0
$\begingroup$

I am confused about union and intersection of sets.

Is my understanding correct that if there are three sets $A,B,C$ and $x \in A$ and $x \notin B$ and $x \notin C$ then we can write $x \in A \setminus (B \cup C)$ or write $x \in A \setminus (B \cap C)$ and both expressions are the same?

Also when $x \notin A$ and $x \notin B$ meaning $x \in A^c$ and $x \in B^c$, can we write $x \in A^c \cup B^c$ or write $x \in A^c \cap B^c$ and they would both mean the same thing?

$\endgroup$
  • $\begingroup$ No, but we do have $A\setminus (B\cup C)\subseteq A\setminus (B\cap C)$ $\endgroup$ – Prasun Biswas Sep 23 '17 at 20:18
1
$\begingroup$

Let $A = \{ 1,2,3\}$, $B=\{ 1 \}, C = \{2 \}$.

$$A \setminus (B \cup C) = \{3\}$$

$$A \setminus (B \cap C) = \{1,2,3\}$$

Try to answer the other question using an example too.

$\endgroup$
  • $\begingroup$ So they are not the same when $x \in A$ and $x \notin B$ and $x \notin c$? I was looking at a proof and in this condition $A \setminus (B \cup C)$ was used and I don't understand how $A \setminus (B \cup C)$ would make sense if both aren't the same because to my understanding $x \in A$ and $x \notin B$ and $x \notin c$ means $A \setminus (B \cap C)$ $\endgroup$ – lakada Sep 23 '17 at 20:23
  • $\begingroup$ @amWhy Doesn't the second also contain elements of A, that remain after discarding elements in B and C? $\endgroup$ – lakada Sep 23 '17 at 20:38
  • $\begingroup$ @lakada the example has illustrated that they are not equal isn't it? $\endgroup$ – Siong Thye Goh Sep 23 '17 at 22:26
  • 1
    $\begingroup$ if $x \in A , x \notin B$, and $x \notin C$, we can write it as $x \in A \cap B^c \cap C^c = A \cap (B \cup C)^c = A \setminus (B \cup C)$ by de Morgan's law. $\endgroup$ – Siong Thye Goh Sep 23 '17 at 22:28
1
$\begingroup$

For the second part, you need to think that if some $x$ is in the intersection of two sets then it is in the union. But if it is in the union not necessarily it is in the intersection.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.