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R=2^\alepha 0 vs Continuum hypothesis! A result in a taste of topology

  1. Jan 24, 2009 #1
    R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

    A year ago or so I read a proof in A Taste Of Topology, Runde that the cardinality of the continuum equals the cardinality of the powerset of the natural numbers. But a few hours ago I found Hurkyl making that statement [tex]|\mathbb{R}| = |\mathcal{P}(\mathbb{N})|[/tex] is undecidable in ZFC, I almost put in this proof but I realized this is the continuum hypothesis (would have been embarrassing!), so there must be something wrong, either in the proof of Runde or a misunderstanding (on by behalf) of the statement! Interestingly though, of all the reviews written on the book, there isn't a single mention of that statement!

    Proposition [tex]c = 2^{\aleph_0}[/tex].
    Here [tex]\aleph_0[/tex] denotes the cardinality of [tex]\mathbf{N}[/tex] and [tex]c[/tex] the cardinality of [tex]R[/tex]
    The proof uses Cantor-Bernstein theorem, basically if [tex]2^{\aleph_0} \leq c[/tex] and [tex]2^{\aleph_0} \geq c[/tex] holds then [tex]2^{\aleph_0} = c[/tex]


    Direction: [tex]2^{\aleph_0} \leq c[/tex].

    Given [tex]S \subset N[/tex], define [tex](\sigma_{n}(S))^{\infty}_{n=1}[/tex] by letting [tex]\sigma_{n}(S) = 1[/tex] if [tex]n \in S[/tex] and [tex]\sigma_{n}(S) = 2[/tex] if [tex]n \notin S[/tex], and let [tex]r(S) := \sum_{n=1}^{\infty} \frac{\sigma_{n}(S)}{10^n}}[/tex]
    Then [tex]P(\mathbf{N}) \rightarrow (0,1)[/tex] defined by [tex]S \rightarrow r(S)[/tex] is injective


    Direction: [tex]2^{\aleph_0} \geq c[/tex]

    For the converse inequality, we use the fact that every [tex]r \in (0, 1)[/tex] not only has a decimal expansion, but also a binary one: [tex]r := \sum_{n=1}^{\infty} \frac{\sigma_{n}(r)}{2^n}}[/tex] with [tex]\sigma_{n}(r) \in \{0,1\}[/tex] for [tex]n \in \mathbf{N}[/tex].
    Hence, every number in [tex](0, 1)[/tex] can be represented by a string of zeros and ones.
    This representation, however, is not unique: for example, both 1000 ... and 0111 ... represent the number [tex]\frac{1}{2}[/tex].
    This, however, is the only way ambiguity can occur. Hence, whenever [tex]r \in (0,1)[/tex] has a period [tex]\overline{1}[/tex], we convene to pick its nonperiodic binary expansion. In this fashion, we assign, to each [tex]r \in (0, 1)[/tex], a unique sequence [tex](\sigma_{n}(r))^{\infty}_{n}=1[/tex] in [tex]\{0, 1\}[/tex].

    The map [tex](0, 1) \rightarrow P(N)[/tex], [tex]r \rightarrow \{n \in N : \sigma_{n}(r) = 1\}[/tex] is then injective.
     
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  3. Jan 24, 2009 #2

    CompuChip

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    Re: R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

    I think by definition of the cardinal numbers [itex]\aleph_n[/itex], [itex]|\mathbb{R}| = |\mathcal P(\mathbb{N})|[/itex]; the continuum hypothesis just asks whether there exists a set A such that |A| lies strictly between those (equivalently, whether [itex]2^{\aleph_0} = \aleph_1[/itex]).
     
  4. Jan 24, 2009 #3
    Re: R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

    That would make perfect sense then. I think I was in a hurry comparing the result to the continuum hypothesis. So in other words, that would make the function Hurkyl defined in https://www.physicsforums.com/showpost.php?p=2046993&postcount=7 decidable and f(0)=1!
     
    Last edited: Jan 24, 2009
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