Talk:Ordered pair

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Unsorted text[edit]

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) = {{x}}, 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 Y1Y2 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 Y1Y2, 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 = {{''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

Extrapolating the tuple sequence[edit]

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)

Other pairs[edit]

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)

Who the hell is Morse?[edit]

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)

Wiener definition[edit]

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}.</nowiki> —Preceding unsigned comment added by 91.59.209.222 (talk) 00:09, 3 December 2007 (UTC)

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

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)

Set vs Multiset?[edit]

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)

Mistake in introduction?[edit]

"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)

Another Definition and Proofs[edit]

It would be nice to provide a proof for the simple treatment [(x,y)={x,{x,y}}] as this is the definition given in many introductory math texts. Also, I remember somewhere seeing another definition of the ordered pair (maybe in Quine's Philosophy of Logic?) such that:

The ordered pair (x,y) = {x,{y}}. I'm probably missing something (and perhaps this definition runs into difficulties with iteration) -- but I don't think I saw that definition in the article. Thoughts? —Preceding unsigned comment added by 169.231.34.169 (talk) 02:33, 30 June 2008 (UTC)

Redundancy[edit]

Can we delete the "reverse definition"? Its completely identical to the "regular" definition, and the distinction is only syntactic —Preceding unsigned comment added by AbnormallyNormal (talkcontribs) 14:49, 3 September 2008 (UTC)

In mathematics, an ordered pair is a collection of two distinguishable objects[edit]

Having the Kuratowski pair <{},{}>, it is totally unclear to me, what does "distinguish" {} and {} mean? Peta 77.104.243.33 (talk) 17:42, 27 March 2009 (UTC)

Wouldn't a changed formulation like:

"In mathematics, an ordered pair is a collection of objects having two coordinates (entries, projections), such that one can always uniquely determine the object, which is the first coordinate (first entry, left projection) of the pair as well as the second coordinate (second entry, right projection). If the first coordinate is a and the second is b, the usual notation for an ordered pair is (a, b). The pair is "ordered" in that (a, b) differs from (b, a) unless a = b."

better communicate, that the coordinates of the pair need not be "distinguishable"? Peta 77.104.243.33 (talk) 09:28, 30 March 2009 (UTC)

