After drawing a diagram of this statement, I believe it to be false. However, I'm having trouble approaching how to disprove this. Do I try to prove the negation? Or what else can I do?
Prove/Disprove: For any sets A and B, $(A \cup B)^c = A^c \cap B^c$
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$\begingroup$
logic
elementary-set-theory
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1The statement is true (if you're talking about complements)... – 2012-09-25
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0Ok, yeah, you're right. I just found out that it is true based on Demorgans law. However, I can't use Demorgans law for the proof. So, how do you suggest approaching proving this? – 2012-09-25
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0There are two ways I'd suggest. One is given by @Babak below, namely show that $(A\cup B)'\subseteq A'\cap B'$ and that $A'\cap B'\subseteq (A\cup B)'$. Another is to use Venn diagrams. Depending on your comfort, you might find one easier than the other. – 2012-09-25
2 Answers
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$x\in (A'\cap B')\to x\in A'$ and $x\in B'$. If $x\in A'$ then $x\notin A$ and if $x\in B'$ then $x\notin B$. In both cases, $x\notin A\cup B$. This means that $x\in (A\cup B)'$. The other direction is analogous. You can do it.
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0Um, how does $x \not\in A$ imply $x \not\in A \cup B$? – 2012-09-25
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0@JohannesKloos: I fixed it. Thanks for noting me on time. :-) – 2012-09-25
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0You are getting very close to $20$K! +1 – 2013-03-22
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0Good answer anyway – 2013-03-22
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$x\in(A\cup B)'$ $\Rightarrow$ $(x\notin A\text{ and } x\notin B)$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $x\in A'\cap B'$ $\Rightarrow (A\cup B)'\subseteq A'\cap B'$
$x\in A'\cap B'$ $\Rightarrow$ $(x\in A'\text{ and } x\in B')$ $\Rightarrow$ $(x\notin A\text{ and }x\notin B)$ $\Rightarrow$ $x\in(A\cup B)'$ $\Rightarrow A'\cap B'\subseteq (A\cup B)'$
And
$(A\cup B)'\subseteq A'\cap B'$ and $A'\cap B'\subseteq (A\cup B)'$ $\Leftrightarrow$ $(A \cup B)' = A' \cap B'$