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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.

h4tter
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  • A note: the objects of complex analysis are extremely constrained, so perhaps this is not quite so surprising as it seems initially. – user3716267 Jul 07 '21 at 14:59
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    Can't you just use the fact that every entire function is continuous (a rather weak deduction) and the well known fact that there are continuum many continuous functions from $\mathbb C$ to $\mathbb R$? (Same proof as for functions from $\mathbb R$ to $\mathbb R,$ namely once the values of a continuous function have been determined on a given countable dense set, such as the rationals, then the remaining values of the function are determined.) – Dave L. Renfro Jul 07 '21 at 15:10
  • @DaveL.Renfro: That is simpler and more general. Why don't you post it as an answer? – Martin R Jul 07 '21 at 16:51
  • @Martin R: Actually, I thought your answer (which I upvoted) was more directly connected to what an entire function is. I usually restrict my answers to things I'd be interested in looking up later (or which I think aren't well known, and thus worth archiving online somewhere), although in the past 2 or 3 years I have been sometimes giving short (and generally well known) answers. And in this case I suppose it complements your answer nicely. – Dave L. Renfro Jul 07 '21 at 17:11

2 Answers2

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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).

Martin R
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    Ah, excellent. I can't believe I used analytic continuation in my argument without thinking about Taylor series and their coefficients – h4tter Jul 07 '21 at 14:30
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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]

Dave L. Renfro
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