Prove that $f(x_n^+)-f({x_n}^-)=2^{-n}$ Let $x_n$ be an enumeration of $\Bbb Q\cap (0,1)$.

Let $\Omega_x=\{n: x_n&lt;x\}$. 

Let $$f:(0,1)\to \Bbb R\\f(x)=\sum_{n\in\Omega_x}2^{-n}$$

Prove that $\lim\limits_{x\to x_n^+} f(x)-\lim\limits_{x\to x_n^-}f(x)=2^{-n}$. I've been trying useless stuff such as writing the limits as $f(x+h), h\to 0$, trying to rewrite the sets in any useful manner, but there's no case, I couldn't solve this. 

Could anyone give me some hints to head for the right direction?
Note that $f$ is an increasing function, so the limits are well-defined.

Fix $n\in \mathbb N$.

Let  $m\in \mathbb N$ be arbitrary.

Note that  $\displaystyle f(x_n+\frac1m)-f(x_n-\frac1m)=\sum_{k\in \Omega_{x_n+\frac1m}\setminus\Omega_{x_n-\frac1m}} 2^{-k}$.

Let $A_m=\Omega_{x_n+\frac1m}\setminus\Omega_{x_n-\frac1m}=\{k, x_k\in \mathbb Q\cap [x_n-\frac1m, x_n+\frac1m(\}$.

Note that $A_m$ is a decreasing sequence, and $\bigcap_mA_m=\{n\}$

Rewriting things a little, we have  $\displaystyle \sum_{k\in \Omega_{x_n+\frac1m}\setminus\Omega_{x_n-\frac1m}} 2^{-k}=2^{-n}+\sum_{k\in A_m\setminus \{n\}}2^{-k}$.

Hence $\displaystyle f(x_n+\frac1m)-f(x_n-\frac1m)=2^{-n}+\sum_{k\in A_m\setminus \{n\}}2^{-k}$.



It remains to prove that $\displaystyle \lim_{m\to \infty}\sum_{k\in A_m\setminus \{n\}}2^{-k}=0$ and we're done.

Since the sets  $A_m\setminus\{n\}$ are decreasing, the sequence $\displaystyle s_m:=\left(\sum_{k\in A_m\setminus \{n\}}2^{-k}\right)_m$ is decreasing, but also bounded below by $0$, hence convergent.

Now, consider the sequence $a_m=\min(A_m\setminus \{n\})$. 

Since $\displaystyle \bigcap_m\left(A_m\setminus \{n\}\right)=\emptyset$ it must be that $\forall k, \exists N, \forall l\geq N, a_l\neq a_k$.

Therefore, $a_m$ is not bounded (otherwise it takes finitely many different values, and one of those values occur infinitely many times, contradicting the previous line).

Consequently, $a_m$ has a monotonic subsequence that goes to $\infty$, say $ a_{m_{p}}\to \infty$.

Finally, notice that $\displaystyle 0\leq s_{m_{p}}\leq  \sum_{k\geq a_{m_{p}} }2^{-k}$.

The sum on the right goes to $0$ as $p\to \infty$ (since it is a subsequence of the sequence $(\sum_{k\geq p} 2^{-k})_p)$.

Therefore, $s_m$ is convergent and $s_m$ has a subsequence that goes to $0$. Hence $s_m$ converges to $0$.
This function $f(x)$ snaps to a value as $x$ hits one of the values in the enumeration, holds that value until we hit the successive value of the enumeration.  So lets focus in on a particular $x_n$ and  pick $h&gt;0$ so that is smaller than the difference between the $n-1$ and $n$ terms AND $n$ and $n+1 $terms :
$$
x_{n}+h &lt; x_{n+1}    \quad    (x_n,x_n+h,x_{n+1})\\
x_{n-1}+ h &lt; x_n - h\quad      (x_{n-1}, x_{n-1} - h, ,x_n )
$$
So we can easily calculate what $f$ is for values over this range. 

Now pick $k_1,k_2$ such that  $0&lt; k_1,k_2 &lt; h$.

Then we have for any such $k_1,k_2$, the difference between the value of the functions is just the last term of the summation $\frac 1 {2^n}$.

$$f(x_n-k_1) - f(x_n+k_2) = \frac 1 {2^n}$$

hence the result .