Talk:Ordered pair

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If (x,y) is {{x},{x,y}}, what is (x,x)? -phma

Obviously, (x,x) is {{x}, {x,x}} which is the same as {{x},{x}}. But then the expression that determines the second element is wrong. Suggestions? --Rade Kutil
No, it's quite right. The ordered pair (x,x) is symmetric -- switching the elements has no effect, and that's exactly what is reflected by the set notation. we get (x,x) = { {x} } after full reduction. -- Tarquin
Yes, but corresponding to what the article says, x would not be the second element of (x,x), which is wrong. --Rade
Then those statements are badly formulated. I've just checked my notes from my university logic class, and it seems we didn't linger much on this. It's not a definition you're meant to work with. The 2nd statement on the main page fails here because it is basically saying "x is the second element IF it's in the pair but not the first element". So if x is both elements, it fails. -- Tarquin
No; the second statement doesn't fail for (x,x) = hammer, and both statements are correct. The second statement says in words "x is the second element IF (i) it's in the pair AND, (ii) if there are distinct elements in p, then x is only in one of them".
(exist Y in p : x in Y) and (for all Y1 in p, for all Y2 in p : if Y1 ≠ Y2 then (x not in Y1 or x not in Y2)).
Since for (x,x), p consists of a single set, Y = {x}, then possibility described in the second clause can never occur - it is never the case that Y1 ≠ Y2, so the second clause of the above statement is trivially true; just as the statement {for all Y in p: if 6 = 9, then Jimi Hendrix lives} is trivially true. Chas zzz brown 06:49 Nov 20, 2002 (UTC)
Yes, it is correct now. It wasn't when this discussion was going on and until AxelBoldt inserted the correct statement. --Rade

Beyond that, though, the definition of (x,y) as {{x},{x,y}} is arbitrary. It's a "dummy definition", if you will, like there are "dummy variables".

Hmmm... putting on my Platonic hat, I have to disagree here that this is an arbitrary "dummy" definition. It's a definition, and a valid one at that; no different than, for example, the definition of ∑i=1 to 5{f(i)} as shorthand for f(1) + f(2) + f(3) + f(4) + f(5). As noted, it's rather clumsy to write in set notation; and thus the usual ordered pair is both clearer and more natural to work with; but I sleep better at night knowing that there is a consistent definition underlying this shorthand! Chas zzz brown 06:49 Nov 20, 2002 (UTC)

It serves only to give a solid set theory foundation to what we do with ordered pairs -- once we've seen it works we can ignore it. Analogously, complex numbers are initially defined as ordered pairs, and once we introduce the notation i = (0,1), we can cleanly forget about the ordered pair and write x+iy -- Tarquin

Why not use the definition (x,y) as {x,{x,y}}? It could avoid the confusion when x=y. -- Wshun

This then requires the axiom of regularity (a.k.a. axiom of foundation) to handle the possibility that one has sets x and z, with x={z} and z={x}, but not x = z. Then (x,x) = {x, {x, x}} = {x, {x}} = {x, z} = {{z}, z} = {z, {z, z}} = (z,z). Bzzzzt!
If one assumes the axiom of foundation (as usual in ZF), then x = [[:Template:x]] is forbidden (sets then can't have themselves as members); and then your definition works. The advantage of (x,y) = {{x}, {x, y}} is that it is more general, as it still works without foundation; so it can even be applied to classes as well as sets, etc.. While there may be some confusion when x=y, the definition still works. Chas zzz brown 22:40 Feb 22, 2003 (UTC)
If (a,b) = (c,d) and we are trying to prove that a=c and b=d (which is the main requirement for a definition of ordered pair), then {a,{a,b}} = {c,{c,d}}. Then either a=c or a= {c,d}. If a=c then {a,b}={a,d}, so that b=d and we are done with the first case. The second case; if a= {c,d} then {a,b} = c. Then there is an infinite recursion: a= {c,d} = {{a,b},d} = {{{c,d},b},d} etc. and then the proof that a=c, b=d cannot be completed. Well, it could still be possible, if a=c then {a,b} = {c,d}, {a,b} = {a,d}, then b=d, QED. But there is no guarantee that a=c in this second case, is there? (NO QED)

On the other hand, letting (a,b) = {{a},{a,b}}, if (a,b) = (c,d) then {a} = {c} OR {a} = {c,d}. If {a} = {c} then a=c. If {a} = {c,d} then card {a} = 1 = card {c,d} so c=d; {a} = {c,c} = {c}, therefore a=c in both cases. Then {{a},{a,b}} = {{a},{a,d}}, thus {a,b}={a,d}, b=d. QED.

