Does the set of entire functions have the same cardinality as the reals? I've recently been thinking about entire functions and the Weierstrass factorisation theorem and it got me thinking about the cardinality of the set of entire functions. Clearly $e^{cz}$ is entire, for all $c \in \mathbb{C}$, so the cardinality of the set of entire functions is at least that of the continuum.
I thought that the set of entire functions would have a cardinality exceeding that of the continuum, but I have what I think is a proof to the contrary, though I feel like I'm missing something.
Let $f$ be an entire function and let $g$ be the restriction of $f$ to the open unit disk. Then $f$ is an analytic continuation of $g$ and by uniqueness of analytic continuations, $f$ is uniquely determined by $g$. But $f$ is holomorphic, so by the Cauchy integral formula, $g$ is completely determined by $f$ on the unit circle. Therefore $f$ is completely determined by its values on the unit circle.
If we let $h: \mathbb{R} \rightarrow \mathbb{C}$ be the polar parameterisation of $f$ on the unit circle, then since $f$ is holomorphic, $\text{Re}(h(z))$ and $\text{Im}(h(z))$ are (periodic) real continuous functions and $f$ is completely specified by them. But real continuous functions are uniquely specified by their values on the rationals, so the cardinality of the set of real continuous functions is that of the continuum. Therefore the set of entire functions also has the cardinality of the continuum.
Am I missing something? I feel like entire functions aren't uniquely determined by a pair of real continuous functions, but I don't see the gap in my reasoning.
 A: Your argument looks good to me. Alternatively you can use that an entire function is uniquely determined by its Taylor coefficients. That gives an injective mapping from the set of entire functions to the set $\Bbb C^{\Bbb N}$, and that has the cardinality of $\Bbb R$ (one can use a similar argument as in cardinality of all real sequences).
A: Since every entire function is a continuous function from $\mathbb C$ to $\mathbb C,$ and there are at most $\mathfrak c = 2^{\aleph_0}$ many continuous functions (each continuous function is determined by its values on a given countable dense subset of $\mathbb C,$ for example complex numbers with rational real and imaginary parts), it follows that there are at most $\mathfrak c = 2^{\aleph_0}$ many entire functions.
Related Stack Exchange question:
Cardinality of set of real continuous functions
In fact, even the collection of Borel measurable functions from $\mathbb C$ (or from $\mathbb R)$ to $\mathbb C$ (or to $\mathbb R)$ has cardinality $\mathfrak c = 2^{\aleph_0}.$ See
Cardinality of the borel measurable functions?
On the other hand, the cardinality of the Riemann integrable functions is $2^{\mathfrak c}.$
Cardinality of the set of Riemann integrable functions on [0,1]
