# Thread: strictly increasing function continous only at irrationals

1. ## strictly increasing function continous only at irrationals

hi!

I'm looking for a strictly increasing function on I = [0, 1] that is continuous only at the irrational numbers in I.

After some thought, I came up with the following arrangement.

Enumerate the rational numbers in I as such: $q_1, q_2, q_3, ... , q_n$ (with n a natural number) such that $q_k < q_{k+1}$

Let $f(x) = 2^k$ if $x = q_k$ (for 1 <= k <= n)
and $f(x) = (2^k + x)$ if x is irrational and $q_k < x < q_{k+1}$

Is the function continuous at x irrational? I think so.
For any Epsilon > 0, choose Delta = $min (x - q_k, q_{k+1} - x) * 0.5$ so that for any x' satisfying $|x' - x| < Delta$ we have $|f(x')-f(x)| < Epsilon$.

Is the function discontinuous at x rational? I think that can be shown using similar reasoning.

Am I close? If not, can I modify my function to meet the requirement?

2. ## Re: strictly increasing function continous only at irrationals

I'm very suspicious of your ability to enumerate the rationals in increasing order but I'm no expert.

3. ## Re: strictly increasing function continous only at irrationals

Thanks for your reply romsek. I'm not an expert either but I think you're right. I have to rethink this.

4. ## Re: strictly increasing function continous only at irrationals

Here is an such a function due to Walter Rudin.
Suppose that $(c_n)$ is a sequence of points in $[0,1]$ such that $\sum\limits_{n = 1}^\infty {{c_n}}$ converges. Note that convergence is absolute so the terms can be rearranged.

Now define $\forall (x \in [0,1])\left[ {{E_x} = \{c_n}|q_n < x\} \right]$ i.e the rational terms so that $q_n.

Then define $f(x) = \left\{ {\begin{array}{rl}{\sum\limits_{t \in {E_x}} t ,}&{{E_x} \ne \emptyset }\\{0,}&{{E_x} = \emptyset }\end{array}} \right.$

Now that function is not quite what you asked for, but is should give you some ideas.

5. ## Re: strictly increasing function continous only at irrationals

Hi again,

In my class, we were shown the following example of a strictly increasing function on (0, 1) that is continuous at the irrationals and discontinuous at the rationals:

Let $g_{1}, g_{2}, g_{3}, ..., g_{m}$ be a listing of all rationals in (0, 1) and
$f(x) = \Sigma_{ \big\{n | g_{n} < x\big\} } \frac{1}{2^{n}}$

I was wondering whether the above could be easily extended to the case of I = [0, 1].
I would need to include the rational end points $g_{1} = 0$ and $1$ in the list.

I think the function would need to be modified because otherwise I would get $f(0) = 0$ and $\lim_{x \rightarrow 0^{+}} f(x) = 0$
and similarly $f(1) = 1$ and $\lim_{x \rightarrow 1^{-}} f(x) = 1$

Will the following function work for the [0, 1] case?

$f(x) =\begin{cases}-2 & x = 0\\ \Sigma_{ \big\{n | g_{n} < x\big\} } \frac{1}{2^{n}} & x \in (0, 1) \\ 2 & x = 1\end{cases}$

Thanks again.