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  • 1 Post By emakarov

Thread: Definable subset

  1. #1
    Oct 2007

    Definable subset

    The set of all n-tuples (b_1,\,b_2,\,\dots,\,b_n)\in(U_M)^n for which M\models\varphi(b_1,\,b_2,\,\dots,\,b_n) is called a definable subset of M. The set U_M is the universe of structure M.

    Sometimes I think I do not understand what a definable subset is.

    Here is an exercise on definable subset:

    Let U_M be the underlying set for structure M. Suppose that A\subset(U_M)^3 is a definable subset of M.
    Suppose we rearrange the order of the n-tuples. Consider the set of all (z,\,x,\,y) such that (x,\,y,\,z) is in A. Show that this set is definable.

    Then I say that the the sought formula is as follows:

    (x=y\wedge y=z\to\varphi(x,\,y,\,z))\wedge(x\not= y\vee y\not= z\to\bigvee_{i}(x=a_{i3}\wedge y=a_{i1}\wedge z=a_{i2}))

    where (a_{i1},\,a_{i2},\,a_{i3})\in A and \varphi defines A.

    While it may be correct, I feel that there is a simpler formula. What do you think?
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  2. #2
    MHF Contributor
    Oct 2009

    Re: Definable subset

    I don't understand what a_{i1}, a_{i2} and a_{i3} are. Suppose \varphi(x,y,z) defines A and let B=\{(z,x,y)\mid (x,y,z)\in A\}. The required formula must have three free variables and no constants, except possibly those that occur in \varphi.

    We have (x,y,z)\in A\iff(z,x,y)\in B, i.e., (y,z,x)\in A\iff(x,y,z)\in B. Therefore, B is defined by \psi(x,y,z) where \psi(x,y,z)\equiv\varphi(y,z,x).
    Thanks from andrei
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