set theory proof

**james121515** · February 25th 2010, 05:12 PM

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 $A \setminus (A\setminus B) = \emptyset$ ?

**Plato** · February 25th 2010, 05:34 PM

$\begin{gathered}<br /> A\backslash \left( {A\backslash B} \right) = A \cap \left( {A \cap B^c } \right)^c \hfill \\<br /> A \cap \left( {A \cap B^c } \right)^c = A \cap \left( {A^c \cup B} \right) \hfill \\<br /> A \cap B \subseteq B \hfill \\ <br /> \end{gathered}$

**Drexel28** · February 25th 2010, 06:11 PM

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?

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