Q:

On the $K$-topology on the real line

Let $X = \mathbb{R}$ and $K = \{ 1/n : n \in \mathbb{N} \}$. If we call $\mathscr{B}$ the collection of all open intervals $(a,b)$ and the sets of the form $(a,b) - K$, then this generates a topology $\mathscr{T}_K$ called the $K$-topology on $\mathbb{R}$.
We want to prove that standard topology on real line is strictly smaller than $K$-topology.
Attempt:
Let $\mathscr{T}$ be the standard topology on line. We show $\mathscr{T} \subsetneq \mathscr{T}_K$
The inclusion is trivial as $\mathscr{T}_K$ contains all opens sets of standard topology. We still need to find some open set in $\mathscr{T}_K$ that is not open in standard topology.
Im trying to understand the k-open sets: Well, if $(a,b)$ does not contain $0$, then this would just be the standard open set. However, if $0 \in (a,b)$, then
$$ (a,b) - K = \{ x \in \mathbb{R} : a<x<b \; \; and \; \; x \neq 1/n \}$$
Now we may write this as $(a,0) \cup \bigcup_{i=1}^{N} \left( \dfrac{1}{i+1}, \dfrac{1}{i} \right) \cup (1/N, b)$
Is this how we can write open sets?
Consider $(0,1)-K = \{0 \} \bigcup (1/[i+1], 1/i)$
So, this is not open in standard topology since $\{ 0 \}$ is not open. Is this correct?

A:

The $K$-topology is the topology where we add to the closed sets of the Euclidean real line a new closed set $K$ (new as it is not Euclidean closed as $0$ is a limit point of $K$, that is not in $K$). So in particular $K^\complement = \Bbb R \setminus K$ is not Euclidian open but is open in the $K$-topology, showing the proper inclusion. 
It's of course not a coincidence that $0 \notin K$ and this new closed set $K$ will precisely be the counterexample to regularity....

