Please Explain Kuratowski Definition of Ordered Pairs

Question

I've seen this Kuratowski definition for ordered pairs, but can't fathom why it implies an order to $x$ and $y$

$(x,y):=\{\{x\}, \{x,y\}\}$

As I understand sets, $\{\{x\}, \{x,y\}\}$ is also $\{\{x,y\}, \{x\}\}$. Only when I think about the Axiom of Union does $\{\{x\}, \{x,y\}\}$ "collapse down" to $S = \{x, y\}$, but that doesn't give me much either. All I can see is some as yet hidden message in the set saying "I am the set $\{x,y\}$ and my order of $x$ first is indicated by having $\{x\}$ along for the ride."

Answer 1

You said:

I can't fathom why it implies an order to $x$ and $y$

It doesn't really. The $x$ and $y$ in the ordered pair $(x, y)$ don't really have an order. Who's to say that the $x$ is first and the $y$ is second? If you read right-to-left, you'd say that the $y$ was first and the $x$ was second.

The important thing isn't which is first. The important thing is that the set we pick to represent $(x,y)$ must be different from the set that represents $(y, x)$, because these are different pairs.

As you pointed out $\{x, y\}$ is the same set as $\{y, x\}$. But let's consider the Kuratowski pairs $(x,y)$ and $(y,x)$:

$$\begin{eqnarray} (x,y) = \{\{x\},\{x,y\}\} \\ (y,x) = \{\{y\},\{x,y\}\} \end{eqnarray} $$

Hey look, they’re different sets. That's what we needed.

Kuratowski's definition was preceded by a number of others. The one by Felix Hausdorff may make you feel more comfortable:

$$\begin{eqnarray} (x,y) = \{\{x, 1\}, \{y,2\}\} \\ (y,x) = \{\{y, 1\}, \{x,2\}\} \end{eqnarray} $$

Now the order you wanted is explicit.

But it is important to realize that the $1$ and $2$ here are completely arbitrary markers! It would have worked just as well for Hausdorff to have used some different markers to indicate which component was first:

$$\begin{eqnarray} (x,y) = \{\{x, \text{potato}\},\{y,\text{banana}\}\}\\ (y,x) = \{\{y, \text{potato}\},\{x,\text{banana}\}\}\\ \end{eqnarray} $$

Now you can recognize the first component of the pair because it is associated with $\text{potato}$.

The point is that the details of the particular representation aren't important. We only care that the representation does what we need it to. For ordered pairs, we need to be able to form the pair $(x, y)$ for any $x$ and $y$; we need to be able to extract the components again, and crucially, we need $(x,y) $ to be equal to $(a, b)$ if and only if $x=a$ and $y=b$. Both Kuratowski's and Hausdorff's definitions do this, and so do many others.

Which definition we pick is not really important. What is important is that the objects we choose to represent ordered pairs must behave like ordered pairs. If we get that much, we are mathematically satisfied. The Kuratowski definition is used not because it captures some basic essence of "ordered pair"-ness, but because it does what we need it to do, which is just enough.

Answer 2

Just how you define order pairs concretely is an "implementation detail". Whichever definition you adopt just has to meet a basic requirement: from $\langle x, y \rangle$, you must be able to uniquely recover each of $x$ and $y$ with (preferably simple) functions $first(z)$ and $second(z)$.

The Kuratowski construction meets this criterion.

$first((x,y)) = x$

Note that $\{x\} = \{x\} \cap \{x,y\} = \bigcap \{\{x\}, \{x,y\}\} = \bigcap z$ where $z = (x,y)$. Now, as for any set, $\bigcup\{x\} = x$. So if $z = (x,y)$ then $x = \bigcup \bigcap z$, so we can define $first$ as: $$ first(z) = \bigcup \bigcap z. $$

$second((x,y)) = y$

Taking the union gives $\{x,y\} = \{x\} \cup \{x,y\} = \bigcup \{\{x\}, \{x,y\}\} = \bigcup z$ where $z = (x,y)$. Consider $\{x,y\} \setminus \{x\}$, which in terms of $z$ is $\bigcup z \setminus \bigcap z$. It's equal to $\emptyset$ if $y = x$, and equal to $\{y\}$ otherwise. So in any case we can recover (return) $y$ by defining $$ second(z) = \begin{cases} \\ &first(z)&\quad\text{if $\bigcup z \setminus \bigcap z = \emptyset$}, \\ &\bigcup (\bigcup z \setminus \bigcap z) &\quad\text{otherwise}, \\ \end{cases}$$

These definitions meet the essential requirement: $$ z = (x, y) \iff [first(z) = x \text{ and } second(z) = y], $$ and all three functions have elementary definitions.

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A fact I used repeatedly and might as well prove: $\bigcup \{x\} = x$.

For any set $A$, $\bigcup A$ is the set of all things $z$ that are members of some thing $y$ in $A$: that is, $\bigcup A = \{ z\mid (\exists y\in A)\,z\in y\}$. In "union of a family of sets" notation, $\bigcup A = \bigcup_{a\in A} a$. So $\bigcup \{x\} = \{z\mid(\exists y\in \{x\})\,z\in y\} = \{z\mid(\exists y = x)\,z\in y\} = \{z\mid z\in x\} = x$.

Answer 3

The point of the ordered is for it to satisfy the following property: $$(a,b) = (c,d) \text{ if and only if }a=c\text{ and }b=d.$$ The definition given satisfies that condition; it doesn’t matter how it satisfies it, just that it does.

It is true that $(a,b)$ is both $\{ \{a\},\{a,b\}\}$ and $\{\{a\}, \{b,a\}\}$ and $\{\{a,b\},\{a\}\}$ and $\{\{b,a\},\{a\}\}$. It doesn’t matter: they are all the same set and they all “are” $(a,b)$. The point is that $(a,b)=(c,d)$ if and only if $a=c$ and $b=d$.

