Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Cardinality of set of real periodic functions

  1. Dec 3, 2008 #1
    what is the cardinality of a set A of real periodic functions ?
    f(x)=x is periodic so R is subset of A but not equal because sin(x) is in A but not in R. hence aleph_1<|A|.
     
  2. jcsd
  3. Dec 3, 2008 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What?

    f(x) = x isn't periodic. And even if it was, that would only tell you A has cardinality at least 1. sin(x) is in A but not R... is R supposed to be the real numbers? How can a function be a number? Also, just because you add a single element to a set doesn't mean the cardinality changes either way
     
  4. Dec 4, 2008 #3
    First, note: [tex]\lvert \mathbb{R} \rvert = 2^{\aleph_0}[/tex], and the statement that [tex]\lvert \mathbb{R} \rvert = \aleph_1[/tex] is the continuum hypothesis holds (which cannot be proven or disproven in ZFC).

    I shall denote [tex]c = 2^{\aleph_0} = \lvert \mathbb{R} \rvert[/tex]. Let C be the set of constant functions from [tex]\mathbb{R}[/tex] to [tex]\mathbb{R}[/tex], and let P be the set of nonconstant periodic functions from [tex]\mathbb{R}[/tex] to [tex]\mathbb{R}[/tex]; then [tex]A = C \cup P[/tex] is a union of disjoint sets. Clearly [tex]\lvert C \rvert = c[/tex].

    Let [tex]P' = \{(p, f') \mid p \in \mathbb{R}^+, f' \colon [0, p) \to \mathbb{R} \}[/tex]; I construct a function [tex]g \colon P \to P'[/tex] by assigning to each periodic function [tex]f \in P[/tex] the pair [tex](p, f')[/tex], where p is the period of f, and f' is the restriction of f to [0, p). It is a bijection; you should check this. Now since [tex]\lvert [0, p) \rvert = \lvert \mathbb{R} \rvert[/tex] for any positive p, [tex]\lvert P \rvert = \lvert P' \rvert = \lvert \mathbb{R}^+ \times \mathbb{R}^{\mathbb{R}} \rvert = c \cdot c^c = c^c (= 2^{\aleph_0 \cdot c} = 2^c = 2^{2^{\aleph_0}})[/tex]. Thus [tex]\lvert A \rvert = \lvert C \rvert + \lvert P \rvert = c^c = \lvert \mathbb{R}^{\mathbb{R}} \rvert[/tex], the cardinality of the set of all functions from [tex]\mathbb{R}[/tex] to [tex]\mathbb{R}[/tex].
     
  5. Dec 4, 2008 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I pick f to be the indicator function over the rationals (which is periodic). What is p?
     
  6. Dec 4, 2008 #5
    Well, that's a pretty embarrassing mistake.

    Well, let's fix that up. I let P instead be the set of periodic functions with a smallest positive period; then [tex]\lvert P \rvert = c^c[/tex] (I think; see below). Then [tex]c^c = \lvert P \rvert \le \lvert A \rvert \le c^c[/tex], so [tex]\lvert A \rvert = c^c[/tex].

    Another embarrassing mistake I made: g isn't really a surjection. I don't know exactly how to handle that, then, but I'm still sure that [tex]\lvert P \rvert = c^c[/tex].
     
  7. Dec 4, 2008 #6

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I remember how to do this. Given A a subset of R... we can map [0,1) onto R, so we can map [0,1) onto A, say by a function f. Then extend f to a function F by F(x) = f([x]) where [x] is the rational part of x. F is periodic and has as its range A. Hence for each subset of R, we have a distinct periodic function, and then just use the fact that the set of periodic functions is a subset of the set of all functions R->R and the cardinality of the latter is cc
     
  8. Dec 4, 2008 #7
    Your idea is a lot simpler; thanks. This is what I thought of when I read your post:

    The mapping from [tex]\mathbb{R}^{[0, 1)}[/tex] to the set [tex]A \subseteq \mathbb{R}^\mathbb{R}[/tex] of periodic functions by periodic extension (as you describe explicitly) is an injection, so [tex]c^c = \lvert \mathbb{R}^{[0, 1)} \rvert \le \lvert A \rvert \le \lvert \mathbb{R}^\mathbb{R} \rvert = c^c[/tex], so [tex]\lvert A \rvert = c^c[/tex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Cardinality of set of real periodic functions
  1. Cardinality of sets (Replies: 1)

  2. Cardinality of sets (Replies: 6)

Loading...