# Axioms

• Sep 10th 2009, 08:48 AM
spaceship42
Axioms
Any ideas on how to do this one?

Let A be a set. Show that a complement of A does not exist. So I need to show that there isn't a set of all x not in A.
• Sep 10th 2009, 08:55 AM
Plato
Quote:

Originally Posted by spaceship42
Any ideas on how to do this one?
Let A be a set. Show that a complement of A does not exist.

Are you quite sure that you have given the statement of that problem correctly?
Is there more to the question that you have omitted?
• Sep 10th 2009, 08:57 AM
spaceship42
Quote:

Originally Posted by Plato
Are you quite sure that you have given the statement of that problem correctly?
Is there more to the question that you have omitted?

After that sentence it just says that the complement of A is the set of all x not in A.
• Sep 10th 2009, 09:13 AM
Plato
Quote:

Originally Posted by spaceship42
After that sentence it just says that the complement of A is the set of all x not in A.

What axioms are you given?
• Sep 10th 2009, 11:15 AM
spaceship42
Oh, I'm sorry. I forgot that I'd called this thread "Axioms". This problem is about sets, not axioms. I mislabeled it accidentally.
• Sep 10th 2009, 11:28 AM
Taluivren
Hi,
$\displaystyle A$ is a set and $\displaystyle B=\{x:\, x \not \in A\}$ ?
Assuming axioms of ZF set theory, $\displaystyle B$ can't be a set: If $\displaystyle B$ were a set, then since $\displaystyle A$ is a set, axiom of pairing gives that $\displaystyle \{A,B\}$ is a set. Axiom of union then tells us that that there exist a set $\displaystyle C$ whose elements are elements of $\displaystyle A$ and elements of $\displaystyle B$. But from the definition of $\displaystyle B$ we get that $\displaystyle C$ is the universal class of all sets, which is a proper class, i.e. is not a set. This is a contradiction, thus $\displaystyle B$ can't be a set.

Can you see why the class of all sets is not a set?