So how do you prove with an example that
$$(A\times B)^c \not\subset A^c\times B^c$$
Let's say I define the Universal set to be $\{1, 2, 3, 4\}$, $A = \{2,4\}$, $B = \{3,4\}$, $A^c = \{1,3\}$, and $B^c = \{1,2\}$.
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2 Answers
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You are basically done. Just compute the cartesian products and observe that they are not equal:
\begin{align*} (A\times B)^c &= \{(2,3),(2,4),(4,3),(4,4)\}^C \\ &= \{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2)\} \end{align*} and $$ A^c\times B^c=\{(1,1),(1,2),(3,1),(3,2)\}. $$
One thing to note is that it is assumed the universal set to which the complement $(A\times B)^c$ is taken is $\{1,2,3,4\}\times\{1,2,3,4\}$.
Also, it just so happened that $A^c\times B^c\subseteq (A\times B)^c$. Is this true in general? It is! To see this, suppose $(x,y)\in A^c\times B^c$. Then $x\in A^c$ and $y\in B^c$, so we cannot have $(x,y)\in A\times B$. Hence $(x,y)\in (A\times B)^c$.
answered Sep 13, 2017 at 2:12
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$\begingroup$ but isn't {(1,1), (1,2), (3,1), (2,3)} in (AxB) complement so doesn't that make A complement x B complement a subset of (AxB) complement? $\endgroup$– user510Sep 13, 2017 at 2:29
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$\begingroup$ @user510 In the particular example, they are subsets. To verify if that is a general fact, one has to prove it. See the end of my answer where I do just that. $\endgroup$ Sep 13, 2017 at 2:31
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$\begingroup$ But I want to show a counterexample that will give me the result to be false and not a a general solution. So do I have to pick other values for A and B? $\endgroup$– user510Sep 13, 2017 at 2:48
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$\begingroup$ No, since you've found an example where the equality does not hold, you've shown that this equality is not generally true (otherwise it would have to hold in your example). $\endgroup$ Sep 13, 2017 at 2:50
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I've included a proof by picture below, for a less formal approach that gives the main idea:
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$\begingroup$ i get they are not equal, I want an counterexample that shows that fact though $\endgroup$– user510Sep 13, 2017 at 2:50
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$\begingroup$ You've found one, as @JohnGriffin showed. $\endgroup$ Sep 13, 2017 at 2:52
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$\begingroup$ but that example is showing that they are subsets. Can you explain me how its not showing that $\endgroup$– user510Sep 13, 2017 at 3:03
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2$\begingroup$ @user510 Why do you keep talking about subsets when the question is about equality? $X$ being a subset of $Y$ is not an obstacle to proving that $X \ne Y$, not in the slightest. $\endgroup$ Sep 13, 2017 at 3:23