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  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?

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  • $\begingroup$ I think i found a counter example to the first one proving its false. There exists sets A,B and C such that C − (A ∪ B) = ∅ but ( A ∩ B) ∪ C is not subset of A ∩ (B ∪ C) . If i take C = {1,2,3} A= {1,2} B= {2,3} . The negation of the statement appears to be true. So i can say the original statement is false. Am i right on this one or am i missing something? $\endgroup$ – Mahir Shahriar Oct 19 '18 at 14:33
  • $\begingroup$ Your counterexample doesn’t satisfy hypothesis of 1. $\endgroup$ – Mayuresh L Oct 19 '18 at 14:35
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Your attempt at 1 isn't correct, because $C-(A\cup B)=\{5\}\ne\emptyset$.


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

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  • $\begingroup$ Thanks! got it now. $\endgroup$ – Mahir Shahriar Oct 19 '18 at 14:46

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