The first sentence is a description of a Cartesian plane geometry, not of an ordered pair; a pair is certainly not a "collection" of pairs. The language here also seems to say that the object is the first coordinate, and the role of the second is left unclear.
Nor is geometry the only application of ordered pairs.
My attempt (five months ago) to improve this muddle was reverted today with this note: "distinguished" was incorrect, both objects can be the same, i.e. indistinguishable. So what? If two coordinates happen to have equal values, they are still distinguished by their function. In the pair (2,2), is there any doubt which is the first coordinate and which is the second? —Tamfang (talk) 18:38, 18 November 2010 (UTC)
My notes and reasons why I reverted your formulation:
  1. In your formulation you replaced "collection of objects" by "combination of objects". That is an error, since according to Combination_(disambiguation) "In mathematics, a combination is a collection of things in no specific order.", while in ordered pair, the order is what matters.
  2. In your formulation you replaced "collection of objects" by "combination of two objects". That is an error, since the "two objects" can in fact be just one object.
  3. In your formulation you wrote that the "two objects" were "distinguishable". That is an error, since as mentioned, there may not be "two objects", in which case, there are no "two objects" to distinguish. To notice, how problematic such a formulation is, notice, that the most commonly used (Kuratowski) definition, gives us \left\{ \left\{ \emptyset \right\} \right\} as an example of an ordered pair. Where you have your "two objects", is a mystery.
  4. Certainly, the coordinates are always distinguishable, in the sense, that you can tell which object is the first coordinate, as well as tell, which object is the second coordinate, but that is a different sense than the one you used in your formulation.Ladislav Mecir (talk)
  • Good point about combination; but doesn't the same objection apply to collection? — How about structure? or an object consisting of two elements?
  • The most natural reading of "a collection of objects having two coordinates" is something with two levels of structure, viz a set of ordered pairs; otherwise either objects or collection is redundant.
  • Thank you for pointing out Kuratowski's pathological example. (In \left\{ \left\{ \emptyset \right\} \right\}, how do you tell which coordinate is which?) You see it as disallowing the obvious operational definition; I see it as a good reason to avoid torturing oneself by hunting for essences.
  • That's exactly the sense of distinguished that I used. Did you confuse it with distinct (i.e. 'unequal')? —Tamfang (talk) 23:58, 18 November 2010 (UTC)
A "collection" is certainly a more "basic" notion, than a "combination", as is demonstrated by the fact, that a combination is defined as a collection of a specific kind.
I tend to agree, that the word "objects" looks redundant in the present formulation, so would it be better if we wrote: "In mathematics, an ordered pair is a collection (of objects) having two coordinates (or entries or projections), such that it is distinguishable, which object is the first coordinate (or first entry or left projection) of the pair and which object is the second coordinate (or second entry or right projection) of the pair." Would that be an improvement?
Re "how do you tell which coordinate is which?" - it is easy, the first coordinate is the \emptyset, which happens to be the second coordinate as well in this specific case.Ladislav Mecir (talk) 00:29, 19 November 2010 (UTC)
,i.e. we are able to "distinguish" the coordinates in the sense, that we can "distinguish", which object is the first coordinate (the \emptyset), as well as "distinguish" which object is the second coordinate (the same \emptyset), while we certainly cannot distinguish \emptyset from \emptyset. Ladislav Mecir (talk) 11:05, 20 November 2010 (UTC)
In fact, the "Kuratowski pathological example" cannot be attributed to Kuratowski, since the ordered pair (a, a) has this property for any definition used. - Let's just for a moment suppose, that a denotes you. It certainly isn't possible to distinguish you from you, while it certainly is possible to distinguish, which object is the first coordinate of such a pair (you) as well as which object is the second coordinate of the said pair (you again).Ladislav Mecir (talk) 00:47, 19 November 2010 (UTC)
It certainly is possible to distinguish two representations of me from each other. —Tamfang (talk) 23:06, 19 November 2010 (UTC)

Is "99" a one-digit number? —Tamfang (talk) 05:54, 19 November 2010 (UTC)

once again, your formulation "two objects ... the first ... is distinguished from the second" is the formulation you wrote, and it is an error, since that is not true. That is like saying that in "99", the first "9" is distinguished from the second "9", while it is not, they are both the same character. What can be distinguished is not "the first from the second", which can not be distinguished, but "what is the first" (between all conceivable objects) and "what is the second" (between all conceivable objects). The fact, that both can be the same clearly disallows to distinguish the first from the second.Ladislav Mecir (talk) 08:16, 19 November 2010 (UTC)
Ordinary people would say "99" consists of two nines, a first "9" and a second "9", distinguished by their position. You insist that two congruent objects are the same object. Such an axiom may be useful for some purposes, but I doubt there's a consensus for applying it universally. It makes your definition obscure, with no apparent benefit.
Could you tolerate a plainer definition (before or after yours) if it were labelled as "informal"? —Tamfang (talk) 20:30, 19 November 2010 (UTC)
Your formulation: 'first "9" and second "9" distinguished by their position' is just an error. It requires all objects to have a position in every pair as their own property. But, e.g. an empty set, or many other mathematical objects simply do not have such a property. In your error, an ordered pair having you as its first as well as second coordinate cannot exist. I can feel sorry for you, but cannot allow you to write such falsities into an encyclopedia article. I cannot help but wonder, how people making more than three serious errors in one sentence can feel to be "chosen" to correct it.Ladislav Mecir (talk) 21:34, 19 November 2010 (UTC)
"You insist that two congruent objects are the same object." - error again. I just insist, that there are such ordered pairs that contain only one object, which plays the role of both their first as well as second coordinate, and that this property is essential for ordered pairs to be what they are.Ladislav Mecir (talk) 21:57, 19 November 2010 (UTC)
Your argument is narrow-minded, if not circular, and your tone is increasingly offensive.
"It requires all objects ... as their own property": No, it requires the representations of the objects to have an attribute which we may call context.
I chose myself to propose language that might make sense to the layman, just as you chose yourself to defend the true religion of obscure mumbo-jumbo. Any other self-nominees are welcome to put their oar in. —Tamfang (talk) 23:10, 19 November 2010 (UTC)
If mentioning one dollar in two different contexts were capable of making two distinguishable dollars from it, I would have been using such a "device" for years.Ladislav Mecir (talk) 10:58, 20 November 2010 (UTC)
Thanks for demonstrating the limits of your understanding. —Tamfang (talk) 17:25, 20 November 2010 (UTC)
Maybe this will help: what is distinguishable, is, which formulation (context, or projection) we use to refer to the object, i.e. whether we say, that "the object is the first coordinate of the pair", or "the object is the second coordinate of the pair". What is not distinguishable, is the object being the first coordinate of the pair, from itself, being the second coordinate of the pair as well.Ladislav Mecir (talk) 11:27, 20 November 2010 (UTC)

