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Math Help - What is an ordered pair?

  1. #1
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    What is an ordered pair?

    The infamous one:

    (a,b) = { {a}, {a,b} }

    I see a lot of net debates on this expression already. While I fully empathised with the laypeople, I still couldn't really grasp what the set-theorist side ppl were trying to say.

    So wary of yet another debate that tends to go nowhere, I would like to structure my post in a series of questions where YES/NO would suffice, but feel free to add anything if you must! Thank you for your time, set theorist.


    1. Can I express it similarly in its 3 other permutations?

    (a,b) = { {a,b}, {a} }
    OR
    (a,b) = { {b,a} , {a} }
    OR
    (a,b) = { {a,} , {a,b} }

    2. Can I just as arbitrarily define an ordered pair (and conveying exactly the same meaning as what Kuratowski had intended) as:
    (a,b) = { {x} , {x,y} }?

    3. Can we deduce from the definition that:
    a = {a}
    b = {a,b}
    ?
    In lay words, a is a singleton. b is a pair (of elements).

    So we can redefine:
    (a,b) = ( any_singleton, any_pair)?
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  2. #2
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    Quote Originally Posted by chopet View Post
    The infamous one:

    (a,b) = { {a}, {a,b} }

    I see a lot of net debates on this expression already. While I fully empathised with the laypeople, I still couldn't really grasp what the set-theorist side ppl were trying to say.

    So wary of yet another debate that tends to go nowhere, I would like to structure my post in a series of questions where YES/NO would suffice, but feel free to add anything if you must! Thank you for your time, set theorist.


    1. Can I express it similarly in its 3 other permutations?

    (a,b) = { {a,b}, {a} }
    OR
    (a,b) = { {b,a} , {a} }
    OR
    (a,b) = { {a,} , {a,b} }

    2. Can I just as arbitrarily define an ordered pair (and conveying exactly the same meaning as what Kuratowski had intended) as:
    (a,b) = { {x} , {x,y} }?

    3. Can we deduce from the definition that:
    a = {a}
    b = {a,b}
    ?
    In lay words, a is a singleton. b is a pair (of elements).

    So we can redefine:
    (a,b) = ( any_singleton, any_pair)?
    An ordered pair is as set with two elements, one of which is singled out.

    Informally this means that it is two elements one of which is first and the other second. That is what the definition:

    (a,b) = { {a,b}, {a} }

    is capturing in the notation of sets. The ordered pair is a set with two elements, the first a set telling you what is "in" the ordered pair {a,b} and the and the second {a} a set telling you which is the "first element".

    Any equivalent definition will do, but I suspect that { {a,b}, {a} } has some advantage over { {a,b}, a } which appears to convey the same information.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    (a,b) = { {a,b}, {a} }

    and the second {a} a set telling you which is the "first element".

    I perfectly understand your explanation, except this part where the so-called "first element" is singled out.

    Is the "first element" arbitrary?
    Could it well be the "second element"?

    Ok. Let me attempt to write a proof with my own Axioms.(please bear with me...I'm but a beginner)

    Proof

    Axiom of Order:A set can either be ORDERED () or UNORDERED {}.

    The relationship between them is:
    {a,b} = (a,b) OR (b,a)

    Trivial:
    {a} = a

    An analogy is with \sqrt9 = -3 \ or \ 3

    Axiom of Inequality:
    A set with 1 element is DIFFERENT from a set with 2 elements.

    This axiom is actually a corollary from :
    (a,b) = (c,d) iff a=c and b=d

    So we dissect the expression {{a,b}, {a}} using the 1st axiom (converting {} to () ).

    { {a,b},{a} }
    = { {a,b} , a }
    = ( {a,b} , a ) OR ( a, {a,b} )
    = ( (a,b), a ) OR ( (b,a),a) OR ( a, (a,b) ) OR ( a, (b,a) )

    Using the 2nd axiom of inequality:
    (a,b) is none of the above.

    So the definition :
    (a,b) = { {a,b},{a} }
    is absurb.
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  4. #4
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    Quote Originally Posted by chopet View Post
    I perfectly understand your explanation, except this part where the so-called "first element" is singled out.

    Is the "first element" arbitrary?
    Could it well be the "second element"?

    Ok. Let me attempt to write a proof with my own Axioms.(please bear with me...I'm but a beginner)

    Proof

    Axiom of Order:A set can either be ORDERED () or UNORDERED {}.

    The relationship between them is:
    {a,b} = (a,b) OR (b,a)

    Trivial:
    {a} = a

    An analogy is with \sqrt9 = -3 \ or \ 3

    Axiom of Inequality:
    A set with 1 element is DIFFERENT from a set with 2 elements.

