4
$\begingroup$

I am reading Charles Pinter's Introduction to Set Theory

Every proper class is in one-to-one correspondence with the universal class $\mathscr{U}$, that is, the class of all sets [emph. added].

He uses the term in the beginning of the book, and then makes a sudden transition to the term "universe of sets," denoted by $V$. What's the difference between $\mathscr{U}$ and $V$? If they are the same,why use different terms?

Any clarification is appreciated.

$\endgroup$
  • $\begingroup$ I have the book in front of me. My guess is they are the same. Another possibility is that V is any subclass of the universal class. Could you give an example of when he uses V? $\endgroup$ – Rachmaninoff Dec 21 '14 at 1:50
  • 1
    $\begingroup$ @Rachmaninoff The first usage of $V$ is definition 11.1; I cannot give a page number; for I have the book as an E-book. $\endgroup$ – Conor O'Brien Dec 21 '14 at 2:45
  • $\begingroup$ You must have a later edition. My Pinter Set Theory ends with chapter 10 with a discussion on inacessible ordinals $\endgroup$ – Rachmaninoff Dec 27 '14 at 1:38
1
$\begingroup$

There are plenty of proper classes which are not the universe of sets, just like how there are many singletons which are not $\{\varnothing\}$.

For example the class of all singletons, or the class of all pairs, or the class of power sets, or the class of ordinals, or the class of successor ordinals, or the class of ... you get the point.

It should be noted, perhaps, that the assumption that every two classes have a bijection between them is equivalent to the axiom of global choice, stating that given a proper class of non-empty sets there is a class-function which chooses one from each of the sets.

$\endgroup$
  • $\begingroup$ So… how do $\mathscr U$ and $V$ fit in? $\endgroup$ – Conor O'Brien Dec 21 '14 at 4:11
  • 1
    $\begingroup$ $\mathscr U$ is a subclass of $V$. It's a collection of some sets, not necessarily all of them. But the working assumption is that it is just too darn big to be a set itself. $\endgroup$ – Asaf Karagila Dec 21 '14 at 4:12
  • $\begingroup$ Then, the difference is that $\mathscr U\subset V$ (because $\mathscr U$ is the set of all elements being considered?)? $\endgroup$ – Conor O'Brien Dec 21 '14 at 4:18
  • 1
    $\begingroup$ The class $\scr U$ is some class being considered. It need not be equal to $V$ itself. $\endgroup$ – Asaf Karagila Dec 21 '14 at 4:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.