# Defintion: Ordered Pairs

1. Sep 24, 2006

### Swapnil

Hi yall, I was just reading a book on set theory and I came across this definition of an ordered pair:
$$\langle a,b\rangle \equiv \lbrace \lbrace a,1 \rbrace, \lbrace b,2 \rbrace \rbrace$$

I think this is a really ingenious way to define an ordered pair but I was wondering are there any other, more intuitive, ways to define an ordered pair?

edit: fixed the pointy brackets thanks to Hurkyl

Last edited: Sep 24, 2006
2. Sep 24, 2006

3. Sep 24, 2006

### Hurkyl

Staff Emeritus
You could just define an ordered pair axiomatically.

Basically, the point of axiomatic set theory is to work with a very minimalist foundation -- so rather than start by assuming the existence of your basic tools (like ordered pairs), it has to actually construct them, and this requires using clever tricks (precisely because you aren't able to use your basic tools).

Incidentally, for your LaTeX, I think you're looking for:

\langle
\rangle

and

\{
\}

(oh, nm, I see you got it)

4. Sep 24, 2006

### mathwonk

that definition fails for the rodered pair <2,1>.

i think the usual definition is <a,b> = {{a}, {a,b}}. as in Kelley's modern algebra, for his continental classroom course on tv, circa 1960.

5. Sep 24, 2006

### Swapnil

So I guess you can define (a,b) the following way: (a,b) := {{a},{a,b}}. Using this definition, the ordered 3-tuple becomes (a,b,c) =
{{a},{a,{b,{b,c}}}. Isn't this way of defining a lot more difficult than the one I mentioned. I mean, the ordered triple would be (a,b,c) = {{a,1},{b,2},{c,3}} using the method I mentioned, right?

6. Sep 24, 2006

Why does it actually fail for (2, 1) ?

7. Sep 24, 2006

### CRGreathouse

Originally I agreed with you, but thinking about it I'm not sure it's a problem, as long as you know the arity beforehand. <2, 1> would be {{1, 2}}, but doesn't that uniquely decompose to <2, 1> as an ordered pair?

It seems to work:

Case 1: Neither element is 1 or 2: {{a, 1}, {b, 2}} has 2 elements with 2 elements each.
Case 2: Both elements are 1 or 2: {{1}, {2}}, {{1, 2}}, {{1}, {1, 2}}, {{1, 2}, {2}}
Case 3: Exactly one element is 1 or 2: {{1}, {b, 2}}, {{1, 2}, {b, 2}}, {{b, 1}, {1, 2}}, {{b, 1}, {2}}.

Case 1 doesn't conflect with case 2 because all elements in case 1 have a number not equal to 1 or 2.
Case 1 doesn't conflect with case 3 because all sets in case 3 that have both elements of cardinality 2 contain {1, 2} which is not in any set in case 1.
Case 2 doesn't conflict with case 3 by checking each case.

8. Apr 28, 2008

### tehminkeh

the definition by $$\langle a,b \rangle = \{\{a,1\},\{b,2\}\}$$ is undesirable mainly due to it's reliance on the existence of 1 and 2. the (arguably) best definition is $$\langle a,b \rangle = \{\{a\},\{a,b\}\}$$

An n-tuple can then be defined as an ordered pair of an element and an (n-1)-tuple. Note that the n-tuple definition doesn't actually rely on the existence of natural numbers, but rather uses natural numbers simply as a naming convention.

I should note that another definition for an n-tuple is in fact a mapping from $$\mathbb{N}$$ to some set, which gives it the structure $$\{\langle 1,a_1 \rangle, \langle 2,a_2 \rangle, \ldots , \langle n,a_n \rangle\}$$

Last edited: Apr 28, 2008
9. Apr 28, 2008

### gel

I'd argue that the best way to define <a,b> is by its properties. That is <a,b> exists for all a,b and <a,b>=<c,d> iff a=c and b=d.

Then the fact that <a,b>=={{a},{a,b}} or <a,b>=={{a,1},{b,2}} satisfies the required properties is a proof that sets exist => ordered pairs exist.

Last edited: Apr 28, 2008