I am doing question 4.6.2. in Understading Analysis. The question goes as follows:
Given a countable set $A = \{a_1, a_2, a_3, .... \}$, define $f(a_n) = 1/n$ and $f(x) = 0$ for all $x\not\in A$ Find $D_f$. Where $D_f$ is the set of points where the function $f$ fails to be continuous.
Proving that every point of A is not continuous is not a problem. My problem is showing that that $x \in A^c$ is either continuous or not. I think my intuition is saying that they should be.
Could anyone point me in some direction on how to prove this?
Here is my suggestion to a proof:
We have the set $A = \{a_1, a_2, ...\}$ and the function $f(a_n) = 1/n$ and $f(x) = 0$ $\forall x \not\in A$. We are going to find $D_f$. We know that A is countable, that tells us that A does not contain any form of intervals from \mathbb{R}. We thus know that $A \subseteq D_f$. Pick any $a_n \in A$ sett $\epsilon < \frac{1}{n} \cdot \frac{1}{2}$ you will never be able to find a $\delta$ s.t. $|x-c| < \delta$ then it follows that $|f(x) - f(c)| < \epsilon$. We must also check whether also the set/subsets of $\mathbb{R}/A$ also belongs $D_f$. Lets check if $x \in \mathbb{R} /A$ is continuous. Let $\epsilon > 0$. Go in each direction of $x$ and search for the first $a_{n_0}$ s.t. $1/n_0 \leq \epsilon$. Set $\delta < |a_{n_0}-x|$. By constructing $\delta$ this way we have found a $\delta > 0$ such that $|f(x)-f(y)| < \epsilon$ for $y \in V_{\delta}(x)$, and we conclude that $D_f = A$.