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Cardinality of sets: prove equality

  1. Apr 29, 2010 #1
    [tex]A=\mathbb{R}-0[/tex]

    [tex](0,1)\subseteq \mathbb{R}-0[/tex]

    Assume A is countable.

    Since A is countable, then [tex]A\sim\mathbb{N}[/tex].
    Which follows that [tex](0,1)\sim\mathbb{N}[/tex].

    However, (0,1) is uncountable so by contradiction, Card(A)=c

    Correct?
     
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  3. Apr 30, 2010 #2

    Hurkyl

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    Re: Card(A)=c

    The fact that A is not countable, together with the fact it is a subset of R only proves that
    [tex]\aleph_0 < \mathop{Card}(A) \leq c[/tex]​
     
  4. Apr 30, 2010 #3

    HallsofIvy

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    Re: Card(A)=c

    So it looks like you will have to use the Continuum Hypothesis!
     
  5. Apr 30, 2010 #4

    Hurkyl

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    Re: Card(A)=c

    Or find a better lower bound. :wink:
     
  6. Apr 30, 2010 #5
    Re: Card(A)=c

    Think about the function that maps [itex]n\mapsto n+1[/itex] for [itex]n\in \mathbb{N}[/itex] where [itex]\mathbb{N}[/itex] is regarded as a subset of [itex]\mathbb{R}[/itex].
     
  7. Apr 30, 2010 #6
    Re: Card(A)=c

    Thanks for the help
     
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