analysis - A Real valued function which is discontinuous **only** on a given specific set.

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...

Answer

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 $y \in (x-\delta, x+\delta)$.