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...


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

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

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


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.