Recent article by Dana Scott and McCarty[edit]

I wish to draw everyone's attention to the following article:

Dana Scott and McCarty, Dominic (2008) "Reconsidering Ordered Pairs," Bulletin of Symbolic Logic 14(3): 379-97.

The entry should mention this article in some way, and incorporate a bit of its rich content.111.69.249.90 (talk) 12:42, 25 October 2010 (UTC)

Math formatting[edit]

There is some mixture of math formatting on this article. For simple ASCII math (anything with just letters, punctuation, and subscript/superscripts), I propose we just use HTML formatting and lose all the "math" tags, many of which are showing up as images for me. — Carl (CBM · talk) 13:58, 4 April 2011 (UTC)

Agree. HTML is preferred where possible. Paul August 14:21, 4 April 2011 (UTC)
I also agree. There is nothing on this page that requires math tags, and that is not likely to change. Hans Adler 14:24, 4 April 2011 (UTC)

intro[edit]

I'd like to nominate the first sentence of this article as the most obscure definition of the most elementary concept. Really, someone who's mathematical knowledge is sufficiently slight that they don't know what "ordered pair" means will not have a chance of understanding the first sentence.

Having said that, I'll try to fix it. McKay (talk) 06:44, 2 September 2011 (UTC)

"Fix"? It contains just your original research "a list of two things" - Where exactly did you find that an ordered pair is a "list"? Where exactly did you find that ordered pair shall contain "things"? Where exactly did you find that ordered pair has to contain "two" "things"?81.200.57.149 (talk) 00:29, 13 October 2011 (UTC)

Ackermann set theory + proper classes[edit]

Does it make sense to use the Kuratowski ordered pair definition for proper classes in Ackermann Set Theory resp. its extension ARC? If so, shouldn't this also be mentioned? -- 132.231.198.38 (talk) 15:32, 22 June 2012 (UTC)

Please "spell out" Kuratowski[edit]

Could someone _please_ do us the favor of spelling out how exactly these definitions, let's pick Kuratowski since it's so common, actually "define" an ordered pair?

Kuratowski says: (a,b) := {{a}, {a,b}}

Let's suppose a and b are very simple objects, say 6 and 13. So from Kuratowski we can supposedly get to the ordered pair (6,13) from this:

{{6}, {6,13}}

... so at the "top level" this is a set that contains two sets, in no particular order. How is this equivalent to a pair which is ordered and contains integers? Clearly I, for one, am missing something here. Either there is more to the notation, or something more is being implied. Thanks for any enlightenment! Gwideman (talk) 11:59, 25 August 2012 (UTC)

