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Prove/disprove:

For every 3 sets $A,B,C$

$A \times ( B \setminus C) = (A \times B) \setminus (A \times C)$

So I separated into 2 cases:

1) $|B \cap C| > 0$

There exist $y$ such that $y\in B$ and $y \in C$

so there exists $x \in A$ such that $(x,y) \in (A \times B)$ and $(x,y) \in (A \times C)$

So $(A \times B) \setminus (A \times C) = (A \times B) \setminus \{(x,y)\}$

And $( B \setminus C) $ is just $B \setminus \{y\}$ so $A \times ( B \setminus C) = (A \times B) \setminus \{(x,y)\}$ as well

Case 2 is when there's no intersection between $B$ and $C$ so it's quite easy to see it's true for this case.

I'm not sure about my answer, can someone please tell me how to solve this right?

Thanks!

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1 Answer 1

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idea:

You should start with $(x,y) \in A \times (B/C)$. By definition this means $x \in A$ and $y\in B/C$. This implies $(x,y) \in A \times B$ And $(x,y) \not\in A \times C$. Consequently $$(x,y) \in (A \times B)/(A \times C)$$

Now try the other set containment.

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