# Does $\mathbb{N} \cap \mathbb{N}^2 = \varnothing$? [closed]

Definitions. A cardinal $$\mathfrak{a}$$ is said to be finite if $$\mathfrak{a} \ne \mathfrak{a} + 1$$. A finite cardinal is also called a natural integer. The natural integers form a set, denoted by $$\mathbb{N}$$. The set $$\mathbb{N}^2$$ is the Cartesian product of $$\mathbb{N}$$ with itself. In other words, $$\mathbb{N}^2$$ is the set of ordered pairs $$(m,n) = \bigl\{\{m\},\{m,n\}\bigr\}$$ where $$m$$ and $$n$$ are natural integers.

• Define $\mathbb N^2$. – Parcly Taxel Jul 4 '19 at 16:47
• Are you defining $\mathbb{N}^2$ as a collection of (some encoding of) two element tuples? Are you defining $\mathbb{N}$ as a collection of (some kind of) tuples? – Eric Towers Jul 4 '19 at 16:47
• @lulu That was originally a different question. I changed to this question after it had been downvoted. – user633691 Jul 4 '19 at 16:50
• Everyone: He said he's taking $\Bbb N$ to be the collection of finite cardinals. Then $\Bbb N^2=\{(n,mm):n,mm\in\Bbb N\}$, where he's presumably using the standard (Kuratowski) definition of $(n,m)$. – David C. Ullrich Jul 4 '19 at 16:51
• I can't see why this question is being downvoted. – Wood Jul 4 '19 at 16:56

In comments you say you're defining cardinals using Scott's trick -- that is, a cardinal is the set of all sets with a certain cardinality of minimal rank.

Then suppose there's a Kuratowski pair $$\{\{a\},\{a,b\}\}$$ that is also a cardinal. If $$a\ne b$$ then $$\{a\}$$ and $$\{a,b\}$$ have different cardinalities, so they're not elements of the same cardinal. Therefore $$a=b$$ and our pair has the form $$\{\{a\}\}$$. Since the element $$\{a\}$$ has one element, we must be looking at the number $$1$$ if we're looking at a cardinal at all.

The smallest-rank singleton is $$\{\varnothing\}$$, and there are no other singletons that has this rank. So under Scott's trick, $$1=\{\{\varnothing\}\}$$. And this is indeed a Kuratowski pair -- but, alas! it is $$\langle\varnothing,\varnothing\rangle$$, and $$\varnothing\notin\mathbb N$$ when we're using Scott's trick.

So with this representation $$\mathbb N\cap \mathbb N^2$$ is still empty.

On the other hand, we might define the natural numbers as $$\varnothing, \{\varnothing\}, \{\{\varnothing\}\}, \{\{\{\varnothing\}\}\},\ldots$$ like Zermelo originally did. (Though it is pretty difficult to imagine how this could be extended to a representation of infinite cardinals without obvious seams showing).

In that representation (but still with Kuratowski's pair definition) we have $$n+2 = \langle n, n\rangle$$ for all $$n$$, and therefore $$\mathbb N\cap\mathbb N^2= \{2,3,4,5,\ldots\}$$

If you use this definition:

\begin{alignat}{2} 0 & {} = \{\} && {} = \emptyset,\\ 1 & {} = \{0\} && {} = \{\emptyset\},\\ 2 & {} = \{0,1\} && {} = \{\emptyset,\{\emptyset\}\},\\ 3 & {} = \{0,1,2\} && {} = \{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\}\} \end{alignat}

then every element of $$\mathbb{N}$$ is either $$\emptyset$$ or contains $$\emptyset$$. This is false for the elements of $$\mathbb{N}^2$$ using the standard definition of ordered pair:

$$(a, b) \triangleq \{\{a\}, \{a,b\}\}$$

since $$\{a\}\neq\emptyset$$ and $$\{a, b\}\neq\emptyset$$. So $$\mathbb{N}\cap\mathbb{N}^2=\emptyset$$.

