# Cardinality of set of real periodic functions

1. Dec 3, 2008

### 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|.

2. Dec 3, 2008

### Office_Shredder

Staff Emeritus
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

3. Dec 4, 2008

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 $$P' = \{(p, f') \mid p \in \mathbb{R}^+, f' \colon [0, p) \to \mathbb{R} \}$$; I construct a function $$g \colon P \to P'$$ by assigning to each periodic function $$f \in P$$ the pair $$(p, f')$$, 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, $$\lvert P \rvert = \lvert P' \rvert = \lvert \mathbb{R}^+ \times \mathbb{R}^{\mathbb{R}} \rvert = c \cdot c^c = c^c (= 2^{\aleph_0 \cdot c} = 2^c = 2^{2^{\aleph_0}})$$. 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 $$\mathbb{R}$$.

4. Dec 4, 2008

### Office_Shredder

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

5. Dec 4, 2008

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 $$\lvert P \rvert = c^c$$.

6. Dec 4, 2008

### Office_Shredder

Staff Emeritus
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. Dec 4, 2008

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 $$\lvert A \rvert = c^c$$.