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Can someone help me better understand the formal definition for ordered pairs?

  1. Apr 29, 2012 #1
    I mean the one saying that:

    (a,b)

    is defined to be the set:

    {{a},{a,b}}

    What exactly does this set definition of an ordered pair mean? Namely, how does it attribute the relevant "order" of terms to the concept of an ordered pair?


    Thanks!
     
  2. jcsd
  3. Apr 29, 2012 #2

    micromass

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    How it is defined doesn't really matter much. What matters is that it has the property:

    [tex]\text{If}~(a,b)=(c,d)~\text{then}~a=c,~b=d[/tex]

    That is essentially what we want ordered pairs to satisfy. So any definition which will make it satisfy this is good.

    The same thing happens everywhere in math. For example: it doesn't matter what the real numbers are as long as it satisfies what we want it to satisfy.
     
  4. Apr 29, 2012 #3
    So what you're saying is that this definition works because

    {{b}, {b,a}} =/= {{a}, {a,b}}

    without even involving the concept of the ordered pair just yet, because it requires that singleton set be different.

    I understand now.

    Thank you.
     
  5. Apr 29, 2012 #4

    micromass

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    Yes, that's one way of looking at it.

    Note that there are other definitions possible. For example, the definition

    [tex](a,b)=\{\{\{a\},\{a,0\}\}~\vert~a\in A\}\cup \{\{\{b\},\{b,1\}\}~\vert~b\in B\}[/tex]

    also works fine as is sometimes preferred for technical reasons.
     
  6. Apr 29, 2012 #5
    Yes, exactly. In effect all we're trying to do is remove the symmetry from the situation so that we can impose an order on a pair of elements without using the concept of order, if you think of it that way.
     
  7. Apr 29, 2012 #6

    AlephZero

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    Playing these sort of games is really to minimize the number of undefined notions at the foundation of math. Another example is defining "a set S containing exactly one element" without using the number "one", by saying "## \exists\, x## such that ##x \in S##, and ##\forall\, y \in S##, ##x = y##".
     
  8. May 2, 2012 #7
    HI,

    A nice and most obvious question, must say. An ordered pair means that the order matters in the same way as 45 and 54 do differ. Similarly, (a,b) and (b,a) are two different elements. and (x,y) is regarded as an element just like it's necessary to uniquely pin-point a point in 2-D coordinate system.

    Hope, I made it clear.

    Akshay
     
  9. May 2, 2012 #8

    HallsofIvy

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    Yes, but that has nothing to do with the question, which was why an ordered pair can be expressed as the set {a, {a,b}}.

    That set obviously shows that there are two things in question, a and b. But "sets" have no natural order defined so we distinguish between a and b by having a and {a,b} separately.
     
  10. May 2, 2012 #9
    Seeking an explanation of how the Kuratowski set definition of ordered pair requires order =/= not knowing what an ordered pair is at all..

    I actually am insulted a bit.
     
  11. May 2, 2012 #10

    HallsofIvy

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    No reason to feel insulted. Akshay simply did not understand the question.
     
  12. May 2, 2012 #11
    Perhaps I am mathematically sensitive. For example, when someone mislabels a large increase as 'exponential,' I am personally offended.

    XD
     
  13. May 2, 2012 #12
    LOL me too. Every time I see a talking head on tv talk about exponential growth, I wish I could be there to ask them, "So, can you please tell me what exponential growth is?" Of course they'd have no idea.

    Glad I'm not alone. There are two of us!
     
  14. May 2, 2012 #13

    mbs

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    You mean the proof that...

    [itex]\left\{\left\{ a_1\right\},\left\{ a_1,a_2\right\} \right\}=\left\{\left\{ b_1\right\},\left\{b_1,b_2\right\}\right\}[/itex] if and only if [itex]a_1=b_1[/itex] and [itex]a_2=b_2[/itex]?

    The "if" result follows formally straight from the extensionality axiom of ZFC set theory. Namely...

    [itex]\forall a,b\left(\forall x\left(x\in a\Leftrightarrow x\in b\right)\Rightarrow a=b\right)[/itex]

    To prove the "only if" part you just have to break the problem up into cases.

    It seems rather bizarre to prove such an intuitively obvious result, but its a fun game for mathematicians to have as few assumptions as possible in their axioms.
     
    Last edited: May 2, 2012
  15. May 2, 2012 #14

    mbs

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    It's even worse than that. What pisses me off the most are popular quotes of statistical factoids from studies. Half the time the statistic is either meaningless or grossly misleading without knowing the context in which it was presented in the actual study. Eventually an extremely distorted and meaningless version turns into something everyone repeats.
     
