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"Where A is an arbitrary set and B is an arbitrary set, when is the statement:

(A ∪ B) ⊆ (A ∩ B)

true? Is it true all the time, sometimes, or is it never true? If it is sometimes true, explain the cases where it is."

Is there any other case aside from when A = B that this is a true statement?

3 Answers 3

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No, there is not. You can prove this as follows. Assume that $A\cup B\subseteq A\cap B$. Then $A\subseteq A\cup B\subseteq A\cap B\subseteq B\;,$ so $A\subseteq B$. Similarly, $B\subseteq A\cup B\subseteq A\cap B\subseteq A\;,$ so $B\subseteq A$. It follows immediately that $A=B$.

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    Ok, we all want to help, the question is what helps people learn best. As a teacher you already know anything I could say, we just have different teaching styles I guess. Peace,2012-09-10
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It is true if and only if $A$ and $B$ are exactly the same set. Any member of $A$ that is not a member of $B$ is a member of $A\cup B$ but not of $A\cap B$, so $A\cup B\not\subseteq A\cap B$ in that case. And similarly if there's any member of $B$ that is not a member of $A$. So $A$ and $B$ must have exactly the same members.

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Really nice counter example in the general case:

Assume $A\ne \emptyset $

$A = \emptyset\cup A \subset A \cap \emptyset = \emptyset$

Therefore $A \subset \emptyset$, which is clearly a contradiction.