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Math Help - set theory ordering question

  1. #1
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    set theory ordering question

    Using an alternative definition of ordered pairs, choose two different sets \triangle and \square, define \langle a, b \rangle  =  \lbrace \lbrace a, \square \rbrace, \lbrace b, \triangle \rbrace \rbrace

    State and prove an analogue of Theorem 1.2

    so Theorem 1.2 states (a, b) = (a' , b') if and only if a = a' and b = b'.

    now the only part I'm not sure about is \triangle and \square, so would the statement be:

    \langle a, b \rangle = \langle a', b' \rangle<br />
if and only if a = a' b = b', \triangle = \triangle ' and \square = \square '

    or simply \langle a, b \rangle = \langle a', b' \rangle<br />
if and only if a = a' b = b'. where the triangle and square are sort of fixed?
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  2. #2
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    Hi ||||| !

    Quote Originally Posted by lllll View Post
    or simply \langle a, b \rangle = \langle a', b' \rangle<br />
if and only if a = a' b = b'. where the triangle and square are sort of fixed?
    This is what you have to do and do not forget [tex]\triangle \neq \square[/Math]. It is important.

    By the way, this definition of an ordered pair was proposed by Felix Hausdorff.

    Best wishes,
    Seppel
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