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I'm not sure how to "show" these two answers. The small group created from the intersection $A\cap B\cap C$ is a subset of $A\cap B$ since abc is a smaller "portion" of the overall sets.
The difference of $(A-B)-C$ is the same as $A-C$ since part of $A$ was removed with the $B$ already.

Let $A, B$, and $C$ be sets. Show that
1) $(A \cap B \cap C) \subseteq (A \cap B)$
2) $(A − B) − C \subseteq A − C$

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The first one is a special case of $S\cap T\subseteq S$, see ProofWiki. The second one can be obtained from $S\setminus T\subseteq T$, see ProofWiki, and monotonicity of set difference, i.e. $X\subseteq Y$ $\Rightarrow$ $X\setminus Z\subseteq Y\setminus Z$. (I did not find this one at ProofWiki, but neither of these three facts is difficult to prove.) – Martin Sleziak Jul 11 '12 at 11:38
up vote 1 down vote accepted

To show prove 1), show that if you choose an arbitrary element $x \in A \cap B \cap C$, then $x \in A \cap B$.

So, let $x \in A \cap B \cap C$. This means that $x$ is contained in all three sets. In particular, $x \in A$ and $x \in B$. Hence, $x \in A \cap B$

Proceed in the same manner for 2). Let $x \in (A - B) - C$. Then $x \in A - B$, and $x \notin C$. Therefore, $x \in A$, but $x \notin C$, so $x \in A - C$.

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Ahh that makes sense now I wasn't thinking tho pick some arbitrary element. The book doesn't have any examples which made it difficult. Thank you. – LF4 Jul 11 '12 at 2:10

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