Re: Does equality between sets contradict the philosophy behind structural set theory?

["Fred E.J. Linton" <fejlinton@usa.net>]

Leopold Schlicht <schlicht.leopold@gmail.com> quotes, in part:

> Notation 1.1: The arrow notation f : A -> B just means the domain of f is A
> and the codomain of f is B, and we write dom(f) = A and cod(f) = B.

and continues:

> Here, the authors talk about the equality of two sets (dom(f) = A).

No. The symbolatry dom(f) = A is NOT expressing the equality of two
sets, dom(f), and A, it is an abbreviation for the assertion that the map
f has domain A, which is itself a gloss on (a part of) the significance 
we may attach to the arrow notation f: A -> B. The remaining part of that
significance is expressed in the symbolatry cod(f) = B. There again it's
NOT an expression of the equality of two sets, cod(f), and B.

Rather, it's as if we split f: A -> B into two chunks,
f: A -> _  (abbreviated dom(f) = A)
and
f: _ -> B  (abbreviated cod(f) = B).

(In the above, please treat the simple underscore _ as just more white space:
  my usa.net mail service will not permit a string of 3 or 4 space characters
  to pass through without condensing them into a single space character.)

Taking a fresh breath of air, now, and asking about objects of a category:
does one not have a right to suppose that, should you and I each focus our
respective attentions on one object in a category, you on A, say, and I on B,
it should be possible to determine when we have chosen to focus on ONE AND
THE SAME object, i.e., to determine when your A is my B, i.e., when A =  B ?

No question on elementhood or membership enters into that question.

And we NEVER actually say that TWO sets are equal -- only that ONE set is  
equal to itself. If you and I have focused on sets A and B, respectively,
and if it happens that A = B, it means exactly that we were NOT focused
on TWO distinct sets, but only on ONE. 

Cheers, -- Fred Linton





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