On cardinality of a set of continuous functions

Solution 1:

The cardinality of this set of functions $\mathcal{F}$ is $\mathfrak{c}$, the continuum.

The cardinality of $\mathcal{F}$ is at least $\mathfrak{c}$ because given a real irrational number $\alpha \in (0,1)$ and any real number $\beta$, we can construct a function with the given properties and having $f(\alpha) = \beta$. In particular, if we fix $\alpha = 1/\sqrt{2}$ or your favorite irrational in the interval, this construction will describe as many different functions as there are real values for $\beta$, and $|\mathbb{R}| = \mathfrak{c}$.

Let $0 = a_1 < a_2 < a_3 < \cdots$ be any strictly increasing sequence of rational numbers converging to $\alpha$, and let $1 = b_1 > b_2 > b_3 > \cdots$ be any strictly decreasing sequence of rational numbers converging to $\alpha$. The function $f$ is:

$$ f(x) = \begin{cases} \frac{\lfloor k \beta \rfloor}{k} + \left(\frac{\lfloor(k+1)\beta\rfloor}{k+1} - \frac{\lfloor k \beta \rfloor}{k}\right) \frac{x-a_k}{a_{k+1}-a_k} & a_k \leq x < a_{k+1} \\ \beta & x = \alpha \\ \frac{\lceil k \beta \rceil}{k} + \left(\frac{\lceil (k+1) \beta \rceil}{k+1} - \frac{\lceil k \beta \rceil}{k}\right) \frac{b_k-x}{b_k-b_{k+1}} & b_{k+1} < x \leq b_k \end{cases} $$

$f(x)$ takes a rational value at every rational $x$, since all floor ($\lfloor \cdot \rfloor$) and ceiling ($\lceil \cdot \rceil$) function values are integers, $k$ is a positive integer, and all $a_k$ and $b_k$ are rational. $f$ is linear on the intervals $(a_k,a_{k+1})$ and $(b_{k+1},b_k)$, and continuity at the corner points $a_k$ and $b_k$ is easy to show. For continuity at $\alpha$, note that if $a_k \leq x < a_{k+1}$, since $f$ is linear on the interval,

$$ |f(x)-\beta| \leq \max(|f(a_k)-\beta|, |f(a_{k+1})-\beta|) $$ $$ |f(a_k)-\beta| = \left| \frac{\lfloor k \beta \rfloor}{k} - \beta \right| = \frac{k \beta - \lfloor k \beta \rfloor}{k} < \frac{1}{k} $$ $$ |f(x)-\beta| \leq \max \left(\frac{1}{k}, \frac{1}{k+1}\right) = \frac{1}{k} $$

Similarly, $b_{k+1} < x < b_k$ also implies $|f(x)-\beta| \leq \frac{1}{k}$. Since $j > k$ implies $1/j < 1/k$ and $f(\alpha)-\beta = 0$, it's also true that $|f(x)-\beta| \leq \frac{1}{k}$ for all $x \in [a_k,b_k]$. So for any real $\epsilon > 0$, choose positive integer $K > 1/\epsilon$, and $|x-\alpha| < \min(\alpha-a_K, b_K-\alpha)$ implies $|f(x)-f(\alpha)| \leq 1/K < \epsilon$; $f$ is continuous at $\alpha$.

The cardinality of $\mathcal{F}$ is at most $\mathfrak{c}$ because we can define an injection from the set of functions to the set of sequences of rational numbers, which has cardinality $\mathfrak{c}$. This mapping is

$$ \Phi : f \mapsto \left(f(0), f(1), f\!\left(\frac{1}{2}\right), f\!\left(\frac{1}{4}\right), f\!\left(\frac{3}{4}\right), f\!\left(\frac{1}{8}\right), \ldots\right) $$

To show this $\Phi$ is injective, suppose $f_1$ and $f_2$ are continuous functions mapping to the same sequence. That is, $f_1\left(\frac{m}{2^n}\right) = f_2\left(\frac{m}{2^n}\right)$ for numbers of the form $\frac{m}{2^n}$ (with $m,n$ non-negative integers) in the domain. Suppose by way of contradiction that $f_1(x_0) \neq f_2(x_0)$ at any real $x_0 \in (0,1)$. Since each function is continuous, there exist a $\delta_1>0$ and $\delta_2>0$ such that

$$|x-x_0| < \delta_1 \implies |f_1(x)-f_1(x_0)| < \frac{1}{2}|f_1(x_0)-f_2(x_0)|$$ $$|x-x_0| < \delta_2 \implies |f_2(x)-f_2(x_0)| < \frac{1}{2}|f_1(x_0)-f_2(x_0)|$$

Let $\delta = \min(\delta_1, \delta_2)$. Since numbers of the form $\frac{m}{2^n}$ are dense, there is such a number $x_1 = \frac{m}{2^n}$ in the interval $(x_0-\delta, x_0+\delta)$. Then $|x_1-x_0| < \delta_1$ and $|x_1-x_0| < \delta_2$ and $f_1(x_1) = f_2(x_1)$, so combining the continuity statements with the triangle inequality gives

$$ \begin{align*} |f_1(x_0)-f_2(x_0)| &= \big|[f_1(x_0)-f_1(x_1)] - [f_2(x_0)-f_2(x_1)]\big| \\ &\leq |f_1(x_1)-f_1(x_0)| + |f_2(x_1)-f_2(x_0)| \\ &< \frac{1}{2}|f_1(x_0) - f_2(x_0)| + \frac{1}{2}|f_1(x_0)-f_2(x_0)| \\ &= |f_1(x_0)-f_2(x_0)| \end{align*} $$

Contradiction: it is not possible that $f_1(x_0) \neq f_2(x_0)$. So if continuous functions have the same sequence $\Phi(f_1) = \Phi(f_2)$, the functions are identical; $\Phi$ is injective.

Solution 2:

Continuous functions on $[0,1]$ are uniquely determined by their values on the countable dense set $D=\mathbb{Q}\cap [0,1]$. That is, if $f$ and $g$ are continuous and if $f(x)=g(x)$ for all $x\in D$, then $f(x)=g(x)$ for all $x\in [0,1]$. Thus, there is an injective function $$\Phi: C([0,1],\mathbb{R}) \to \mathbb{R}^D,$$ where $\mathbb{R}^D$ is the set of all functions $D\to\mathbb{R}$. Finally, $$card(\mathbb{R}) \leq card(C([0,1],\mathbb{R})) \leq card(\mathbb{R}^D) = card(\mathbb{R})^{card(D)} = card(\mathbb{R}).$$ Consequently, $card(\mathbb{R}) = card(C([0,1],\mathbb{R}))$.

Edit: I missed a crucial part of the question at the first look. Let $A\subseteq C([0,1],\mathbb{R})$ be the set of all continuous functions that map rationals to rationals.

Every continuous function defined on $I=[0,1]$ (generally on a compact set $I$) attains its minimum on $I$. It is not difficult to show that for every $a\in\mathbb{R}$, there exists $f\in A$ with minimum value $a$. (Just consider polynomials with rational coefficients, or piecewise linear functions with a single kink.) Thus, there exists a surjective map $\Psi:A\to\mathbb{R}$. Consequently, $$card(\mathbb{R})\leq card(A).$$ Also, from above, $card(A)\leq card(C([0,1],\mathbb{R})) = card(\mathbb{R})$. Together, $card(\mathbb{R}) = card(A)$.