) } =df ($F)((x) F!x equ Gx ) &

f{ F! zc}

*20.02 x e (F ! zc) =df F ! x

My understanding is as follows (I found it difficult

to figure out & think

Whitehead and Russell should and could have

explained it better.)

*20.01 says (the class of z such that Gx) has the

property f.

*20.02 says x is a member of F ! zc

I think *20.02 is intended really to be part of a

special case of *20.01

where fx = x e (F ! zc)

So *20.02 could be stated better y e (zc(Gz)) =df

($F)((x) F!x equ G!x) & G!(F! y))

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"*20.01 f {zc (Gz) } =df ($F)((x) F!x equ Gx ) & f{ F! zc}"

*20.01 f{zc(Gz)} =df ($F)((x)(F!x <- Gx) & f(F!zc)), is better.

*20.02 x e (F!zc) =df F!x.

"I think *20.02 is intended really to be part of a

special case of *20.01

where fx = x e (F ! zc)

So *20.02 could be stated better y e (zc(Gz)) =df

($F)((x) F!x equ G!x) &

G!(F! y))"

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Owen

?? Not so. G!(F!y) has no sense, nor does fx = x e(F!zc). f applies to

functions or classes not to individuals.

Dennis

I don't see how *20.02 can be applied except as taking

it as an fx in *20.01 (imagine fx as a function of f

not x.) So you could write

(x e (G!zc)) =df ($F)((x) F!x equ G!x) & G!(F! y))"

I know Russell wouldn't put G!(F!zc)) but that's how

it seemed to work out when I substituted in. I think

'predicative' functions cannot do the work (block the

paradoxes) that they were created for. The axiom of

reducibility doesn't tell you anything. There is no

way to know what functions are predicative (or have

predicative equivalents).

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Owen:

y e {zc(Gz)} <- ($F)((x)(F!x <- Gx) & y e (F!zc)),

by *20.01.

But, y e (F!zc) has not been defined. See: *20.3

x e {zc(Gz)} <- Gx.

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Dennis:

*20.3 is a theorem not a definition

Owen:

How do you show that ($F)((x)(F!x <- Gx) & y e

(F!zc)) <- Gy ? ..without

*20.02 x e (F!zc) =df F!x.

Dennis:

I think you *could* use just that definition instead

of *20.02. (not prove it)

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Owen

We can show later that: {zc(F!x)}= (F!zc), ie. both

functions and classes

are extensional.

Dennis:

I show 'belief-r' & 'tom_believes_now' are

intensional.

see

http://dennisdarland.com/philosophy/Class_Definitions.html

but extensional classes can be defined from them.

And see PM pp. 72-73.

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