For all sets $ A $, $B $, $ C $, if $ C - (A \cup B) = \emptyset $ then $ (A \cap B) \cup C \subseteq A \cap (B \cup C) $?

Question

1. I need to prove or disprove the question in the title. I figured out the conclusion isn't true.

   Since when I take $ A = \{ 1, 2 \} $, $ B = \{ 2, 3 \} $, and $ C = \{ 5 \} $, $ (A \cap B) \cup C = \{ 2, 5 \} $ and $ A \cap (B \cup C) = \{ 2 \} $. So $ \{ 2, 5 \} $ is not a subset of $ \{ 2 \} $ and the conclusion of that statement is false. So is the statement as a whole true or false? If I had to prove it how would I prove it?

2. What if the statement was reversed, saying for all sets $ A $, $ B $, and $ C $, if $ (A \cap B) \cup C \subseteq A \cap (B \cup C) $, then $ C − (A \cup B) = \emptyset $. Would that be true or false?

Answer 1

Your attempt at 1 isn't correct, because $C-(A\cup B)=\{5\}\ne\emptyset$.

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Saying that $C-(A\cup B)=\emptyset$ is the same as saying that $$ C\subseteq A\cup B $$ On the other hand, $A\cap B$ could be empty. In this case the given inclusion would read $$ C\subseteq A\cap(B\cup C) $$ If we take $C=A\cup B$, the main hypothesis is certainly satisfied. With these additional assumptions, $$ A\cap(B\cup C)=A\cap C=A $$ and you just need to take $B\ne\emptyset$ to find an explicit counterexample. So $A=\emptyset$, $B=C=\{1\}$. Then $$ (A\cap B)\cup C=\{1\} $$ but $A\cap(B\cup C)=\emptyset$.

If you don't like the empty set, take $C=\{1,2\}$, $A=\{1\}$, $B=\{2\}$. Then $$ (A\cap B)\cup C=\{1,2\} $$ whereas $$ A\cap(B\cup C)=\{1\}\cap\{1,2\}=\{1\} $$