Prove that $S = \{ f: [0,1]\rightarrow \mathbb{R} \ \text{continuous} : x\in\mathbb{Q}\implies f(x) \in \mathbb{Q}\}$ is. uncountable Prove that
$$S = \{ f: [0,1]\rightarrow \mathbb{R} \ \text{continuous} : x\in\mathbb{Q}\implies f(x) \in \mathbb{Q}\}$$ is uncountable.
I know that reals are uncountable,  so I look for injection from reals to set $S$, i.e., if we can define a function as above for each real number, then we are done.
Am I going in the right direction?
Please help with some hint/solution.
 A: Let  $X $ be the set of all sequences r = $\{r_n \}_1^{\infty}$ such that  $r_n=\{ 1,-1\}$ it is an uncountable !!
For $r$ in $X$  let  $f_r$ be function defined on [0, 1]  such that

*

*$f_r (0) =0$


*$ f_r(1/n) = r_n / n$  for n positive integer ;


*on each interval  $ [ 1/(n+1), 1/n] , f_r$  is linear function whose values at the endpoints agree with those given by (2)
Each function $f_r $ is continuous at point of $(0, 1] $ is obvious, and continuity at $0$ follows  because $|f(x) | \leq x$
Each $f_r$ takes rational values at rational points :
if $x$ is rational point between  $ a=1/(n+1) $ and   $b = 1/n$ , then
$x= (1-t) a+tb $  with  $t$ a rational points of [0, 1] , so
$f_r(x)= (1-t) f_r (a) + tf_r(b) $ is rational because  $ f_r( a) $ and  $ f_r(b)$ are.
The functions $ f_r$ thus form an uncountable subset of S, showing that S is uncountable
A: I would probably start with the uncountable set of all increasing sequences of integers and assign $f(1/n) = 1/a_n$ for a sequence $a_n$.
Extend this step-wise linearly; it should map $\mathbb{Q}$ to $\mathbb{Q}$.
A: Given a real number $a\in [0,1]$ with decimal expansion $a=0.a_1a_2a_3a_4...$ consider the piecewise linear function defined by
$f\left(\frac{1}{i}\right) = a_i/i \in S$.
This gives you an injection of the uncountable set $[0,1]$ in $S$, therefore $S$ also uncountable.
EDIT: Thanks for the commenters pointing out that the function needs to be continuous at $0$ as well.
A: HINT:
For every $a\in \{0,1\}^{\mathbb{N}}$ consider
$f_a(\frac{1}{n}) = \frac{1}{2n + a_n}$, $1\mapsto 1$, $0\mapsto 0$, and extend by linearity.
A: *

*For any real function $f,$ if $a,b,f(a),f(b)\in \Bbb Q$ and $f$ is linear on $[a,b]$ then $\{f(x):x\in \Bbb Q\cap [a,b]\}\subset \Bbb Q.$


*Let $T$ be the set of strictly increasing sequences of $positive$ rationals that converge to $1$. Then $T$ is uncountable. Because (by a typical diagonal method) if $A=\{(q_{n,j})_{n\in \Bbb N}: j\in \Bbb N\}\subset T,$ then let $r_1\in \Bbb Q\cap (q_{1,1},1)$ and for $n\in \Bbb N$ let $r_{n+1}\in \Bbb Q\cap (\max (r_n,q_{n+1,n+1}),1).$ Then $(r_n)_{n\in \Bbb N}\in T$ \ $A.$


*For $t=(q_n)_{n\in \Bbb N}\in T$ and $n\in \Bbb N$ let $f_t(1-2^{-n})=q_n$ and let $f_t$ be linear on $[1-2^{-n}, 1-2^{-n-1}].$ Let $f_t(0)=0$ and let $f_t$ be linear on $[0,1/2].$ And let $f_t(1)=1.$  Then $\{f_t:t\in T\}$ is an uncountable subset of $X.$
