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Math Help - Another Axiom of Separation Qns

  1. #1
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    Another Axiom of Separation Qns

    Let \phi (v) be a formula of the languages of set theory. Let X be a set. Then the following is a set: \{a \in X|\phi [a] \} , ie all the elements of X with the property \phi form a set.

    Write down a formula \phi (v) saying that "v is a set which has exactly three elements".

    I would like to check whether it is possible to represent the above statement as this: \exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge \forall y (y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3 ))

    Thanks in advance.
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  2. #2
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    Re: Another Axiom of Separation Qns

    No, that only says that v has at least 1 element and at most 3 elements. You need to include formulation that x, y, and z are different from one another.

    Also, you're missing a right parenthesis on the end.
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  3. #3
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    Re: Another Axiom of Separation Qns

    Quote Originally Posted by MoeBlee View Post
    No, that only says that v has at least 1 element and at most 3 elements. You need to include formulation that x, y, and z are different from one another.

    Also, you're missing a right parenthesis on the end.
    Hey,

    Can I write
    \exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge (x_1 \not= x_2 \not= x_3) \wedge \forall y (y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3)))?

    Thanks in advance.
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  4. #4
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    Re: Another Axiom of Separation Qns

    Now you've only said that v has at least two elements and at most three elements. But you're getting close.
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  5. #5
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    Re: Another Axiom of Separation Qns

    Hey,

    Am I right to represent it as such:
    \exists x_1 \exists x_2 \exists x_3 ((x_1 \in v \wedge x_2 \in v \wedge x_3 \in v) \wedge (x_1 \not= x_2 \wedge x_2 \not= x_3 \wedge x_1 \not= x_3) \wedge \forall y(y \in v \longrightarrow (y = x_1 \vee y = x_2 \vee y = x_3)))?

    Thanks in advance.
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  6. #6
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    Re: Another Axiom of Separation Qns

    Yep. That's it.
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