-- AugPi

Here's a proof written out on Metamath: http://us.metamath.org/mpegif/opthreg.html -- Sekkusu —Preceding unsigned comment added by 67.201.195.107 (talk) 22:02, 29 January 2008 (UTC)


If it is required that card (x,y) = 2 even when x=y or there is any dependence between x and y, define (x,y) as

(x,y) = \{\{\,0,x\},\{\,1,x,\{y\}\}\}

Then to define an ordered triple so that card (a,b,c) = 3, let (a,b,c) = { (0,a), (1,b), (2,c)} where 0,1, and 2 are Von Neumann cardinals: 0 = {}, 1={{}}, 2={{},{{}}}. Likewise, for an n-tuple, let (a1, a2, ..., an) = {(0,a1), (1,a2), ... , (n-1,an)}, then the cardinality of such n-tuple will be n. -- AugPi

Contents

[edit] Extrapolating the tuple sequence

Can the sequence triple, quadruple, etc. be extrapolated?? Let me see:

  • 2. pair
  • 3. triple
  • 4. quadruple
  • 5. quintuple
  • 6. sextuple
  • 7. septuple
  • 8. octuple
  • 9. nonuple
  • 10. decuple
  • 11. undecuple
  • 12. duodecuple
  • 13. tredecuple
  • 14. quattuordecuple
  • 15. quindecuple
  • 16. sexdecuple
  • 17. septendecuple
  • 18. octodecuple
  • 19. novemdecuple
  • 20. vigintuple
  • 30. trigintuple
  • 40. quadrigintuple
  • 50. quinquagintuple
  • 60. sexagintuple
  • 70. septuagintuple
  • 80. octogintuple
  • 90. nonagintuple
  • 100. centuple
  • 125. quasquicentuple
  • 150. sesquicentuple
  • 175. terquasquicentuple
  • 200. bicentuple
  • 300. tercentuple
  • 400. quatercentuple
  • 500. quincentuple
  • 1000. milluple
  • 10^6. milliontuple

66.245.6.12 00:50, 25 Aug 2004 (UTC)

[edit] Other pairs

I replaced the text In the usual Zermelo-Fraenkel formulation of set theory including the axiom of regularity, ordered pairs (a, b) can also be defined as the set {a, {a, b}}. However, the axiom of regularity is required, since without it, it is possible to consider sets x and z such that x = {z}, z = {x}, and x ≠ z. Then we have that

(x, x) = {x, {x, x}} = {x,{x}} = {x, z} = {z, x} = {z, {z}} = {z, {z, z}} = (z, z)

although we want (x,x) ≠ (z,z).


by a few examples of possible other definitions, plus reasons why they are not used. Aleph4 13:44, 18 Feb 2005 (UTC)

I added a reference to the first definition of the ordered pair using sets alone (due to Norbert Wiener, 1914) and attributed the Rosser pair correctly to Willard van Orman Quine.

Randall Holmes 21:39, 14 December 2005 (UTC)

[edit] Who the hell is Morse?

Who the hell is Morse?

Anthony Perry Morse (1911-1984), Professor of Mathematics, University of California at Berkeley. He ought to get a Wikipedia article. photo bio advisor and students --Hoziron 14:00, 26 February 2006 (UTC)

[edit] Wiener definition

Is the Wiener definition actually

(x,y)=\{\{\{x\},\emptyset\},\{\{y\}\}\}?
Yes.

Why did he not use

(x,y)=\{\{\{x\},\emptyset\},\{y\}\}? --NeoUrfahraner 06:13, 22 March 2006
Because he introduced his definition the the framework of Russell and Whitehead's theory of types. There all elements in a class must be of the same type, namely one less than the type of the class (they are elements of). Now if x and y are of, say, type n, then {x} and {y} would be of type n+1, and {{x}, 0} would be of type n+2. Now putting {{x}, 0} and {y} in the same class would mean a "type clash". Hence we have to use { {y} } --which is also of type n+2-- instead of {y}. 00:09, 3 December 2007 (UTC)

[edit] definition of caartesian product uses ordered pair, but Morse definition of ordered pair uses cartesian product

in the 2nd paragraph it says:

"The notion of ordered pair is crucial for the definition of Cartesian product"

in the paragraph "morse definition" it then says

(x,y)=\{x \times \{0\}\}\cup\{y \times \{1\}\} which uses a cartesian product. so an ordered pair is defined in terms of a cartesian product which is defined in terms of an ordered pair which is defined in terms of a cartesian product etc 12:38, 7 April 2007 (UTC)

[edit] Set vs Multiset?

As the Kuratowski definition defines an ordered pair (a,b) as the set (a,b)= {{a}, {a,b}}, what if a=b? Quote: "However, for purposes of foundations of mathematics it has been considered desirable to express the definition of every type of mathematical object in terms of sets". That means the 'set' {a,b} isn't a set but a multiset. Am I correct or just running in circles?

If a=b, then (a,b) = {{a}}, i.e. if the set defining the pair has only one element, then both components of the ordered pair are equal. I see no need for a multiset. --NeoUrfahraner 10:28, 28 September 2007 (UTC)

[edit] Mistake in introduction?

"In mathematics, a pair (a,b) is ordered if (a,b) is not equivalent to (b,a)."

Isn't that only true if it is also true that a is not equivalent to b?- (User) WolfKeeper (Talk) 00:15, 24 February 2008 (UTC)

The person who made this edit appeared to be trying to clarify the introduction, which currently seems to be rather technical. The addition doesn't seem to handle the possibility that a and b are the same thing. I've reverted the edit, since I don't see an obvious way to make the introduction clearer. Michael Slone (talk) 03:26, 24 February 2008 (UTC)
I attempted some rewording to the idea clearer and less technical (or at least easier to read...). Although probably more understandable, I'm not sure if it's rigorous enough. We might want a multi-paragraph lede to explain the concepts in more detail. —AySz88\^-^ 08:21, 2 March 2008 (UTC)