Why is the $K$-topolgy finer than the standard? I've already see another questions about this but i still dont get it.
Take $x= 1/2$ then the interval $A= (0,1)$ contains $x$ and $A$ is a element of the basis for the standard topology, but how can i get a interval on $K$-topology such that $A$ is on it? if we take any interval that contains (0,1) in this topology $x$ will never be there.
 A: You seem to be confused about the definition of the $K$-topology on the reals.
By definition all open subsets of $\mathbb{R}$ are open in the $K$-topology. We only add one new open set $K^c = \mathbb{R} \setminus K$, where $K = \{\frac{1}{n}: n \in \mathbb{N}\}$ .As $K$ was not closed in the usual topology ($0 \in \overline{K}\setminus K$), this is a new open set, so the $K$-topology is strictly finer than the usual topology. If $\frac{1}{n} \in K$, then the only type of open set it is contained in are the usual open sets of the reals, so $(0,1)$ will do fine as a neighbourhood of any $\frac{1}{n}$. 
As $\mathcal{T} \cup \{K^c\}$ is a subbase for the $K$-topology, where $\mathcal{T}$ is the usual topology of $\mathbb{R}$ (generated by the open intervals, or metric balls, whatever suits you), a base for the open sets are all open sets $O$ of the reals, and all sets of the form $O \cap K^c = O \setminus K$, where $O$ is again open in the reals. This is already closed under unions, so it actually describes the whole topology, so $$\mathcal{T}_K = \mathcal{T} \cup \{O \setminus K: O \in \mathcal{T}\}$$
In terms of local bases, the only point which has different neighbourhoods from the ones in the reals is $0$. It's the raison d'être of this topology. It allows us to see the $K$-topology is not regular, as $0 \notin K$ and $K$ is closed, yet any open neighbourhood of $K$ will intersect any open neighbourhood of $0$.
