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Construction of bijection

  1. Sep 18, 2008 #1
    Can you prove that [tex]\mathbf{R}[/tex] and [tex]\mathbf{R}-\mathbf{Q}[/tex] have same cardinality?

    One way would be to say that [tex]\mathbf{R}-\mathbf{Q}[/tex] is not countable and must have cardinality <= [tex]\mathbf{R}[/tex] and invoke the Continuum Hypothesis to conclude that its cardinality is aleph-1 same as that of [tex]\mathbf{R}[/tex]..

    Somehow this does not look appealing...

    Can you explicitly construct a bijection and help me to visualise the situation better??

  2. jcsd
  3. Sep 18, 2008 #2
    Well the union of Q and R\Q is R right? |Q| = \aleph_0 and |R| = c so if |R\Q| was aleph_0 then you have a contradiction i.e. isn't the union of countable sets, countable?
  4. Sep 18, 2008 #3


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    Yes, that proves that R\Q is not countable. But it does not prove that card(R\Q)= card(R). As soumyashant said, You would need the contiuum hypothesis, that there is no set of cardinality strictly between that of R and that of Q, to finish that proof.
  5. Sep 19, 2008 #4


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    Try to prove the following fact: If A is an infinite set and B is a countable set, then [itex]|A \cup B| = |A|[/itex]. You shouldn't need to invoke the CH.
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