Cardinality of set of real continuous functions I believe that the set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$, the cardinality of the continuum. However, I read in the book &quot;Metric spaces&quot; by Ó Searcóid that set of all $[0, 1]\to\mathbb{R}$ continuous functions is greater than $\mathfrak c$:

&quot;It is demonstrated in many textbooks that $\mathbb{Q}$
is countable, that $\mathbb{R}$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions deﬁned on  $[0,1]$ is of a greater cardinality than $\mathbb{R}$, and that there are sets of greater and greater cardinality.&quot;

I understand that (via composition with the continuous function $\tan$ or $\arctan$) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this?
The cardinality is at least that of the continuum because every real number corresponds to a constant function.  The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $\mathbb R^N$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.
The Schroeder-Bernstein theorem now implies the cardinality is precisely that of the continuum.
Note that  then the set of sequences of reals is also of the same cardinality as the reals.  This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.
I have a somewhat simple answer to this question. It is not as elaborate as the other but perhaps it will add some intuition.

let's look at the number of functions from $\mathbb R^n \to \mathbb R$.

For every element in $\mathbb R^n$ we need to choose a corresponding image in $\mathbb R$. There are $c$ elements in $\mathbb R$, and so if there are $\alpha$ elements in $\mathbb R^n$, there are $\alpha c$ functions (not continuous! just functions) from $\mathbb R^n \to \mathbb R$.

Let's find alpha: For the first element in an $n$ vector we have $c$ choices, for the second one $c$ choices, for the third one $c$ choices etc...overall $c^n=c$ choices.

So overall there are $c^{n+1}=c$ functions from $\mathbb R^n \to \mathbb R$.

since the continuous functions are a subset of this set, there are AT MOST $c$ continuous functions.

Now let's look at the function $f(x)=\xi ||x||_2$ where $\xi$ is some real number, $x$ is an $n$ vector, and $||x||_2$ is the euclidean norm of the vector $x$.

This function $f$ is continuous no matter which $\xi$ we pick. We have $c$ options to choose from, and so the set of continuous functions includes all the functions of the form $f(x)=\xi ||x||_2$ and so it is AT LEAST $c$.

Since it is at most $c$, and at least $c$, the conclusion is that it is exactly $c$.