1

Sets $X,Y,Z$ are subsets of a universal set $S$. If $Z \subseteq (X \cap Y^c) \cup (Y \cap X^c)$, then $Z \cap (X \cap Y) = \emptyset$

I started with Allow $k\in Z$ such that $k\in (X\cap Y^c)$ or $k\in (Y\cap X^c)$, therefore $k\notin (X\cap Y)$ but I don't know where to go from here. Thanks

lostintranslation
  • 35
  • 2
    You proved that $Z$ and $X\cap Y$ have no elements in common. This is exactly the statement that $Z\cap(X\cap Y)=\varnothing$ so you are ready. – drhab Apr 11 '20 at 15:16
  • We have the two cases, (1) $k \in (X \cap Y^c)$ or (2) $x \in (Y \cap X^c)$. Case (1): By the Set Difference Law, $k \in (X - Y)$. And by definition of Set Difference, this means $k$ is in $X$ but not $Y$. So that tells us, $k$ cannot be in both $X$ and $Y$, so $k \notin X \cap Y$. Case (2), same as case 1. and then bring it all together, not sure exactly how to do the final piece? Maybe the final piece is something simple like: In both cases, we have shown $k \notin X \cap Y$. Thus, since $k \in Z$, the intersection of $Z$ and $X \cap Y$ would not contain any elements, $ = \emptyset$ – lostintranslation Apr 11 '20 at 18:20
  • In short: $x\in Z$ implies that $x\in X\cap Y^{\complement}$ or $x\in Y\cap X^{\complement}$. In the first case we have $x\notin Y$ and consequently $x\notin X\cap Y$. In the second case we have $x\notin X$ and consequently $x\notin X\cap Y$. So the assumption that $x\in Z$ leads to the conclusion that $x\notin X\cap Y$. This proves that $Z$ and $X\cap Y$ have no elements in common: $Z\cap(X\cap Y)=\varnothing$. – drhab Apr 12 '20 at 07:27

1 Answers1

0

$Z \cap (X \cap Y) \subseteq$
$(X \cap Y) \cap (X \cap Y^c) \cup (Y \cap X^c)$ = $(X \cap Y \cap X \cap Y^c) \cup (X \cap Y \cap X^c)$ =
empty set $\cup$ empty set = empty set.

William Elliot
  • 17,598