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I tried many sets for this and I can't find an example when this statement is false.

$$ (X \times Y )-(B\times D)= [(X-B) \times Y] \cup [X \times (Y -D)] $$

where $B \subseteq X$ and $D \subseteq Y$.

Some help please, just hint me the sets please. Thanks for your time and help.

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    just sets sorry i edited. – 2017-02-01
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    What makes you think it's false? If you can't find a counterexample, maybe you should try to prove that it's always true? – 2017-02-01
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    This thing is indeed true. Why do you want a counterexample? – 2017-02-01
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    if $(x,y)$ in the first set either x is not in B or y is not in D. So (x,y) is in the second set. If (x,y) is in the second set either x is not in B or y is not in D. So (x,y) is in the first set. There .... now you don't have to read my long tedious answer. – 2017-02-01

1 Answers 1

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Let $(x,y) \in (X\times Y)-(B \times D)$

The $x \in X$ and $y \in Y$.

Case 1: $x \not \in B$ then $x \in (X-B)\times Y\subset (X-B)\times Y \cup X \times (Y-D)$

Case 2: $x \in B$. Then $y \not \in D$ else $(x,y) \in B \times D$ which is a contradiction. So $(x,y) \in X \times (Y-D)\subset (X-B)\times Y \cup X \times (Y-D)$

So $(X\times Y)-(B \times D) \subset (X-B)\times Y \cup X \times (Y-D)$....

If $(x,y) \in (X-B)\times Y \cup X \times (Y-D)$ then either $(x,y) \in (X-B)\times Y$ or $(x,y) \in X \times (Y-D)$ and either way $x\in X$ and $y \in Y$.

If $(x,y) \in (X-B)\times Y$ then $x \in X - B$ so $x \not \in B$ so $(x, y) \not \in B \times D$ so $(x,y) \in (X\times Y)-(B \times D)$

If $(x,y) \in X \times (Y-D)$ then $y \in Y-D$ and $y \not \in D$ so $(x,y) \not \in B\times D$ so $(x,y) \in (X\times Y) - (B-D)$.

So $(X-B)\times Y \cup X \times (Y-D) = (X\times Y) - (B\times D)$.

So the sets are equal.