K topology: Examples

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.)

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.