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Let A, B and C be sets. Show that it is always true that A \ (B \ C) ⊆ (A \ B) ∪ C. Give an example of specific sets A, B and C to show that it is not necessarily true that A \ (B \ C) = (A \ B) ∪ C.

I don't know where to start and I have never written a proof before.

I have tried: Let x be an element of A \ (B \ C). By the definition of set difference, we have: x is an element of A and x is not an element of (B \ C)

I don't know if this is right or not. If it is right, where do I go from there. If not, how do I do this proof? My professor is not teaching this class face to face. All he is giving us is notes that we copy. No videos or conference calls at all.

Thanks!

Clay
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  • you're right so far $x\in A$ and $x \not \in B\setminus C$. Go further. can $x$ be in $B$ ? in $C$. if so, under what circumstances? If $x$ is in $B$ then as $x\not \in B\setminus C$ so $x$ must be in $C$. But if $x$ isn't in $B$ then $x\in A\setminus B$. – fleablood Sep 24 '20 at 04:35

2 Answers2

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It never hurts to make a picture...

enter image description here

Now can you describe in words what the picture shows?

Every element of $A (\backslash B\backslash C)$ is also a member of $(A \backslash B) \cup C$

Doug M
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It's okay to break into cases and say something such as "Either $x$ is an element of $B$... or it isnt".

So if $x \in A\setminus (B\setminus C)$ then $x$ is in $A$. And $x$ is not in $B\setminus C$.

Now $x \in A$ and either $x \in B$ or .... it isn't. If $x$ is not in $B$ then $x \in A\setminus B$.

And if $x$ is in $B$ then we also have $x$ is not in $B\setminus C$. So IF $x \in B$ but not in $B\setminus C$ then $x$ can not be excluded excluded for $C$ (because that would mean $x$ is in $B\setminus C$. So if $x \in B$ then $x \in C$.

so if $x\in A\setminus(B\setminus C)$ then either $x \in A \setminus B$ or $x \in C$

so $x \in (A\setminus B)\cup C$.

.....

Now for an example $A\setminus (B\setminus C)\ne (A\setminus B) \cup C$ simply note, that if $x \in A\setminus (B\setminus C)$ then $x$ must be in $A$. But if $y\in (A\setminus B) \cup C$ we don't need $y$ to be in $A$-- we could have $y\in C$.

Let $A,B,C$ but any sets where there is a $y\in C\setminus A$. Then $y\not \in A\setminus (B\setminus C)$ but $y \in (A\setminus B) \cup C$.

For a simple example let $C = \{y,1,2,3\}$ and $A=\{2,3,4,5\}$ and $B=\{y,2,4,w,z\}$. (I pretty much made those up off the top of my head; the only thing I needed was $y\in C$ but $y\not \in A$. I could have chosen $C= \{y\}; A=\emptyset; B=\emptyset$. That would make for a more basic example; but I don't think it'd be as illustrative example).

Then $A \setminus (B\setminus C) = A\setminus \{y,4,w,z\}=\{2,3,4,5\}\setminus \{y,4,w,z\}= \{2,3,5\}$.

But $(A\setminus B)\cup C =\{3,5\} \cup C = \{3,5\} \cup \{y,1,2,3\} = \{y,1,2,3,5\}$.

Notice that $\{2,3,5\} \subset \{y,1,2,5\}$ but $\{2,3,5\} \ne \{y,1,2,5\}$

fleablood
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