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I need someone to assess whether the proof I attempt to provide for the following statement is correct.


Statement:

CORRECTED

$$ \left( A \cup B \right)\cap C \neq A \cup \left(B \cap C \right)$$


My approach: use indicator functions.

Two indicator functions $I_{S_1}$ are $I_{S_2}$ are identical if and only if the two sets, $S_1$ and $S_2$, are identical.

This means that $ \left( A \cup B \right)\cap C = A \cup \left(B \cap C \right)$ if and only if $I_{\left( A \cup B \right)\cap C} = I_{A \cup \left(B \cap C \right)}$.

$I_{\left( A \cup B \right)\cap C} = \left(I_A+I_B-I_A I_B\right)I_C = I_A I_C + I_B I_C - I_A I_B I_C$
$I_{A \cup \left(B \cap C \right)}= I_A + I_B I_C - I_A I_B I_C $

These 2 expressions are not always equal, therefore I conclude that these sets are not always identical.


I know, another approach I may try is to show that either $\left( A \cup B \right)\cap C \not\subset A \cup \left(B \cap C \right)$ or $\left( A \cup B \right)\cap C \not\supset A \cup \left(B \cap C \right)$. I am curious whether the first one above is rigorous enough?


I wrote wrongly the statement, which means some of the solutions provided below don't answer it precisely. Anyway I wasn't looking for a solution but rather for a proving method. Thanks anyone for suggestions!

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    $\begingroup$ If the question is that there exists some sets where $(A\cup B)\cap C\neq (A\cap B)\cup C$ then okay. If you've been taught indicator functions and it is an appropriate tool to use in your class, then fine. I personally would prefer just seeing an example. Many times for questions like these, examples and counterexamples can be formed using just $\emptyset$'s and $\{1\}$'s. I will point out also that in your final statement, we don't actually care that $(A\cup B)\cap C\subset (A\cap B)\cup C$ (although it is in fact a true statement), just that they aren't equal. $\endgroup$ – JMoravitz Aug 25 '17 at 15:09
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    $\begingroup$ If the question is that for all sets $A,B,C$ we have $(A\cup B)\cap C\neq (A\cap B)\cup C$, then think again and look for counterexamples. What happens if $A,B,C$ are all the same set? $\endgroup$ – JMoravitz Aug 25 '17 at 15:10
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    $\begingroup$ To show there exists at least one collection of sets $A,B,C$ where the statement is true, I'd show the example where $A=B=\emptyset$ and $C=\{1\}$. To show there exists at least one collection of sets $A,B,C$ where the statement is false (and therefore it isn't true that the statement is true for all sets, only for some), I'd show the counterexample where $A=B=C=\emptyset$. $\endgroup$ – JMoravitz Aug 25 '17 at 15:20
  • $\begingroup$ $(A\cup B)\cap C$ are elements that must be in $C$ and must also be in one of $A$ or $B$. $(A\cap B)\cup C$ are elements that might be in $C$ or could be in both $A$ and $B$. So if $a \in C$ but $a \not \in A\cup B$ that is a counter example. So Let $a \in C; a \not \in A; a\not \in C$ then $a \not \in (A\cup B)\cap C$ but $a \in (A\cap B)\cup C$. So that sets can not be equal. Far simpler and obvious. $\endgroup$ – fleablood Aug 25 '17 at 15:27
  • $\begingroup$ Let $A\ne \emptyset =C$. $\endgroup$ – DanielWainfleet Aug 25 '17 at 16:14
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Just take $A = B = \{0\}$ and $C = \{1\}$. Then $A \cup B = A \cap B = \{0\}$ and $$\left( A \cup B \right)\cap C = \{0\} \cap \{1\} = \{\} = \emptyset$$ while $$\left( A \cap B \right)\cup C = \{0\} \cup \{1\} = \{ 0, 1 \}.$$

This simple examples shows that in general, $$\left( A \cup B \right)\cap C \neq \left( A \cap B \right)\cup C.$$

There are cases where equality holds. Just take $A = B = C.$

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The elements of $(A \cup B) \cap C$ must be in $C$ and the elements of $(A \cap B) \cup C$ might be in $C$. Obviously these are different concepts and we can come up with a counter example but finding and element in the latter that is not in $C$.

Say $1 \not \in C$ but $1 \in (A\cap B) \cup C$ so $1 \in (A\cap B)$.

So let $1 \in A; 1 \in B; 1\not \in C$. Then $1 \not \in (A\cup B)\cap C \subset C$. But $1 \in (A\cap B) \subset (A\cap B) \cup C$. So $(A\cup B)\cap C\ne C \subset (A\cap B) \cup C$.

If you want to be even more specific: Let $A = B = \{1\}$ and $C = \emptyset$. Then $(A\cup B) \cap C = \emptyset$ while $(A \cap B) \cup C = \{1\}$.

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  • $\begingroup$ Would you still use this approach if there were more caps, cups and more sets involved? $\endgroup$ – Sandu Ursu Aug 25 '17 at 15:43
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    $\begingroup$ Mmm, maybe. Actually, I'd probably break down $(A \cup B) \cap C = (A\cap C) \cup (B\cap C) = (A\cap B \cap C) \cup (A\cap B^c \cap C) \cup (A^c \cap B \cap C)$ whereas $(A\cap B)\cup C$ is $(A\cap B \cap C^c)\cup (A\cap B \cap C) \cup (A^c\cap B \cap C)\cup (A\cap B^c \cap C))\cup (A^c\cap B^c \cap C)$ which are different unions of disjoint sets. ... which I guess is the actual purpose of "indicator functions" which is a concept I have not seen taught much. $\endgroup$ – fleablood Aug 25 '17 at 15:56
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You wrote down that

$$ \tag 1 I_A + I_B I_C - I_A I_B I_C \neq I_A I_C + I_B I_C - I_A I_B I_C $$

but did not explain why the LHS is not always equal to the RHS.

You need to finish off the argument with some simple observation. Well, if $C$ is the empty set, then $I_C = 0$ and (1) becomes

$$ \tag 2 I_A \ne 0$$

and I can think of some situations where (2) is true.

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