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Given two sets$$A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.$$ Then I compute the $A\setminus B=\{5\}.$

But my book$^\dagger$said: ... . Also, observe that $x\notin A\setminus B$ does not mean that $x\notin A\lor x\in B$. Why?

I don't know how to explain the Why, please give me some example.

$^\dagger$a friendly introduction to analysis, second edition, page $4$.

EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set.

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    Right -- Lela Hatami's answer gives you the right concept; ignore the book's question -- it's either a typo or worse. – 2017-02-13
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    angryavian was the first user to comment on your question, but apparently, that comment was deleted (presumably by angryavian). I edited my comment to point to Lela Hatami's answer instead. – 2017-02-13
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    @quasi Thank you, But my name is Leila with $i$ – 2017-02-13
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    @Leila Hatami -- sorry; I was careless. – 2017-02-13
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    @quasi Your welcome. Thanks again... – 2017-02-13
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    @Leila: Do not add irrelevant tags. – 2017-03-27

1 Answers 1

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It seems wrong:

$x \in A\setminus B \iff x\in A \cap B' \iff x \in A \wedge x\notin B$

So

$x \notin A\setminus B \iff x \notin A \: \vee \: x\in B$