Is that true to say: For every set A defined by A={ A : ∅∉A } then the Cardinality of A equals to |A|-1 ?

1 Answer
Dec 16, 2017

A few thoughts...

Explanation:

First let us ask what the meaning of the expression { A : O/ !in A }{A:A} is.

It seems to be saying something like "The set of sets AA such that the empty set O/ is not an element of AA".

From what collection of sets are we allowed to choose candidate elements for { A : O/ !in A }{A:A} ?

Here are a few ideas:

  • The "set of all sets". The idea of the "set of all sets" leads to an inconsistent system. For example, we would have problems deciding whether { A : A !in A }{A:AA} is an element of itself or not.

  • The "class of all sets". If we recognise that the collection of all sets is too big to count as a set, we can then happily write { A : O/ !in A}{A:A}, meaning the collection of all sets that do not contain the empty set as a member, but this is too big to be a set too and is therefore a proper class (which has no defined cardinality).

  • What if we restrict the choices to some set Omega of sets? Then we might more properly write: { A in Omega : O/ !in A }. This is then a perfectly good set, but what can we say about its cardinality? We can say abs({ A in Omega : O/ !in A }) <= abs(Omega). If this subset of Omega is also infinite then removing any finite number of elements from it will not change its cardinality. Hence:

    abs({A in Omega : O/ !in A}) - 1 = abs({A in Omega : O/ !in A})

That might look like a false assertion, but it's really just a property of infinite cardinals and their arithmetic.

Notes

I suspect the back story of this question relates to attempting to find paradoxes like { A : A !in A } which was an historic example that drew attention to the need for formal set theory.