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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

19

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 $A=B$.

  • 1
    Nice, but should you (and everyone else) be posting complete answers to homework problems?2012-09-10
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    @alexis: This one’s borderline; I gave Jessica the benefit of the doubt on the grounds that she’d made useful progress. You’ll find that I gave only a hint for her next question.2012-09-10
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    I'd say you're well over the line :-) I've taught math and if she was my student and turned in this proof without attribution, it would most certainly be cheating. So you're putting her in a difficult position: She was honest enough to flag it as homework, but now how can she admit she got help when you didn't leave _anything_ for her to do herself?2012-09-10
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    @alexis: I spent my entire working life teaching mathematics. I’m more interested in helping people learn than in whether they got too much help on one question out of many. If someone has several related questions, I’m perfectly willing to make the first one a lesson.2012-09-10
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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.