K topology: Examples Why would the interval $(-3,1)$ be open in the $k$-topology? (I'm using Munkres).
Can I have some other examples of intervals in $k$-topology?
What exactly does $(a,b)$ $\cup$ $(a,b)-k$ for $k=\left\{\frac{1}{n}, n\geq1\right\}$ mean?
 A: The K-topology is defined to be the topology on $\mathbb{R}$ generated by the following base:
$$ \mathcal{B}_K = \{ ( a , b ) : a < b \} \cup \{ ( a , b ) \setminus K : a < b \}$$ where $K = \{ \frac 1n : n \geq 1 \}$.  As such, the K-topology is finer than the usual topology, which means that every open set in the usual topology on $\mathbb{R}$ is open in the K-topology.  (Every open set in the usual topology is a union of sets/intervals from the first collection in the union above.)
It follows that all open intervals are open in the K-topology.
Furthermore, one can show that the open sets in this topology are exactly the sets which can be expressed as $U \setminus L$, where $U$ is open in the usual topology on $\mathbb{R}$ and $L \subseteq K$.
In particular the set $\mathbb{R} \setminus K$ is open, meaning that $K$ is itself closed. Note that $K$ is not closed in the usual topology (since $0$ is a limit point).  (You will probably later see that $\mathbb{R}_K$ is not regular, while the usual real line is.)
(I'm really uncertain where you are seeing "$(a,b) \cup ( a,b ) - k$" in Munkres, and cannot really discern what you mean by your last question.)
A: To answer your first question, the $K$-topology on $\mathbb{R}$ takes the basis for the standard topology and appends extra set to get a finer topology. Recall that the basis for the standard topology on $\mathbb{R}$ is the intervals of the form $(a, b)$ for $a, b$ rational. (You can let $a, b$ be real instead but it's nice to have countable bases, and taking $a,b$ rational generates the whole standard topology.) 
So, to answer your second question, every open interval in the standard topology is open in the $K$-topology.
For your third question, note that you're unioning a set $A$ with a set that $A$ contains, so you get the original set $A$ back.