If you use an alternative definition, like $$(a, b) \triangleq \{a, \{a, b\}\}$$, then we can have $$a = b = \emptyset \Rightarrow (a, b) = \{\emptyset, \{\emptyset,\emptyset\}\} = \{\emptyset, \{\emptyset\}\} = 2 \neq \emptyset$$. Under this definition of ordered pair, we have $$\mathbb{N}\cap\mathbb{N}^2 \neq \emptyset$$.

• What about the definition I gave? – user633691 Jul 4 '19 at 17:12
• I actually posted the same argument 15 minutes earlier. Btw, as one sees from what you wrote, it's not true that every element of $\Bbb N$ has $\emptyset$ for an element. – David C. Ullrich Jul 4 '19 at 17:24
• Do you know that that "alternative definition" works? (Ie, that it gives $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$?) – David C. Ullrich Jul 4 '19 at 17:31
• @Wood Thank you for the thoughtful answer. – user633691 Jul 4 '19 at 17:32
• @DavidC.Ullrich No, I'm trusting Wikipedia on this. They give the link to this proof, which I haven't checked: us.metamath.org/mpegif/opthreg.html – Wood Jul 4 '19 at 17:33

I think that the answer to this question is $$\mathbb{N}\cap\mathbb{N}^2=\varnothing$$.

$$\mathbb{N}$$ is a set of numbers: $$\mathbb{N} = \{0,1,2,\ldots\}$$. $$\mathbb{N}^2$$ is a Cartesian cross product of elements of $$\mathbb{N}$$. That is, an element of $$\mathbb{N}^2$$ has the form $$(a,b)$$ where $$a,b\in\mathbb{N}$$. An ordered pair $$(a,b)$$ is defined as $$\{\{a\},\{a,b\}\}$$. Since no element of $$\mathbb{N}$$ takes this form, the intersection must be $$\varnothing$$.

• Can you provide a more formal proof? – user633691 Jul 4 '19 at 17:07
• This amounts to just saying the intersection is empty because it's empty! The point is to prove that no element of $\Bbb N$ has the form $(a,b)$. – David C. Ullrich Jul 4 '19 at 17:27
• If the elements of the intersected sets aren't the same type of element, then no element can belong to both sets. – NicNic8 Jul 4 '19 at 17:51
• [sigh] Right. Except you have to define "same type" and then prove that $n$ is not the same type as $(a,b)$. (See Wood's answer - there's a reasonable definition of $(a,b)$ where $(0,0)=2$.) – David C. Ullrich Jul 4 '19 at 17:56
• @DavidC.Ullrich I see. I've added a definition of ordered pair to my answer. – NicNic8 Jul 4 '19 at 18:27

Note: No, you absolutely cannot take $$(\forall x)(\forall x^\prime)(\forall y)(\forall y^\prime)\big((x,y) = (x^\prime,y^\prime) \implies x = x^\prime \text{ and } y = y^\prime\big).$$ as the definition of ordered pairs! The above is a property of ordered pairs, but it does not specify what $$(x,y)$$ is. In fact there's a standard definition of $$(x,y)$$ in set theory.

Hint: If $$x$$ and $$y$$ are sets then $$(x,y)\ne\emptyset$$ and also every element of $$(x,y)$$ is nonempty.

(Of course the answer depends on the definition of $$\Bbb N$$.)

• I did not downvote but I also do not understand how to proceed from the hint. – user633691 Jul 4 '19 at 17:01
• @simplejack Have you thought about it? If $n\in\Bbb N$ you want to show that $n\ne(j,k)$. If you believe the assertion in the hint it's enough to show that $n=\emptyset$ or $\emptyset\in n$. – David C. Ullrich Jul 4 '19 at 17:13
• So it is the same idea as the answer by @Wood? I don't see how it can be used with the definitions I added to the question. – user633691 Jul 4 '19 at 17:15
• @simplejack Yes, it's the same idea as the other answer. It's also correct, which the other answer is not, quite (it's not true that every $n\in \Bbb N$ has the empty set as an element. – David C. Ullrich Jul 4 '19 at 17:21
• @simplejack It's impossible to do this without a definition of $(x,y)$. There is no definition of $(x,y)$ in your question. – David C. Ullrich Jul 4 '19 at 17:22