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I'm trying to answer a question regarding the set of discontinuities for a function, but I want to make sure that I understand the setup correctly. Here it is:

Let {$x_n$} be a sequence of real numbers defined by: $$ x_n = \begin{cases} \frac{1}{2k}, & \text{if $n=2k-1$} \\[2ex] \frac{1}{2k-1}, & \text{if $n=2k$} \end{cases} k\ge1 \text{ and } n\ge1 $$

Set $\epsilon_n$ $=$ $(\frac{1}{2})^n$ and define $f:\mathbb R\to \mathbb R$ by

$$ f(x)= \sum_{n: \text{ }x_n \le x}\epsilon_n $$

  1. Compute $f(0), \text{ }f(-1), \text{ }f(1), \text{ }f(\sqrt2),\text{ }f(\frac{1}{2})$

  2. Determine the set of discontinuities for the function and justify your claim.

To answer #1 I began by creating a table for the first six values of the sequence:

$$ \begin{array}{c|lcr} n & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6}\\ \hline x_n & \frac{1}{2} & 1 & \frac{1}{4} & \frac{1}{3} & \frac{1}{6} & \frac{1}{5}\\ \epsilon_n &\frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \frac{1}{64} \\ \end{array} $$

If I am approaching this correctly, $f(0)=0$ since there are no $x_n \le 0$ (although the sequence appears to converge to $0$). Likewise: $f(-1)=0$. However, things become unclear with $x=\sqrt2$ since all of the $x_n$ appear to be less than $\sqrt2$. Is there something amiss in this approach?

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    Your approach is fine. As you correctly state, all $x_n$ are $\le \sqrt{2}$, so the sum in the definition of $f(x)$ will be taken over all $n\ge 1$. Can you calculate that sum? – 2017-02-20
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    @Ingix: Let me know if you agree: f(0)=0, f(-1)=0, f(1)=1, f(sqrt(2))=1. However, I am unclear about the case where x= 1/2; here we exclude the value 1/4 from the summation since 1>1/2, however that series still presumably converges to 1. – 2017-02-25

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