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A very probably flawed attempt at CH

  1. Feb 8, 2012 #1
    [There is no set whose cardinality is strictly between that of the integers and that of the real numbers.]

    Lets suppose we have the set F. We define F as the set of irrational numbers, which are greater than 0 and less than 1. (if x is a member of F, then 0<x<1). Now it is obvious, that FR is uncountable, since Cantor's diagonal argument holds here. And since it is uncountable, its cardinality is bigger than the set of the integers.

    (FR is strictly a subset of ℝ, since a real number may be either rational or irrational; either algebraic or transcendental; and either positive, negative, or zero.)

    Suppose we attempt a bijection, as in ( f:F→R ). We only need to pair all of the reals, that are greater than 0 and less than 1, to F. This is possible, since F is strictly a subset of ℝ. But there are infinitely more members in ℝ, in fact for any member of F, there are infinitely more members:
    As we defined F, we stated that for any member x of F, x is greater than 0, but less than 1. Thus any x of F can be defined as the fractional part of an irrational number.
    Say for example π : 3,14159265...
    The fractional part of π is in F, since pi is irrational.
    But in ℝ, we can find such numbers :
    for all n, such that n is a integer.
    And this is true for any member of F.
    That is, for any x of F, there exists a subset of ℝ, whose members, have the aforementioned x as the fractional part:
    say x is a member of F, then Ax={x + n | n∈Z, n≠0}.

    Now we can biject Ax, for any x of F, to the set of all integers, that is f:AxZ . This is possible, because if we were to remove the fractional parts of all of the members of a Ax, we would get the set of all integers.

    The set of all Ax, is uncountable, since for every x of F, there exists a Ax.

    Any Ax is a subset of the real numbers.

    Thus, the cardinality of F is greater than the set of integers, and less than the set of real numbers.

    I doubt, that there aren't any flaws in this, or that this "proof" could in anyway be defined as formal or rigorous. But in both cases, I would like to know where I have gone wrong.
  2. jcsd
  3. Feb 8, 2012 #2
  4. Feb 8, 2012 #3


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    Gold Member

    The problem with your argument: You never show that there is no bijection [itex]F \to \mathbb{R}[/itex]. Nothing in your argument supports the claim that there is no bijection. In fact, it is not difficult to show that there is a bijection between the irrationals in [itex][0,1][\itex] and [itex]\mathbb{R}[\itex].

    The CH is independent of ZFC anyway, so there is no way you can actually settle the question using only the usual axioms for set theory.
  5. Feb 8, 2012 #4
    Actually, you argument also doesn't explain why the set F is uncountable. You quote Cantor's argument, but you don't explain why it works for this set.
    (For example, a diagonalization might give you a rational number not in the list, but that is not in F so the argument doesn't work)

    This is a good proof that [itex]|F|\times|\mathbb{Z}|=|\mathbb{R}|[/itex]. But due to funny facts on multiplication of cardinalities, this does not mean [itex]|F|<|\mathbb{R}|[/itex].
  6. Feb 8, 2012 #5
    Let F = the irrationals between 0 and 1. Let G = the rationals between 0 and 1.

    Then [itex]F\cup G=]0,1[[/itex] and thus (recall that ]0,1[ has the cardinality of the reals):

    [tex]|F\cup G|=|\mathbb{R}|[/tex]

    We know that G is countable, thus [itex]|G|<|\mathbb{R}|[/itex]. If also [itex]|F|<|\mathbb{R}|[/itex], then [itex]|F\cup G|<|\mathbb{R}|[/itex]. Contradiction.
  7. Feb 8, 2012 #6
    To show that one set has smaller cardinality than another, you need to show that there is NO POSSIBLE bijection between the sets.

    It's not enough to show that one particular ATTEMPT at a bijection seems to fail.
  8. Feb 9, 2012 #7
    Oh crap, you're all right. I am done here, as in terribly wrong.

    Thank you all for taking the time to show me the flaws.
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