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

Dec 16, 2017

A few thoughts...

#### Explanation:

First let us ask what the meaning of the expression $\left\{A : \emptyset \notin A\right\}$ is.

It seems to be saying something like "The set of sets $A$ such that the empty set $\emptyset$ is not an element of $A$".

From what collection of sets are we allowed to choose candidate elements for $\left\{A : \emptyset \notin A\right\}$ ?

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 $\left\{A : A \notin A\right\}$ 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 $\left\{A : \emptyset \notin A\right\}$, 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: $\left\{A \in \Omega : \emptyset \notin A\right\}$. This is then a perfectly good set, but what can we say about its cardinality? We can say $\left\mid \left\{A \in \Omega : \emptyset \notin A\right\} \right\mid \le \left\mid \Omega \right\mid$. If this subset of $\Omega$ is also infinite then removing any finite number of elements from it will not change its cardinality. Hence:

$\left\mid \left\{A \in \Omega : \emptyset \notin A\right\} \right\mid - 1 = \left\mid \left\{A \in \Omega : \emptyset \notin A\right\} \right\mid$

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 $\left\{A : A \notin A\right\}$ which was an historic example that drew attention to the need for formal set theory.