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When defining the following Cartesian product:

$P(\mathbb{N})\times\mathbb{N}$

Do we get sets that includes the empty set? Such as:

$\{\emptyset\}\times\{2\}=(\emptyset,2)$

Or, because of the definition of Cartesian product with empty set we get:

$\emptyset\times2=\emptyset$

Any explanation would be great, Thank you.

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Let $A$ and $B$ be sets.

$$A\times B:=\{(a,b):a\in A\;\text{and}\; b\in B\}$$

Remark:

$$\varnothing \times A=\varnothing$$ for every set $A$.

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  • $\begingroup$ As simple as it gets... Thanks! $\endgroup$
    – GuD
    Dec 29 '16 at 12:40
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$\mathcal{P}(\mathbb{N}) \times \mathbb{N}$ is the set of all 2-tuples such that the first element is a set of naturals and the second is a natural. So, for example, it contains $(\emptyset, 2)$.


It's often unhelpful to think of a tuple conceptually as anything other than a tuple. To implement a tuple in set theory, of course it has to be a set, but we've chosen our definitions so that we can just treat them however we would intuitively treat a tuple. There's no way we would intuitively take the tuple $(\emptyset, 2)$ and turn it into $\emptyset$.

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The notation is unfortunate in this case. Nowhere here do we take the Cartesian product with anything except $P(\mathbb N)$ and $\mathbb N$. Thus we have $\emptyset\in P(\mathbb N)$, $2\in \mathbb N$, so

$$"\emptyset\times 2"=(\emptyset,2)\in P(\mathbb N)\times \mathbb N$$ and $(\emptyset,2)\neq \emptyset$. I used scare quotes because you should never write it in that confusing way.

Note that $\{\emptyset\}\notin P(\mathbb N)$ and $\{2\}\notin \mathbb N$, so your even more unfortunate looking example doesn't come up.

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  • $\begingroup$ Isn't $\emptyset\times(anything)=\emptyset$? $\endgroup$
    – GuD
    Dec 29 '16 at 12:33
  • $\begingroup$ @MathStudent but the expression in quotes is not the Cartesian product of anything with the empty set. With the Kuratowski construction, it is $\{\{\emptyset\},\{\emptyset,2\}\}$. $\endgroup$ Dec 29 '16 at 12:34
  • $\begingroup$ @MathStudent: In other words, you've confused yourself by writing $\times$ for something that isn't the Cartesian product. $\endgroup$
    – user14972
    Dec 29 '16 at 12:36

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