On page 287 of the article On the Axiom of Extensionality, Part II, The Journal of Symbolic Logic, Vol. 24, No. 4 (Dec., 1959), pp. 287- 300, the author R. O. Gandy writes:

"But, in the absence of the axiom of extensionality, there may be several distinct classes having the same extension; and in order to develop the theory we need to single out one of them. It would be possible to introduce a special descriptions operator for this purpose; but it seems simpler, and closer to the spirit of Part I, to introduce abstracts. This represents a move back towards the set theory of Zermelo, but for our purpose that is not a disadvantage."

It is obvious that the set brackets ({ and } used for set formation) can be eliminated from classical set theory with extensionality. But how and when, i.e. by what authors and publications, did the elimination come about? Are there some published accounts of this, perhaps?

  • 2
    $\begingroup$ In what sense "disappeared"? They seem to me alive and well .. $\endgroup$ Commented Jul 7 at 13:57
  • $\begingroup$ @MauroALLEGRANZA In the sense that the language of set theory is now almost canonically defined as first order logic, with or without identity, plus $\in$. $\endgroup$ Commented Jul 7 at 15:00
  • 2
    $\begingroup$ Ok, you mean the set abstraction operator $\endgroup$ Commented Jul 7 at 15:13
  • 1
    $\begingroup$ But if we use it, we have to define it. I think that most set theory textbooks define it someway. $\endgroup$ Commented Jul 7 at 15:19
  • 1
    $\begingroup$ The "abstraction notation" evolved from Peanesque original through W&R's Principia (1910-13) where we have: $(\alpha = \hat z (\phi z)) \equiv \forall x( x \in \alpha \equiv \phi x)$. See page 188: "treating $\hat z (\psi z)$ as the class determined by [the propositional function] $\psi \hat z$". $\endgroup$ Commented Jul 8 at 12:40


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.