# set theory proof

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

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

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

Originally Posted by james121515

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

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