It's pretty clear that all by itself, just from the symbol string {{a}, {a,b}}, a mechanism (e.g. a Turing machine) cannot determine the first element and then the the second element and then properly write the symbol string (a, b). The TM will need a set of rules embedded in its "program" to guide it. (Intuitively for folks who like puzzles, it should be suspicious that there's a singleton {a} and then a pair {a, b} and the element in the singleton, a, appears in as an element in the pair {a, b}.) By the time Suppes 1970:22-23 introduces the ordered pair (Kuratowski version) he has introduced 5 axioms of set theory:
  • 1 Axiom of extension (set A equals set B if and only if they have the same elements)
  • 2 Axiom of specification (a "condition" S(x) creates a set B from the set A for all elements of A for which S(x) holds)
  • 3 Axiom of pairing (creates unordered pairs)
  • 4 Axiom of union (combining sets)
  • 5 Axiom of powers (power set E contains all subsets of the given set)
So by the time we reach his Section 6 Ordered Pairs there's powerful set of rules available to define the notion of ordering. By use of the definition (a, b) = {{a}, {a,b}}, Suppes' proof proceeds by cases and it has to do both the if and the only-if parts, so unfortunately it is rather long (about 200 words) and somewhat tedious. It begins with the statement "We have to show that if (a, b) and (x, y) are ordered pairs and if (a, b) = (x, y), then a = x and b = y." It proceeds through all the various singleton cases, then arrives at the x <> y case.
Maybe someone knows a shorter proof somewhere. Wvbailey (talk) 16:22, 25 August 2012 (UTC)Bill
Wvbailey: Thanks for your response. Your "Turing machine" comment succinctly captures where I was stuck, and indeed the "program to guide it" is instrumental in understanding this ostensibly simple piece of set theory.
The crux of what stumped me (and evidently was what Peta and Tamfang discussed earlier on this talk page) is this: (a,b) := {{a}, {a,b}} does *not* define a pair with an ordering per se (in the sense of identifying that a is first and b is second). It only defines a way to use the mechanism of sets which distinguish items only by value, to implement two distinct variables -- that is to say, two distinct slots in which to store items, and later retrieve the items from each distinct slot.
The "distinct slot" idea is more or less just a restatement of the "defining property" of the ordered pair, and a reader who neglects to note that the defining property of "ordered pair" omits any mention of "order" will be vulnerable to confusion downstream. It remains for additional "logic" to actually perform the retrieving of one item or the other, and attribute to them the order (first and second). And thanks *very* much to the author who explicitly included that logic (Pi1 and Pi2) that actually performs that job. That crucial piece was missing from Hrbacek "Introduction to Set Theory", which is how I got here in the first place.
For the sake of others who might be similarly stuck: In order to understand how "{{a}, {a,b}}" corresponds to "ordered pair", it's crucial to understand how "first item" and "second item" are picked out of "{{a}, {a,b}}". A key challenge is that "{{a}, {a,b}}" has no ordering to it, so, for example, the {a} element is just an element in the set, not the first element per se. Indeed, so far as I can see, Karatowsk could just as well be written: (a,b) := {{a,b}, {a}} (not to be confused with (a,b)reverse, mentioned later in the article).
The answer lies in the two Pi equations, which in English are:
first: take the element that appears in both members.
second: take the element that appears in only one of the members. If there is none, the elements must be the same, so use that.
Hope that this is all correct, and might help other folks who stumble in here! Gwideman (talk) 00:51, 26 August 2012 (UTC)

Couple of questions/suggestions[edit]

1. Intro: "and hence the ubiquitous functions". Here "ubiquitous" looks suspiciously like a technical word. Is there some distinction between ubiquitous functions and non-ubiquitous functions? Or is this simply pointing out that functions are of broad importance?

2. In "Defining the ordered pair using set theory": "The above characteristic property of ordered pairs is all that is required to understand the role of ordered pairs in mathematics. Hence the ordered pair can be taken as a primitive notion, whose associated axiom is the characteristic property."

The first sentence here seems fatuous. Even if true, the second sentence doesn't follow from it.

I suspect this is trying to say something like: the ordered pair is a powerful concept, and consequently its axiom (the characteristic property) has been chosen as a primitive upon which to build further set theory.

Gwideman (talk) 01:08, 26 August 2012 (UTC)

Coordinate extraction formulas[edit]

In the formulas given under Kuratowski definition for extracting the first and second coordinates, what is the purpose of the outermost Union? I had difficulty finding a definition of what unary union does in all cases. Is the idea to "remove an outer layer of set enclosure"? Ie: {a} --> a? But if a is not a set, but a value such as 6, will this succeed? I expected the result of a union operation to be a set. Gwideman (talk) 13:02, 28 August 2012 (UTC)

Everything, including "6", that can be put in a set is a set. --JBL (talk) 13:45, 28 August 2012 (UTC)