0
$\begingroup$

Why is the set of all continuous functions (from the reals to the reals) of size Beth one? Doesn't that mean that there is a bijection between the real numbers and continuous functions?

  • 3
    Indeed there is such a bijection. Note that all continuous functions are determined by their values at $\mathbb Q$, so there are at most $|\mathbb R|^{|\mathbb Q|}=\beth_1$ continuous functions. – 2017-02-03
  • 0
    Well, I don't think it is *in* $\beth_1$ (seen as an initial ordinal), but indeed continuous (and, more in general, Borel) functions $\Bbb R\to\Bbb R$ are in bijection with $\Bbb R$. – 2017-02-03
  • 0
    Also, as the other comment states - the set is not "in" $\beth_1$, it has size/cardinality $\beth_1$. – 2017-02-03
  • 0
    @Wojowu Thanks, I'll update the question. – 2017-02-03
  • 0
    @Wojowu Okay, I see it now. If you wouldn't mind adding that as an answer, I'll accept it. – 2017-02-03

1 Answers 1

3

Recall that $\mathbb Q$ is a countable, dense subset of $\mathbb R$. Because of this, every continuous function $f:\mathbb R\rightarrow\mathbb R$ is determined by its values at $\mathbb Q$. More precisely, if $f,g$ are two such functions and their restrictions to $\mathbb Q$ are the same, then $f=g$. Therefore the operation of taking $f$ to $f\mid_{\mathbb Q}$ is an injection from the set of all continuous functions $\mathbb R\rightarrow\mathbb R$. The latter set has cardinality precisely $|\mathbb R|^{|\mathbb Q|}$, and it's a matter of simple cardinal arithmetic to see this is equal to $\beth_1$ (which, by the way, is more commonly denoted by $\frak c$ or $2^{\aleph_0}$ and is called continuum). So there are at most $\frak c$ continuous functions.

Conversely, clearly there are at least $|\mathbb R|=\frak c$ continuous functions (just take the constant functions). Hence there are precisely $\frak c$ many continuous functions.