# I need to disprove an alternate definition of an ordered pair. Why is $\langle a,b\rangle = \{a,\{b\}\}$ incorrect?

So we know that the an ordered pair $(a,b) = (c,d)$ if and only if $a = c$ and $b = d$. And we know the Kuratowski definition of an ordered pair is: $(a,b) = \{\{a\},\{a,b\}\}$

http://en.wikipedia.org/wiki/Ordered_pair#Kuratowski_definition

The proof for the Kuratowski definition is in the wikipedia link.

Now, why is this alternate definition, $(a,b) = \{a,\{b\}\}$ incorrect?

I'm trying to follow the proof for $(a,b) = (c,d)$ iff $a = c$ and $b = d$ as given in the wikipedia link, only for this alternate definition for an ordered pair, in order to search for a contradiction. But I don't think I'm dong it right.

I started with...

• $(a,b) = (c,d)$
• Then $\{a,\{b\}\} = \{c,\{d\}\}$ based on the alternate definition

Now...

• Suppose $a \neq b$
• $\{a,\{b\}\} = \{c,\{d\}\}$
• But since it's an ordered pair, either of the following can be true?
• $a = c$ and $\{b\} = \{d\}$ ?
• OR $a = \{d\}$ and $\{b\} = c$ ?

Yeah I have no idea where to reallly go from here. Is that a contradiction in itself? I can't tell.

Thank you for the help.

Edit: Alright, I have developed a counter example based mostly off of Asaf Karagila answer (Thanks Asaf!). Essentially what I needed to do was prove that, by this definition, a != c or b != d, even when (a,b) = (c,d).

So using what Asaf told me, I set a = {x} and b = y. Which by the incorrect definition gives... (a,b) = {{x},{y}}

Then I set c = {y} and d = x, which gives (c,d) = {{y},{x}} which is equivalent to {{x},{y}}

So, (a,b) = (c,d) even though a != c and b != d, which is a contradiction. I cleared this method with my professor.

Thanks for the help everyone!

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Think of the case where $a$ and $b$ are sets, maybe this will give you the idea of a counterexample... – Lukas Geyer Oct 5 '12 at 19:20
What exactly does $\{a, \{b\}\}$ mean to you? It means the set that consists of the element $a$ and the element $\{b\}$. So, if we're dealing with real numbers, $a$ is a real number and $\{b\}$ is a set that contains the real number $b$. Is this really what you want to say? Do you mean $\{\{a\}, \{b\}\}$? – Graphth Oct 5 '12 at 19:24
One point being, $a = \{d\}$ makes no sense. A real number equals a set that contains a real number? So, maybe you could distinguish the first element by having it be the one that is not inside set brackets. I mean, all you're doing is rewriting the ordered pair in a more complicated way. But, I guess you could do it. – Graphth Oct 5 '12 at 19:29
@Graphth: Probably not -- $\{\{a\},\{b\}\}$ is obviously not a viable ordered-pair construction, because it is symmetric in $a$ and $b$. Further, $a=\{d\}$ makes perfect sense -- it says that $a$ is the singleton set whose element is $d$. Who said $d$ was a real number? – Henning Makholm Oct 5 '12 at 19:30
@Graphth: But the point in set theory is not to build upon the real numbers, instead it builds the real numbers. Everything is a set. Besides, you want set theory for more than just real numbers, you want it for its own sake. – Asaf Karagila Oct 5 '12 at 19:30

The decoding of the first and second elements is not unique. Suppose that $c\neq d$, now we have:

$$(\{c\},d)=\{\{c\},\{d\}\}=(\{d\},c)$$

Recall that the definition of ordered pairs should not only hold for pairs of numbers. It should allow set theory be adequate for expressing a lot. Using this definition, the above shows that the two functions:

1. $f(x)=\{x\}$
2. $g(\{x\})=x$

Are both represented by the same set. Which is usually a sign of an inadequate representation of an ordered pair.

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Technically, if we said whichever one is not inside set brackets is the first element, then it does just fine. – Graphth Oct 5 '12 at 19:28
@Graphth: But... what if you want the function which sends an element to its singleton; and the function which sends a singleton to its element, the sets would be the same. – Asaf Karagila Oct 5 '12 at 19:29
Oh, that seems like an ok counter example. I don't see why he wouldn't accept that. – Joey Oct 5 '12 at 19:47
@Joey: Who wouldn't accept this? – Asaf Karagila Oct 5 '12 at 19:50
@Graphth You're forgetting that you can have sets of sets so both elements of the ordered pair could be "inside set brackets". – David Richerby Apr 6 '14 at 21:34

The ordered pair definition is meant to work for any kind of set, so assuming you work with numbers is really not the right way to go.

There is a reason why the "second coordinate" must contain two elements. Here's why.

Say you define ordered pairs $(a,b) = \{a, \{b\}\}$. Is there anyway you can decide "which one is the first coordinate"? a priori both elements of the ordered pair are sets, and you have no reason to believe that you can distinguish between one of the two by the fact that they belong to some other set. For instance if you take an ordered pair $(a,b) \in \mathcal P(X) \times X$, then for some element $x \in X$ you can consider the ordered pair $(\{x\},x)$, which under your definition gives the set $\{\{x\},\{x\}\}$. Therefore there is no way to distinguish between both coordinates.

Kuratowski's definition ensures that we can know which coordinate is the first /second one because there are well-defined operators that output the right coordinate. Your website linked explains it well.

Hope that helps,

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+1 It helps me! – Graphth Oct 5 '12 at 19:41
Sorry about this but I'm still relatively new at set theory. What is P(X)×X supposed to represent? I do understand why this definition is wrong, I'm just having trouble showing a counterexample or contradiction. Your counterexample makes sense to me, except for P(x)×X. – Joey Oct 5 '12 at 19:44
@Joey : $\mathcal P(X) \times X$ stands for the cartesian product of the set of all subsets of $X$ with $X$ itself. $\mathcal P(X)$ is the set of all subsets of $X$, also called the "power set" of $X$. – Patrick Da Silva Oct 6 '12 at 7:39