# Proof: Cardinality Question

• Apr 11th 2010, 08:44 PM
firebio
Proof: Cardinality Question
If S is an uncountable set and T is a countable subset of S. Show that cardinality of |S\T|= |S|.

S\T is a subset of S, so |S\T|<= |S|.
I was going to use the cantor-schroder-bernstein theorem, but i dont know how to proof that |S\T|=> |S|.

If there is other ways to proof this, please post it.
Any help would be nice.
• Apr 12th 2010, 12:51 AM
emakarov
You can prove this by contradiction: if both $\displaystyle S\setminus T$ and $\displaystyle T$ are countable, then $\displaystyle S\subseteq (S\setminus T)\cup T$ is countable.
• Apr 12th 2010, 03:37 PM
Drexel28
Quote:

Originally Posted by firebio
If S is an uncountable set and T is a countable subset of S. Show that cardinality of |S\T|= |S|.

S\T is a subset of S, so |S\T|<= |S|.
I was going to use the cantor-schroder-bernstein theorem, but i dont know how to proof that |S\T|=> |S|.

If there is other ways to proof this, please post it.
Any help would be nice.

Quote:

Originally Posted by emakarov
You can prove this by contradiction: if both $\displaystyle S\setminus T$ and $\displaystyle T$ are countable, then $\displaystyle S\subseteq (S\setminus T)\cup T$ is countable.

Haven't you only technically shown that $\displaystyle S-T$ is uncountable? But there are more than one cardinal number bigger than $\displaystyle \aleph_0$...
• Apr 13th 2010, 12:35 AM
emakarov
Quote:

Haven't you only technically shown that http://www.mathhelpforum.com/math-he...a32fe830-1.gif is uncountable?
You are absolutely right.

Wikipedia says that $\displaystyle \kappa_1+\kappa_2=\max(\kappa_1,\kappa_2)$ for infinite cardinals $\displaystyle \kappa_1, \kappa_2$ using the axiom of choice, but I don't know off top of my hat either how to prove this or if there is an easier proof.