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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.

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  • $\begingroup$ Define $\mathbb N^2$. $\endgroup$ – Parcly Taxel Jul 4 '19 at 16:47
  • $\begingroup$ 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? $\endgroup$ – Eric Towers Jul 4 '19 at 16:47
  • $\begingroup$ @lulu That was originally a different question. I changed to this question after it had been downvoted. $\endgroup$ – user633691 Jul 4 '19 at 16:50
  • $\begingroup$ 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 $(n,m)$. $\endgroup$ – David C. Ullrich Jul 4 '19 at 16:51
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    $\begingroup$ I can't see why this question is being downvoted. $\endgroup$ – Wood Jul 4 '19 at 16:56
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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 $$\mathbb N\cap\mathbb N^2= \{2,3,4,5,\ldots\} $$

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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 $\mathbb{N}\cap\mathbb{N}^2 \neq \emptyset$.

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  • $\begingroup$ What about the definition I gave? $\endgroup$ – user633691 Jul 4 '19 at 17:12
  • $\begingroup$ 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. $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:24
  • $\begingroup$ Do you know that that "alternative definition" works? (Ie, that it gives $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$?) $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:31
  • $\begingroup$ @Wood Thank you for the thoughtful answer. $\endgroup$ – user633691 Jul 4 '19 at 17:32
  • $\begingroup$ @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 $\endgroup$ – Wood Jul 4 '19 at 17:33
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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 $\varnothing$.

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  • $\begingroup$ Can you provide a more formal proof? $\endgroup$ – user633691 Jul 4 '19 at 17:07
  • $\begingroup$ 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 $(a,b)$. $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:27
  • $\begingroup$ If the elements of the intersected sets aren't the same type of element, then no element can belong to both sets. $\endgroup$ – NicNic8 Jul 4 '19 at 17:51
  • $\begingroup$ [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 $(0,0)=2$.) $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:56
  • $\begingroup$ @DavidC.Ullrich I see. I've added a definition of ordered pair to my answer. $\endgroup$ – NicNic8 Jul 4 '19 at 18:27
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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 $\Bbb N$.)

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  • $\begingroup$ I did not downvote but I also do not understand how to proceed from the hint. $\endgroup$ – user633691 Jul 4 '19 at 17:01
  • $\begingroup$ @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 $\emptyset\in n$. $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:13
  • $\begingroup$ 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. $\endgroup$ – user633691 Jul 4 '19 at 17:15
  • $\begingroup$ @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. $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:21
  • $\begingroup$ @simplejack It's impossible to do this without a definition of $(x,y)$. There is no definition of $(x,y)$ in your question. $\endgroup$ – David C. Ullrich Jul 4 '19 at 17:22