Why is the set of all continuous functions of size Beth one?

Question

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?

Answer 1

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.