Cardinality of set of real periodic functions

[TTob]

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|.

 

[Office_Shredder]

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

 

[adriank]

First, note: \lvert \mathbb{R} \rvert = 2^{\aleph_0}, and the statement that \lvert \mathbb{R} \rvert = \aleph_1 is the continuum hypothesis holds (which cannot be proven or disproven in ZFC).

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

Let P&#039; = \{(p, f&#039;) \mid p \in \mathbb{R}^+, f&#039; \colon [0, p) \to \mathbb{R} \}; I construct a function g \colon P \to P&#039; by assigning to each periodic function f \in P the pair (p, f&#039;), 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 \lvert [0, p) \rvert = \lvert \mathbb{R} \rvert for any positive p, \lvert P \rvert = \lvert P&#039; \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}}). Thus \lvert A \rvert = \lvert C \rvert + \lvert P \rvert = c^c = \lvert \mathbb{R}^{\mathbb{R}} \rvert, the cardinality of the set of all functions from \mathbb{R} to \mathbb{R}.

 

[Office_Shredder]

I pick f to be the indicator function over the rationals (which is periodic). What is p?

 

[adriank]

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 \lvert P \rvert = c^c (I think; see below). Then c^c = \lvert P \rvert \le \lvert A \rvert \le c^c, so \lvert A \rvert = c^c.

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 \lvert P \rvert = c^c.

 

[Office_Shredder]

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 c^c

 

[adriank]

Your idea is a lot simpler; thanks. This is what I thought of when I read your post:

The mapping from \mathbb{R}^{[0, 1)} to the set A \subseteq \mathbb{R}^\mathbb{R} of periodic functions by periodic extension (as you describe explicitly) is an injection, so c^c = \lvert \mathbb{R}^{[0, 1)} \rvert \le \lvert A \rvert \le \lvert \mathbb{R}^\mathbb{R} \rvert = c^c, so \lvert A \rvert = c^c.