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Set Theory - Uncountable Sets

  1. May 11, 2014 #1
    Set Theory -- Uncountable Sets

    1. The problem statement, all variables and given/known data
    Prove or disprove.
    There is no set A such that ##2^A## is denumberable.

    3. The attempt at a solution
    A set is denumerable if ##|A| = |N|##

    My book shows that the statement is true.
    If A is denumerable, then since ##|2^A| > |A|, 2^A ## is not denumerable.


    I don't understand why they state that 2^A > A? I thought that in denumerable sets you can't really say that there is one set greater than other? I thought that there is denumerable or uncountable and that's it.
    But, there is not a denumerable set greater than another denumerable set.

    Help please.
     
  2. jcsd
  3. May 11, 2014 #2

    haruspex

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    That's true, but 2A is not countable - that's the point.
    Well, no - there are infinitely many orders of infinity. If B is uncountable, there is still no bijection between B and its power set, so |2B| is a higher order of infinity.
    Your book appears to be using the general theorem that |2A|>|A|. I suggest you locate that in your book and study it.
     
  4. May 12, 2014 #3

    HallsofIvy

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    Some text books use the term "denumerable" to mean "finite or countable".

    If A is a finite set then [itex]2^A[/itex] is also finite so "denumerable".

    If, however, you are using "denumerable" to mean "countably infinite" then the statement is true.
     
  5. May 12, 2014 #4
    So, for a countably(denumerable) infinite A, ## |2^A| ## is an uncountably infinite?
     
  6. May 12, 2014 #5

    micromass

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    Yes.
     
  7. May 12, 2014 #6

    HallsofIvy

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    Since "infinite" is an adjective, the word "an" should not be there!:tongue:
     
  8. May 12, 2014 #7
    Yes.
    :/
     
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