10
$\begingroup$

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.

$\endgroup$
0

5 Answers 5

6
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Thankyou for detailed answer , $\endgroup$
    – Mathsingh
    Oct 17, 2020 at 3:35
  • $\begingroup$ Your 3rd "bullet" refers to (2) but I don't see what (2) is. $\endgroup$ Oct 17, 2020 at 5:35
  • $\begingroup$ it refers to 2nd bullet $\endgroup$
    – ਮੈਥ
    Oct 17, 2020 at 6:31
4
$\begingroup$

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 $\mathbb{Q}$.

$\endgroup$
0
4
$\begingroup$

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.

$\endgroup$
3
  • 4
    $\begingroup$ How do you define $f$ at $0$ so that the function is continuous? $\endgroup$
    – Wojowu
    Oct 15, 2020 at 9:55
  • 2
    $\begingroup$ There is a small problem about the decimal expansion not being unique.. $\endgroup$ Oct 15, 2020 at 10:02
  • 2
    $\begingroup$ Come on, I am sure we can fix this answer. :-( $\endgroup$
    – nicomezi
    Oct 15, 2020 at 10:03
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Your function maps the interval 0 to 1 into the interval 0 to 1/2 $\endgroup$
    – Kosh
    Oct 15, 2020 at 10:23
  • 1
    $\begingroup$ @Kosh: typo corrected, thanks $\endgroup$
    – orangeskid
    Oct 15, 2020 at 10:32
0
$\begingroup$
  1. For any real function $f,$ if $a,b,f(a),f(b)\in \Bbb Q$ and $f$ is linear on $[a,b]$ then $\{f(x):x\in \Bbb Q\cap [a,b]\}\subset \Bbb Q.$

  2. 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 $(r_n)_{n\in \Bbb N}\in T$ \ $A.$

  3. 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 $X.$

$\endgroup$
1
  • $\begingroup$ To be pedantically rigorous, in 2. I should have noted that $T\ne \emptyset.$ $\endgroup$ Oct 17, 2020 at 5:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .