$\begingroup$

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) $(A − B) − C \subseteq A − C$

$\endgroup$ 1

1 Answer

$\begingroup$

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 $x \in A - C$.

$\endgroup$ 1

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy