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Suppose I have $B=\{w,x\}$. I am asked to find $|\varnothing\setminus B|$ and $|B\times\varnothing|$.

Since $\varnothing$ is the empty set, I would imagine $\varnothing-B=\varnothing$. Thus, I would figure $|\varnothing\setminus B|=1$. Is this right?

What about for $|B\times\varnothing|$? Is that even possible to compute or is it undefined? I'd have $\{w,x\}\times\{\}$. So would the ordered pairs be $(w,)$ and $(x,)$? That doesn't really seem to make sense.

Edit: Based on an answer here, it would seem that $|B\times\varnothing|=|\varnothing|=1$. Is this correct?

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    $\begingroup$ Wait, how is $|\varnothing|=1$? $\endgroup$ – Asaf Karagila Feb 15 '15 at 9:40
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Suppose $$|\varnothing\setminus B|=1,$$ Then there exist $$x\in\varnothing$$ such that $$x\not\in B.$$

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The size of $\emptyset$ is defined to be $0$. Also $B\times\emptyset=\{(x,y):x\in B \wedge y\in \emptyset\}=\emptyset$. In this sort of problems do use the definition since it clears your mind.

You don't have to imagine that $\emptyset \setminus B = \emptyset$. In fact, $\emptyset \setminus B = \{x:x\in\emptyset\wedge x \notin B \}$. Is there any $x$ satisfying the conditions to be in the last set I mentioned?

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  • $\begingroup$ So you are, in effect, saying that the answer to both of my problems is $|\varnothing|=0$. Is that right? $\endgroup$ – snake Feb 15 '15 at 3:59
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    $\begingroup$ @snake yes, you can see it here $\endgroup$ – Vladimir Vargas Feb 15 '15 at 4:01

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