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Equality and identity of sets

  1. Jul 9, 2009 #1
    Two sets are equal iff they contain the same elements.

    I would argue that two sets that have the same elements are identical as well as equal and that there is a difference between identity and equality. In general {2,3}={3,2} if neither set is defined to be ordered. However obviously {5} [tex]\neq[/tex] {2,3}. Under addition 2+3=5 but I would argue this is an equality but not an identity. I'm not sure distinction is really observed in mathematics.

    EDIT: Two formulas in a formal language are equivalent iff one can be substituted for the other in a sentence. I don't think there is a necessary distinction between 'equal' and 'equivalent' unless you consider 'identical' and 'equal' to be synonymous. If you do, than this would lead to a lot of problems with the usual descriptions of equations. 3+2=5, but the two formulas are not identical. If they were, we wouldn't need to solve equations.
    Last edited: Jul 9, 2009
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  3. Jul 9, 2009 #2


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    There is a difference between an object and the method we use to write an object.

    The 3-character string "2+3" and the 1-character string "5" are indeed different character strings, but that does not imply that the integers they name should be different in any way.

    That said, I would agree with a claim of the form "identity tends to be overemphasized", however you shouldn't underemphasize it either.
  4. Jul 9, 2009 #3
    I'm not sure what you mean by "...does not imply that the integers they (the characters strings) name should be different in any way." No integer is identical to any other. They are all unique. However formulas constructed from integers and operators can express the equivalence (or equality) of two character strings. There are an infinite number of finite character strings that equal 5 under addition if you include all integers; none these strings identical to each other or to 5 (except 5=5).
    Last edited: Jul 10, 2009
  5. Jul 9, 2009 #4


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    How would you define the difference between identity and equality? The string of characters {2,3} is not identical to the string of characters {3,2}.
  6. Jul 10, 2009 #5
    Two sets are equal if they contain the same elements. These sets would therefore be equal. They would also be identical sets unless ordered sets were specified. At least that's how I would define it. I don't of know anything in Set Theory that defines identical sets (as opposed to equal sets) without a specification of order (such as permutation groups).

    In addition, order doesn't matter in the examples I gave: 2+3=3+2=5 so I would say 3+2 is identical to 2+3 under addition with positive integers but neither is identical to 5. If the operation is subtraction, or addition with negative integers clearly 3-2 [tex]\neq[/tex] 2-3.

    Hurkyl introduced the term 'character strings'. Certainly character strings in computer programs or representing addition of integer strings (positive and negative integers) are identical iff they contain exactly the same characters in the same order.
    Last edited: Jul 10, 2009
  7. Jul 10, 2009 #6
    To clarify, would you consider [tex]\left\{ 2, 3 \right\}[/tex] and [tex]\left\{ n \in \mathbb{Z} \; | \; 1 < n < 4 \right\}[/tex] to be equal, but not identical as sets?
  8. Jul 10, 2009 #7
    Assuming Z is the set of all integers, and n is an integer greater than one and less than four, than n is either 2 or 3. I don't read your formula as specifying more than one number, therefore it is not identical or equal to {2,3}. However {all x in Z: 1<x<4} would be identical to {2,3} or {3,2} provided no order is specified.

    EDIT: The way I'm thinking about this is that the issue between equality and identity arises with compositions of elements (binary operations with numbers, etc), not sets of elements with unspecified order. Also, I have to clarify what I said before (post 5). Otherwise unordered strings of commutative operations like addition on the same numbers are all identical. So for integers: (3)+(-2) is identical to (-2)+(3) but neither is identical to 1. That is, the set {1} is not same as {-2,3} or {3,-2}

    EDIT: If x:1<x<4 and we accept the notion of implicit order for all subsets of Z, than only {2,3} is identical to {all x in Z: 1<x<4}.
    Last edited: Jul 11, 2009
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