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This is a problem in a test I found online: Prove that: $(A\setminus B)\cup(B\setminus C)=A\setminus C‎$

When I express it in a defenition of a set, I get stuck.

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What do you mean by group? The word group has a specific meaning and I don't think it coincides with your meaning. –  Michael Albanese Sep 2 '13 at 15:09
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You mean sets, right? Not groups. –  MyUserIsThis Sep 2 '13 at 15:11
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There is a problem... let $A=\{1,2,3,4\}$, $B=\{3,4,5,6\}$ and $C=\{2,4,6,7\}$. Now $A\backslash B=\{1,2\}$, $B\backslash C=\{3,5\}$ but $A\backslash C=\{1,3\}\neq (A\backslash B)\cup (B\backslash C)$. –  Jp McCarthy Sep 2 '13 at 15:14
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Supposedly $C\subseteq B\subseteq A$? –  walcher Sep 2 '13 at 15:16

2 Answers 2

This has no more meaning for groups than it does for sets -- unions and set differences do not respect group operations.

That being said, this result is not true unless there's more to the problem statement than you are giving us.

Take, for instance, $A=\{1,2,4\}$, $B=\{1,2,3\}$, $C=\{1,3,4\}$. Then $A\setminus B=\{4\}$, $B\setminus C=\{2\}$, and $A\setminus C=\{2\}$... and $\{2,4\}\neq\{2\}$.

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When confronted with such a problem, you might sketch a Venn diagram to get an idea of what’s going on, especially if you’re visually oriented. In this case you get the following picture, where I’ve labelled the parts of the diagram in $A\setminus B$ with $1$, the parts in $B\setminus C$ with $2$, and the parts in $A\setminus C$ with $3$. As you can see, $(A\setminus B)\cup(B\setminus C)$ corresponds to the four labelled regions, while $A\setminus C$ corresponds to only two of them. Thus, we’ll have

$$(A\setminus B)\cup(B\setminus C)=A\setminus C\tag{1}$$

if and only if the regions labelled with just $1$ and just $2$ are both empty. The first of these is empty precisely when $A\cap C\subseteq B$, and the second is empty precisely when $A\cap B\subseteq C$. Suppose that these are both true. Certainly $A\cap C\subseteq A$, so $A\cap C\subseteq A\cap B$. Similarly, $A\cap B\subseteq A\cap C$, and therefore $A\cap C=A\cap B$. Thus, $(1)$ is true if and only if $A\cap C=A\cap B$.

This occurs, for instance, if $C\subseteq B$ or $B\subseteq C$. More generally, you might try to show that it occurs if and only if $B\mathbin{\triangle}C\subseteq A$, where $B\mathbin{\triangle}C=(B\setminus C)\cup(C\setminus B)$ is the symmetric difference of $B$ and $C$.

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