To verify that this happens, note that if $a=c$ and $b=d$, then $$(a,b) = \{\{a\},\{a,b\}\} = \{\{c\},\{c,d\}\} = (c,d).$$

Conversely, suppose that $(a,b)=(c,d)$. That means that the sets $\{\{a\},\{a,b\}\}$ and $\{\{c\},\{c,d\}\}$ are equal as sets.

To verify that this implies that $a=c$ and $b=d$, consider two cases:

Case 1. $a=b$. Then $(a,b) = \{ \{a\},\{a,b\}\} = \{\{a\},\{a,a\}\} = \{ \{a\}\}$.

As this is equal to $\{\{c\},\{c,d\}\}$, then $\{c\}\in\{\{a\}\}$, which means $\{c\}=\{a\}$, which means $c=a$. And $\{c,d\}\in\{\{a\}\}$, so $\{c,d\} = \{a\}$, hence $d=a=b$. Thus, $a=c$, and $b=d$ in this case.

Case 2. $a\neq b$.

Recall that is $X$ is a set, whose elements are sets, then $$\begin{align*} \cup X &= \bigcup_{S\in X}S\\ \cap X &= \{s\in\cup X\mid s\in S\text{ for all }S\in X\}. \end{align*}$$

If $(a,b) = (c,d)$, then, as sets, $\cap (a,b) = \cap (c,d)$. We have $$\begin{align*} \cap (a,b) &= \cap\{ \{a\},\{a,b\}\}\\ &= \{a\}\cap\{a,b\} = \{a\}\\ \cap (c,d) &= \cap \{ \{c\},\{c,d\}\}\\ &= \{c\}\cap\{c,d\} = \{c\}. \end{align*}$$ Therefore, $\{a\}=\{c\}$, so $a=c$.

And $\cup (a,b) - \cap (a,b)$ must be equal to $\cup(c,d)-\cap(c,d)$, so $$\begin{align*} \cup(a,b)-\cap(a,b) &= \cup\{ \{a\},\{a,b\}\} - \{a\}\\ &= (\{a\}\cup\{a,b\})- \{a\}\\ &= \{a,b\} - \{a\} = \{b\};\\ \cup(c,d)-\cap(c,d) &= \cup\{ \{c\}, \{c,d\}\} - \{c\}\\ &= (\{c\}\cup\{c,d\})-\{c\} = \{c,d\}-\{c\} = \{d\}, \end{align*}$$ so we must have $\{b\}=\{d\}$, and therefore $b=d$.

In both cases, if $(a,b)=(c,d)$, as sets, then $a=c$ and $b=d$.

Which is what we want from an ordered pair. So now that we know this definition will guarantee this, we can safely forget about it in general and just use the “defining property” of the ordered pair: $(a,b) = (c,d)$ if and only if $a=c$ and $b=d$.

Answer 4

Say you have the ordered pairs $(x,y),(a,b)$, i.e, $\{\{x\},\{x,y\}\}$, idem for the other.

What do you do if I ask them if they're equal? You test equality as sets. It turns out that they're equal iff $x=a$ and $y=b$ (why?).

Answer 5

We define $(x, y)$ as $\{\{x\}, \{x, y\}\}$. This absolutely implies an order: the first element of the ordered pair is the one that appears on its own in a singleton. $(y, x)$ would be $\{\{y\}, \{x, y\}\}$.

Take $\{\{5\}, \{3, 5\}\}$, for example. Is that $(5, 3)$ or $(3, 5)$? The correct answer is completely unambiguous.

Answer 6

We have that $$(a,b)=\{ \{a\},\{a,b\}\}$$ and that $$(b,a)=\{ \{b\},\{b,a\}\}=\{ \{b\},\{a,b\}\}$$ The two sets differ by a single element, namely $\{a\}$ and $\{b\}$.

Answer 7

The order in sets usually has no impact. {a,b} = {b,a}

But the number of members in a set DOES have an impact. {a,b} $\neq$ {a,b,c}

The reason they are not equal is because {a,b} contains 2 members, while {a,b,c} contains 3 members. And if two sets have a different number of members then they are not equivalent to each other. They are different sets.

Kurtovsky uses this fact in an ingenious way. First he takes the ordered pair and puts it in a set which has two members. But, as opposed to the ordered pair, each of these members is a set in itself. Lets call them member-set-1 and member-set-2.

(a,b) = { member-set-1, member-set-2 }

Of course this in itself is no good. Because in set theory {a,b} = {b,a}. But watch closely:

Next he brings the members of the ordered set into his groups.

member-set-1 = {a}

This means that the first member of the Kurtovsky Pair Set contains the first member of the paired couple: $a$ in it. And only the first member. (The "Abscissa").

Inside the second member of the Kurtovsky Pair Set he brings BOTH members of the paired couple, both $a$ and $b$!

member-set-2 = {a,b}

Notice that $a$ - the "abscissa" - is in both the first and second members of the Kurtovsky Pair Set, but $b$ - the "ordinate" is only in the second member.

Now we'll see how Kurtovsky forced the mathematical notation to be used "against itself" so that any (x,y) where x is different from y, is not equal to (y,x).

According to the formula: (x,y) = { {x}, {x,y} } we have x twice and y once in the set. Right?

But according to the formula: (y,x) = { {y}, {y,x} } now we have y twice and x only once. Clearly NOT the same as (x,y).

Looking at it another way: (x,y) = { {x}... etc.

Meaning there is a set with x and only x in it.

While (y,x) = { {y}... etc. Here there is no set with x and only x in it. So clearly the two sets are different.

Nice.