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Math Proof: Uncountable binary sequence and a bijection from R to R-{0}

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Question 1:
    Prove that the cardinality of R (the set of all real numbers)is the same as the cardinality of R-{0} by constructing a bijective function from R to R-{0}

    Question 2: Let A be the infinite sequence of binary numbers as follows:
    A={(a1,a2,a3...)|ai= 0or 1 for all i in the natural numbers}

    Show that A is uncountable


    2. Relevant equations



    3. The attempt at a solution

    For question 2 I think I have to use a proof similar to Cantor's diagonalization argument for proving that the set of real numbers is uncountable. I think I have to use contradiction and assume that the set is countable.
     
  2. jcsd
  3. Dec 3, 2011 #2

    micromass

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    Yes, for (2) you need to do something very similar to Cantor diagonalization.

    For one, you'll want to find a bijection [itex]f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}[/itex].

    Hint, if [itex]x\notin \mathbb{N}[/itex], define f(x)=x.
     
  4. Dec 3, 2011 #3
    I think Imay have misunderstood the second part. Isnt f(x) still f(x)=x?
     
  5. Dec 3, 2011 #4

    micromass

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    You define f(x)=x in (1) for all x not in [itex]\mathbb{N}[/itex].
     
  6. Dec 3, 2011 #5
    so {(x| x =/= 1,2,3,4....)} ?
     
  7. Dec 3, 2011 #6

    micromass

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    Uuh, what do you mean with that??

    Also note that I consider 0 to be in [itex]\mathbb{N}[/itex].
     
  8. Dec 3, 2011 #7
    gahh im so sorry I dont understand..am I looking for a function that hits all numbers except for 0,1,2,3.....?
     
  9. Dec 3, 2011 #8

    micromass

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    No, that's not what I stated. I just said you had to define f(x)=x for [itex]x\notin\{0,1,2,3,...\}[/itex]. You still need to define f(0), f(1), f(2), ...

    But you have to end up with a bijection [itex]f:\mathbb{R}\rightarrow \mathbb{R}\setminus \{0\}[/itex].
     
  10. Dec 3, 2011 #9
    so how about f(x)={x+1 for the natural numbers and x otherwise) would that make it so that x is not in the natural numbers but f(x) exists for the natural numbers?
     
  11. Dec 3, 2011 #10

    micromass

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    That's a nice proposal!!
     
  12. Dec 3, 2011 #11
    yay...i cant believe it took me that long to understand what you were trying to say..its obvious now though :)
     
  13. Dec 17, 2011 #12
    Just wondering if under that function, the preimage of 1.
     
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