Cardinality of set of real periodic functions

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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|.
 

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  • #2
Office_Shredder
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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|.
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
 
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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].
 
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Office_Shredder
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I pick f to be the indicator function over the rationals (which is periodic). What is p?
 
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I pick f to be the indicator function over the rationals (which is periodic). What is p?
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].
 
  • #6
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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
 
  • #7
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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].
 

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