  16. May 3, 2012 #15

    mathwonk

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    you know I kind of see where the doubters are coming from on this definition. i mean if this is valid definition of an ordered pair it ought to come with a definition of the first element and the second element as well, but i have never seen this done. i.e. i claim this is not really equivalent to an ordered pair unless you also say which element is first because with the usual definition that information is given as well. i.e. if AxB is the set of all ordered pairs of elements with first element from A and second element from B, we should have maps to both A and B, but I have never seen these maps also defined in this abstract setting. No doubt those of you conversant with this stuff can easily do it, just saying it seems it should be done for completeness, shouldn't it?. presumably the first element is just the unique element of the intersection of both elements of the "pair"? and the second element is trickier, i.e. there seem to be two cases based on whether the union of the two elements has one or two elements..,...oink....now that I'm old, this kind of seems like silly business. interesting though to have something new to say/see about ordered pairs after all these years.
     
  17. May 4, 2012 #16

    mbs

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    Maybe I don't quite understand, but I don't really see where the problem is. In most texts on ZFC set theory pairs are defined first, then ordered pairs, then relations are defined as sets containing ordered pairs. Then the Cartesian product [itex]A\times B[/itex] is defined as the set [itex]\left\{\left\langle x,y \right\rangle~|~x\in A~\text{and}~y\in B\right\}[/itex] which is really a specific relation that simply connects every member of A to every member of B. The Cartesian product is really a maximal type of relation since all other maps and relations between A and B are subsets of the Cartesian product.

    It is possible to go about it a different way, having mappings as fundamental undefined objects with certain axiomatic properties, just like sets. I think this is how von Neumann originally constructed his set theory, but later the Zermelo-Fraenkel construction became the standard for set theory.
     
    Last edited: May 4, 2012
  18. May 4, 2012 #17

    HallsofIvy

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    Do you see that if every ordered pair, (a, b), were replaced with (b, a), nothing would change? It is not necessary to state which is "first" and which "second". It is only necesary to distinguish between the two objects.
     
    Last edited: May 4, 2012
  19. May 4, 2012 #18

    mathwonk

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    the question is very simple, how do you define the map from AxB to B? i.e. how do you define the "second" element of the pair {{a}, {a,b}}?

    I am just saying that merely defining an ordered pair, and then proving that (a,b) equals (x,y) if and only a=x and b = y, is not all you want from a definition of ordered pairs,

    since that theorem would also hold if we defined (a,b) as {{b},{a,b}}.

    You also want the maps AxB-->A, and AxB-->B.


    Halls, I disagree. In a genuine ordered pair, one can distinguish between the two elements.

    I am just wondering if not reproducing all the data of an everyday intuitive ordered pair may be partly responsible for the questioner not understanding it. The OP asked us to help understand them.


    "Do you see that if every ordered pair, (a, b), were replaced with (b, a), nothing would change?"

    In other words, yes I see that, and that is exactly the basis for my question.
     
    Last edited: May 4, 2012
  20. May 4, 2012 #19

    mathwonk

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    I am just trying to take the gentleman's question seriously. If we want the set AxB of ordered pairs to satisfy the usual requirements of a categorical product, then we need to define not just ordered pairs, but also maps AxB--A and AxB-->B so that there is a one-one correspondence between maps X-->AxB and (ordered) pairs of maps X-->A and X-->B. To do this we have to show how to recover the element of A from (a,b) and the element of B. Since A and B may overlap this is equivalent to recovering the first, and second elements.

    I am not saying this is hard or abstruse, just that it is part of the data of ordered pairs, and is usually ignored.

    I.e. the "first" element of {{a},{a,b}} can be recovered as the unique element of the intersection of two "elements" {a} and {a,b} of (a,b).

    Then there are two cases for recovering the second element. If the union of the two elements {a} and {a,b} equals the intersection, the second element also equals the unique element of that intersection. If the union {a,b} of the two elements of (a,b), itself has two elements, then the second element of (a,b) is the unique element of the difference, i.e. of

    {a,b} - {a}, the union minus the intersection.

    This then defines both maps AxB-->A, and AxB-->B and makes AxB into a categorical product.

    To the OP, sorry if this does not help, just trying to give some honest thought to your question of how to recover the familiar concept from the abstract. The set theory experts are of course not challenged by this. I just enjoy it when someone makes me think of something I hadn't thought about before, especially in regard to a familiar object.
     
  21. May 6, 2012 #20

    mbs

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    Okay, I think I might understand better now. I'm not familiar with category theory.

    You can also, given the ordered pair <a,b> = {{a},{a,b}}, define a second type of ordered pair (a,b) = <<0,a>,<1,b>> where 0 = {} and 1 = {{}}. Then AxB defined as...

    [itex]\left\{(a,b)~|~a\in A~\text{and}~b\in B \right\}[/itex]

    is just a special case of a generalized product...

    [itex]\prod_{i\in I}X_i\equiv\left\{(x_i)_{i\in I}~|~x_i\in X_i~\forall (i\in I)\right\}[/itex]

    with the indexing set I chosen to be the set {0,1}.

    Then it's the map from the indexing set to each set comprising the product that provides the actual "order" to the product with subscripts interpreted as the defining maps for individual elements of the product set.
     
    Last edited: May 6, 2012
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