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Thread: set theory proof

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    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?
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    \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}
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by james121515 View Post

    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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