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I need to find three sets for both statements I have above. I have tried drawing Venn diagrams and shading appropriately then adding numbers in each shaded region to try and guess and check for sets $A$, $B$ and $C$ but am unsure how to find an appropriate solution.

Rodrigo Dias
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Bruh
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    I'm feeling blind, but what is the difference between the two statements in the title? – Michael Burr Jun 10 '20 at 03:28
  • As an aside, if all you care about are quick examples and all you care about for the restriction is that $A,B,C$ are non-empty and you don't care about repetition... then just have as many of them be equal to ${1}$ or ${2}$ as possible. For the first, ${1}=A=B=C$ works to have the expressions be equal. For the second, ${1}=A,~{2}=B=C$ works to have the expressions be unequal. – JMoravitz Jun 10 '20 at 03:30
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    @MichaelBurr equality versus inequality. As for the left side versus the right, it is effectively showing that intersection and union are not associative with one another in general. – JMoravitz Jun 10 '20 at 03:30
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    @MichaelBurr One is equal, and one is not equal. The exercise is to show that $(A \cap B) \cup C$ and $A \cap (B \cup C)$ are sometimes the same, and sometimes different. – user797616 Jun 10 '20 at 03:31
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    @JMoravitz , user797616 That makes sense. I find it confusing because the OP asks for three sets, not six sets and the two statements appear to be connected by an and, but this is not a mathematical and. – Michael Burr Jun 10 '20 at 03:39

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I think you're on the right track with the Venn diagrams. If you plot $(A \cap B) \cup C$ and $A \cap (B \cup C)$ on separate diagrams, you'll see that they're not the same set in general. How? Because, out of the $8$ regions shaded in each diagram, there are some regions shaded in each diagram that are not shaded in the other diagram.

To find an example of a set where they are the same, start writing some distinct numbers (or whatever other objects) into the regions they have in common. You're also allowed to put numbers in regions which are unshaded in both diagrams. Leave any regions blank that belong to one but not the other. Make sure, by the end, that there's at least one number in each circle, so that your sets are non-empty. Then, collect the elements in each circle, and they represent your sets!

To get an example where they're not equal, add another element to a region that is in one Venn diagram, but not the other, and do the same stuff.

Hope that helps!

user797616
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$$B \subset C \subset A$$ $$A \subset B \subset C$$

For general view is good to write right hand side as: $$(A\cap B)\cup C=(A \cap B) \cup(A \cap C)$$ now you see, that difference between sides is difference between sets from right hand of union: on left you have $C$ on right $(A \cap C)$ and on this way you can create any example you would like.

zkutch
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Consider the Venn diagram above. Let $$\begin{align*}A&=\text{Yellow}\cup \text{Green}\cup \text{Red}\cup \text{Dark},\\ B&=\text{Cyan}\cup \text{Blue}\cup \text{Green}\cup \text{Dark},\\ C&=\text{Magenta}\cup \text{Blue}\cup \text{Red}\cup \text{Dark}\end{align*}$$ then: $$A\cap B=\text{Green}\cup \text{Dark}\\ (A\cap B)\cup C=\text{Magenta}\cup \text{Blue}\cup \text{Green}\cup \text{Red}\cup \text{Dark}\\ B\cup C=\text{Cyan}\cup \text{Magenta}\cup \text{Blue}\cup \text{Green}\cup \text{Red}\cup \text{Dark}\\ A\cap (B\cup C)=\text{Green}\cup \text{Red}\cup \text{Dark}$$ We see if $\text{Magenta}\cup \text{Blue}$ is empty, then equality holds, if not empty, the inequality holds.
In other words, it depends on whether $((A\cap B)\cup C)\setminus(A\cap (B\cup C))=C\setminus A$ is empty or not,
i.e. if $C\subseteq A$ then equality holds, if $C\not\subseteq A$ the inequality holds.
From here the examples (if not to consider the diagram above as the example of sets of points on a plane) may be derived rather fast, e.g. $$A=\{1,2\},\,B=\{3\},\,C=\{2\}\hbox{ for the first case and}$$ $$A=\{1\},\,B=\{3\},\,C=\{2\}\hbox{ for the second one.}$$

Alexey Burdin
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Let's abuse notation and say that we are allowed to designate (we are NOT allowed to do this) nothing math objects as $k = $nothing at all. and allowed to list them in sets so that the set $\{1,2,3,k,4\} = \{1,2,3,4\}$ because $k$ is nothing at all.

We AREN'T allowed to do this but lets pretend we can.

Let's suppose we have the following seven items. $a, ab, ac, abc, b, bc,c$ but some of them might not be any thing at all. And lets assume no two of them are the same.

Now suppose $A=\{a,ab,ac, abc\}$ and $B= \{b,ab,bc,abc\}$ and $C=\{c,ac,bc,abc\}$.

The $(A\cap B) \cup C = \{ab,abc\}\cup\{c,ac,bc,abc\}=\{c,ab,ac,bc,abc\}$.

And $A\cap (B\cup C) = \{a,ab,ac, abc\}\cap \{b,ab,bc,abc, c,ac\}=\{ab,ac,abc\}$.

For these two sets to be equal we must have $c$ and $bc$ not exist.

So we can have $a=1, ab=2, ac=3,b=4;abc=5$ and $A=\{1,2,3,5\}, B=\{2,4,5\}$ and $C= \{3,5\}$ and $(A\cap B) \cup C=A\cap (B\cup C) = \{2,3,5\}$.

Not all of them have to exist. At a bare minimum we can have $A=C=\{ac\} =\{3\}$ and $B=\{b\}= 4$ and then $A\cap (B\cup C)=A\cap (B\cup A)=A=\{3\}$ and $(A\cap B)\cup C = (A\cap B)\cup A = A=\{3\}$

And for the sets to not be equal we just need to have either $c$ or $bc$ exist.

We could have $a=1, ab=2, ac=3,b=4;abc=5; c= 6; bc=7$ then $A=\{1,2,3,5\}, B= \{2,4,5,7\}, C=\{3,5,6,7\}$ and $(A\cap B)\cap C = \{2,3\}\cup C=\{2,3,5,6,7\}$ and $A\cap (B\cup C)= \{2,3,5\}$.

Or we could minimize $A= \{a\}, B=\{b\}, C=\{c\}$ then $(A\cap B)\cup C = \emptyset \cup C = \{c\}$ while $A\cap (B\cup C)=\{a\} \cap \{b,c\} = \emptyset$.

fleablood
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