# set theory proof

• Feb 25th 2010, 05:12 PM
james121515
set theory proof
Prove or disprove that $\displaystyle A\setminus(A \setminus B) \subseteq B$

$\displaystyle \textbf{Proof. }$ Let $\displaystyle x \in A \setminus (A \setminus B)$. Then $\displaystyle x \in A$ and $\displaystyle x \notin A \setminus B$. Since $\displaystyle x \notin A \setminus B$, $\displaystyle x \notin A$ or $\displaystyle x \in B$. Since it can't be the case that $\displaystyle x \notin A$, we must have that $\displaystyle x \in B$. Therefore, $\displaystyle A \setminus (A \setminus B) \subseteq B$

The proof seems to make sense (to me at least), but what if $\displaystyle A$ and $\displaystyle B$ are disjoint? Wouldn't that mean that $\displaystyle A \setminus (A\setminus B) = \emptyset$?
• Feb 25th 2010, 05:34 PM
Plato
$\displaystyle \begin{gathered} A\backslash \left( {A\backslash B} \right) = A \cap \left( {A \cap B^c } \right)^c \hfill \\ A \cap \left( {A \cap B^c } \right)^c = A \cap \left( {A^c \cup B} \right) \hfill \\ A \cap B \subseteq B \hfill \\ \end{gathered}$
• Feb 25th 2010, 06:11 PM
Drexel28
Quote:

Originally Posted by james121515

The proof seems to make sense (to me at least), but what if $\displaystyle A$ and $\displaystyle B$ are disjoint? Wouldn't that mean that $\displaystyle A \setminus (A\setminus B) = \emptyset$?

And saying that $\displaystyle \varnothing\subseteq B$ bothers you?(Wink)