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Let $F$ denote the set of all functions $f: \mathbb{R} \to \mathbb{R}$, and let $C \subset F$ denote the subset of all continuous functions. Prove that $|\mathbb{R}| = |C| < |F|$. (Hint: use the fact that a continuous function on $\mathbb{R}$ is determined by its values on the rational numbers $\mathbb{Q} \subset \mathbb{R}$.

I've done some searching on this website about this proof and almost every answer seems to involve arithmetic of infinite cardinals, but this is not something I am familiar with, so I am hoping there is a proof that doesn't require it, even if it's longer and has more moving parts.

Here is a sketch of what I have so far.

By the Cantor-Schroeder Bernstein theorem, we can prove $|\mathbb{R}| = |C|$ by finding injections $f: \mathbb{R} \to C$ and $g: C \to \mathbb{R}$. For $f$, we can take $c \mapsto f(x) = c$, the constant function, which is of course continuous (take $\delta = \epsilon$). For $g$, we enumerate the rationals (which are countable), $q_1, \ldots, $ and map a function $f$ to the sequence $(f(q_1), f(q_2), \ldots)$. This map is injective because a continuous function from $\mathbb{R}$ to $\mathbb{R}$ is uniquely determined by how it acts on the rationals (by density). If I could prove that the reals have the same cardinality as the sequences of reals, I can compose injections to conclude that, by Schroeder Bernstein, $|\mathbb{R}| = |C|$.

I don't know how to prove that $|C| < |F|$. I believe the definition of a strictly larger cardinality is that there is an injection from $C$ to $F$ but not surjection. (Is this correct?) If $so$, I can definitely inject $C$ into $F$ by the identity function. I would need only show there is no surjection from $C$ to $F$, which I don't know how to do.

2 Answers2

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To finish your argument for the first part I’m going to assume that you know that $|\Bbb R|=|\wp(\Bbb N)|=\left|\{0,1\}^{\Bbb N}\right|$, and that $|\Bbb N\times\Bbb N|=|\Bbb N|$. Thus,

$$\left|\Bbb R^{\Bbb N}\right|=\left|\left(\{0,1\}^{\Bbb N}\right)^{\Bbb N}\right|=\left|\{0,1\}^{\Bbb N\times\Bbb N}\right|=\left|\{0,1\}^{\Bbb N}\right|=|\Bbb R|\,.\tag{1}$$

To rephrase that in terms of mappings, let $f:\Bbb R\to\{0,1\}^{\Bbb N}$ be a bijection. Then

$$g:\Bbb R^{\Bbb N}\to\left(\{0,1\}^{\Bbb N}\right)^{\Bbb N}:\langle x_n:n\in\Bbb N\rangle\mapsto\langle f(x_n):n\in\Bbb N\rangle$$

is a bijection, establishing the first equality in $(1)$. The second follows from the general result that for sets $A,B$, and $C$ there is a bijection between $\left(A^B\right)^C$ and $A^{B\times C}$; it’s a little messy to write down the details, but it’s not too hard and is a good exercise. The third follows easily from the existence of a bijection between $\Bbb N\times\Bbb N$ and $\Bbb N$, and I’ll leave that one to you.

For the inequality note that each $A\subseteq\Bbb R$ has an indicator (characteristic) function

$$\chi_A:\Bbb R\to\Bbb R:x\mapsto\begin{cases} 1,&\text{if }x\in A\\ 0,&\text{if }x\notin A\,. \end{cases} $$

The map from $\wp(\Bbb R)$ to $\Bbb R^{\Bbb R}$ that takes $A$ to $\chi_A$ is easily seen to be injective, so $|\wp(\Bbb R)|\le\left|\Bbb R^{\Bbb R}\right|$. (In fact they are equal, but you don’t need that.) Finally, Cantor’s theorem says that $|\Bbb R|<|\wp(\Bbb R)|$, so $|\Bbb R|<\left|\Bbb R^{\Bbb R}\right|$.

Brian M. Scott
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The cardinality of the set of all functions $f:\mathbb{R} \to \mathbb{R}$ is simply $|\mathbb{R}|^{|\mathbb{R}|}$ which by cardinal arithmetic can be looked at as $2^{| \mathbb{R}|}$ or denoted as $2^{\mathfrak{c}}$, $\mathfrak{c}$ being the cardinality of the continuum.

Now concerning the cardinality of the set of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$, we must be careful. let's use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. Let's go ahead and apply the Cantor-Schroder-bernstein theorem as you tried to.

  1. The cardinality of the set of all continuous functions is at least that of the continuum $\frak{c}$ since we can associate injectively to each constant function a real number.

  2. The cardinality of the set of all continuous functions is at most that of the continuum $\mathfrak{c}$ since we can inject this set into the set $\mathbb{R}^{\mathbb{N}}$. Essentially one associates to each continuous function its values on all the rational points (infinite sequence of integers).

  3. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ this uniquely determines the continuous function.

Hence nay the Bernstein-Cantor-Schroder theorem, $$|\{f: f:\mathbb{R} \to \mathbb{R} \text{ is continuous}\}|=\mathfrak{c}$$

Rachid Atmai
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