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Are there classes with different sizes ? I will put a precise statement of my question below:

Are there two well formed formulas $P,Q$ each with one free variable such that there is no well formed formula $F$ with two free variables such that the following holds:

1)$\forall x \forall y [F(x,y)\rightarrow[P(x)\land Q(y)]]$

2)$\forall x\exists!y F(x,y)$ (Analogue of the condition imposed on a relation to make it a function)

3)$\forall y \exists!x F(x,y)$ (Analogue of the condition imposed on a function to make it injective and surjective)

(The well formed formulas $P,Q,F$ are expressions of the language $L_1Set$)

Thank you

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up vote 8 down vote accepted

Trivially if we allow to consider that sets are classes. But let's assume that you really just meant proper classes.

If we only assume $\sf ZFC$, then it is consistent that yes, and consistent that no.

  • If one assumes that global choice holds, then there is a definable bijection between the class of ordinals and the universe. It follows that there is a bijection between any two proper classes.

    For example if $V=L$ holds, or even $V=HOD$, or $V=L[x]$, all those imply that there is a definable (perhaps with parameters) bijection like that.

  • On the other hand, it is also consistent that there is no bijection between the class of ordinals and the universe. If use class forcing, an Easton product of forcing which add two sets to every regular cardinal, then we can prove that the universe of the extension cannot be linearly ordered (in $\sf ZFC$), let alone well-ordered.

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Hi. I have a probably trivial question:Why is the question of my post trivial if we allow to consider that sets are classes ? –  Amr Feb 14 at 18:16
    
Because there are trivially sets with different cardinalities, and no set has the cardinality of the universe... –  Asaf Karagila Feb 14 at 18:17
    
Wouldn't this choice of $P,Q$ work (I am not sure if it works) let $a,b$ be two sets of different cardinality. Then set $P(x)$ to be $x\in a$ and $Q(x)$ to be $x\in b$... –  Amr Feb 14 at 18:20
    
Yes, but if you want to avoid using parameters (the sets $a$ and $b$) you can use something which is definable, like the natural numbers and its power set. –  Asaf Karagila Feb 14 at 18:22
    
Hence, the suggestion of my last comment still works even if we don't allow the collection of all sets to be a class, no ? –  Amr Feb 14 at 18:24
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