    This axiom is actually a corollary from :
    (a,b) = (c,d) iff a=c and b=d

    So we dissect the expression {{a,b}, {a}} using the 1st axiom (converting {} to () ).

    { {a,b},{a} }
    = { {a,b} , a }
    = ( {a,b} , a ) OR ( a, {a,b} )
    = ( (a,b), a ) OR ( (b,a),a) OR ( a, (a,b) ) OR ( a, (b,a) )

    Using the 2nd axiom of inequality:
    (a,b) is none of the above.

    So the definition :
    (a,b) = { {a,b},{a} }
    is absurb.
    Remember that you are starting only from a universal set and subsets you have no concept of order until you define it, so saying {a,b}=(a,b) or (b,a) is playing with undefined concepts.

    RonL
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  5. #5
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    Quote Originally Posted by chopet View Post
    Axiom of Order:A set can either be ORDERED () or UNORDERED {}.
    A set is never ordered. You are given sets A and B you are defining (A,B) = {A,{A,B}}. Because with such a definition it might makes a difference if you have (A,B) or (B,A) (assuming A and B are different sets). Why do you say "axiom" it is not, it is a definition.


    Axiom of Inequality:
    A set with 1 element is DIFFERENT from a set with 2 elements.
    That does not make sense. Remember you never defined what "1" means and what "2' means. Instead you should say if A and B and distinct and C is an arbitrary set then {A,B} is not equal to {C}.
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  6. #6
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    I have not joined this discussion before because I think it pointless to argue with someone who thinks that \sqrt 9  =  - 3 about the rather abstract set notation for ordered pairs. However, I will point out why {{a}.{a,b}} is preferred over {a,{a,b}}. It strictly is that sticking with sets of sets, we can get A \times B \subseteq \wp \left( {\wp \left( {A \cup B} \right)} \right).
    That way we have functions as members of sets.
    This is important in model theory.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I have not joined this discussion before because I think it pointless to argue with someone who thinks that \sqrt 9  =  - 3 about the rather abstract set notation for ordered pairs. However, I will point out why {{a}.{a,b}} is preferred over {a,{a,b}}. It strictly is that sticking with sets of sets, we can get A \times B \subseteq \wp \left( {\wp \left( {A \cup B} \right)} \right).
    That way we have functions as members of sets.
    This is important in model theory.
    Hi Plato, one of the reasons why I couldn't fathom the definition was that I thought of set ordering not in terms of abstraction, but in the definite sense of functions. I thank you for your comment, but I take umbrage at the word "pointless", which suggests that people are beyond redemption. You may still be right though, for I am still not getting it!
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    A set is never ordered. You are given sets A and B you are defining (A,B) = {A,{A,B}}. Because with such a definition it might makes a difference if you have (A,B) or (B,A) (assuming A and B are different sets).
    1. Can you demonstrate how it MIGHT make a difference?
    (and how it MIGHT not)?

    2. What's the significance in having the 1st element {A} appear in the second element {A,B}?
    >Are they just trying to make 2 sets dissimilar?

    Once again, any input is much appreciated.
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  9. #9
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    Quote Originally Posted by chopet;105652I
    take umbrage at the word "pointless", which suggests that people are beyond redemption. You may still be right though, for I am still not getting it!
    Please do not take offence at someone who is dismayed at a child who tries to walk before he knows how to crawl. From the nonsense that you have posted on this topic if is painfully obvious that you have a lot of crawling to do before walking in mathematical circles.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    Please do not take offence at someone who is dismayed at a child who tries to walk before he knows how to crawl. From the nonsense that you have posted on this topic if is painfully obvious that you have a lot of crawling to do before walking in mathematical circles.
    hehe, you are so harsh, Plato
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    (a,b) = { {a,b}, {a} }

    is capturing in the notation of sets. The ordered pair is a set with two elements, the first a set telling you what is "in" the ordered pair {a,b} and the and the second {a} a set telling you which is the "first element".

    RonL

    OK guys! I suspect I have it figured it out, but I need your confirmation.

    The RHS are like snapshots of the ordered pair which was being defined in real time.
    That is, as "a" was written in (a...., a snapshot {a} as taken.
    Then, as "b" is written in (a,b..... , another snapshot {a,b} was taken.

    Then we group all the snapshots together in one big set {}, safe in the knowledge that the orders have been captured, through the varying sizes of the snapshots.


    So I can define similarly:
    {a,b,c} = { {a,} , {a,b} , {a,b,c} }

    and so forth.

    Conversely, a definition like:
    {a,b,c} = { {a}, {a,b} , {a,c} } makes little sense as we do not know if b or c comes first.
    Last edited by chopet; February 9th 2008 at 11:59 PM.
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