Let $\mathbb{L}=\{x_n \ |\ n=1,2,3 \dots\}$ be a countable subset of $\mathbb{R}$.

My aim is to construct a real valued function $f$ on $\mathbb{R}$ such that $f$ is discontinuous at every point from $\mathbb{L}$ and continuous at all the other points.

I don't know how to proceed. Hints will be appreciated.

Thanks in advance...

  • $\begingroup$ I don't see why you would need $\mathbb L $ to be finite. $\endgroup$ – Surb Sep 17 '14 at 10:00
  • 1
    $\begingroup$ A characteristic function won't necessarily work if $L$ is infinite, e.g. $\chi_\mathbb{Q}$. $\endgroup$ – Josh Keneda Sep 17 '14 at 10:03
  • $\begingroup$ @surb.I do not mean that $L$ is necessarily finite. I think if $L$ is finite then characteristic function on $L$ satisfies the requirement. $\endgroup$ – yazhini Sep 17 '14 at 10:07
  • $\begingroup$ Here i'm posting two problems.. First one is " how to construct a function which is discontinuous only on $$\mathbb L," where $\mathbb L$ is countably infinite. $\endgroup$ – yazhini Sep 17 '14 at 10:11
  • $\begingroup$ If $L$ is finite, you're right, you can just take a characteristic function. $\endgroup$ – Josh Keneda Sep 17 '14 at 10:14

As @Josh Keneda points out, a characteristic function won't work in general if $L$ is infinite.

But we can use the following slight modification:

$$ f(x) := \sum_{n=1}^\infty \frac{1}{n} \cdot \chi_{x_n} (x). $$

It is clear that $f$ is discontinuous at every $x_n$, because the set where $f(x) = 0$ is dense.

Below is a proof that $f$ does what you want (first try it yourself).

To see that $f$ is continuous at every $x \notin \{x_n \mid n\}$, let $\varepsilon > 0$ be arbitrary. Choose $\delta > 0$ such that $x_n \in (x-\delta, x+\delta)$ only holds for $n \geq \frac{1}{\varepsilon}$. Hence, $|f(y)| < \varepsilon$ for all $y \in (x-\delta, x+\delta)$.

EDIT: If $L$ is finite, it is clear (as above) that $f = \chi_L$ is discontinuous at every $x = x_n$, because the set where $f = 0$ is dense in $\Bbb{R}$ (it has countable/finite complement).

Conversely, $f = \chi_L$ is continuous at every $x \in \Bbb{R} \setminus L$, because we can take $\delta > 0$ with $(x - \delta, x + \delta) \cap L = \emptyset$ (take $\delta = \min\{ |x - y| \mid y \in L\}/2$), so that $|f(y) - f(x)| = 0 <\varepsilon$ holds for all $y \in (x-\delta, x+\delta)$.

  • 1
    $\begingroup$ Replace 1/n by 1/n^2 (or anything summable), otherwise the series defining f(x) may diverge. $\endgroup$ – Did Sep 17 '14 at 10:20
  • $\begingroup$ This example is similar to Riemann's example (see en.wikipedia.org/wiki/Thomae%27s_function), an example of a function with discontinuity set the rationals. $\endgroup$ – Orest Bucicovschi Sep 17 '14 at 11:13
  • 1
    $\begingroup$ @Did: It's OK, it's enough that the sequence $\frac{1}{n}$ converges to $0$. $\endgroup$ – Orest Bucicovschi Sep 17 '14 at 11:15
  • $\begingroup$ @Did: We do not even need $\frac{1}{n} \to 0$. Note that $\chi_{x_n} (x)$ is not null only for $x = x_n$. Hence, only (at most) one summand of the series does not vanish for every $x$ (here, I assume $x_n \neq x_m$ for $n\neq m$). So, we could even use $n$ instead of $1/n$, but then $f$ would not necessarily be continuous on $\Bbb{R} \setminus \{x_n \mid n\}$. $\endgroup$ – PhoemueX Sep 17 '14 at 11:21
  • $\begingroup$ Right, I mistook your chi function at $x_n$ for the indicator function of the set $(-\infty,x_n]$ (don't ask me why I went astray like that...). Sorry for the noise. $\endgroup$ – Did Sep 17 '14 at 11:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.