Proving a distribution function is continuous at irrational points Let $\{x_1, x_2, \ldots\}$ be a collection of rational points from the interval $[0, 1]$. A random variable $X$ takes on $x_n$ with probability $1/2^{n}$. Prove the distribution function $F_{X}(x)$ of $X$ is continuous at every irrational point $x$.

I am pretty sure this type of question will involve a density argument by using the fact that the irrationals are dense in $\mathbb{R}$ (I'm not completely sure -- this is just my guess from other problems).
I'm really not too sure how to proceed though because they also gave us a collection of rational points as well. I don't really see how to combine both of these facts to show continuity. 
Also, some other problems I've solved use inequalities like Markov's and Chebyshev's, but I can't really do that here since there's no expectation. 
I would greatly appreciate your help. I am trying to get better at these sort of problems.
 A: The argument here is actually analysis, not probability at all. 
$$F_X(x):= P(X \leq x)=\sum_{n \geq 1}{2^{-n}1(x \geq x_n)}.$$
The series is normally convergent, and, if $x$ is irrational, each term is continuous at $x$. So the sum of the series is also continuous at $x$. 
A: $F_X$ is an increasing right-continuous function. So it is continuous at a point $x$ iff $F_X(x)=F_X(x-)$  where $F_X(x-)$ is the left hand limit at $x$. Also, $P(X=x)=\lim P(X-\frac 1 n < X \leq x)=\lim [F_X(x)-F_X(x-\frac 1 n)]=F_X(x)=F_X(x-)$. Hence $F_X$ is continuous at $x$ iff $P(X=x)=0$. For $x$ irrational $P(X=x)=0$ so $F_X$ is continuous at $x$. 
A: Let $A=\{r_{n}\mid n\in\mathbb{N}\}$ be a subset of $\mathbb{Q}\cap[0,1]$.
Let $F:\mathbb{R}\rightarrow[0,1]$ be the cumulative distribution
function of $X$. Note that $F(x)$ is explicitly given by 
$$
F(x)=\sum\left\{ \frac{1}{2^{n}}\mid r_{n}\leq x\right\} .
$$
Let $x_{0}\in[0,1]\setminus\mathbb{Q}$. Let $\varepsilon>0$ be arbitrary.
Choose $N\in\mathbb{N}$ such that 
$$
\sum_{n=N+1}^{\infty}\frac{1}{2^{n}}<\varepsilon.
$$
Let $\delta=\min\{|r_{n}-x_{0}|\mid1\leq n\leq N\}>0$, then $(x_{0}-\delta,x_{0}+\delta)\cap\{r_{n}\mid1\leq n\leq N\}=\emptyset$.
For any $x_{1},x_{2}\in(x_{0}-\delta,x_{0}+\delta)$ with $x_{1}<x_{2}$,
we have that 
\begin{eqnarray*}
0 & \leq & F(x_{2})-F(x_{1})\\
 & = & \sum\{\frac{1}{2^{n}}\mid r_{n}\in(x_{1},x_{2}]\}\\
 & \leq & \sum\{\frac{1}{2^{n}}\mid r_{n}\in(x_{0}-\delta,x_{0}+\delta)\}\\
 & \leq & \sum_{n=N+1}^{\infty}\frac{1}{2^{n}}\\
 & < & \varepsilon.
\end{eqnarray*}
It follows that $F$ is continuous at $x_{0}$. 
A: Pick any irrational point and look at its neighborhood containing none of the $x_i$ (why must this exist). $F_X$ is constant in this neighborhood and hence is continuous
Edit: This also means $F_X$ is continuous at every point except the $x_i$
