Are there sets in the K-Topology that aren't open in the standard topology? It seems to me that the basis for the K- Topology and the basis for the standard topology generate the same open sets.  For instance, the open sets in the K-topology's basis that are different from the standard topology's basis are the sets which are of the form $(a,b)\backslash K $.  This seems to be just the union of a countable number of disjoint intervals (which we can generate from the basis elements of the standard topology).  To be more explicit, suppose we have a set in the K topology $(0,1)\backslash K$.  We can generate this open set with  $\cup_{n\in\mathbb{Z+}} (\frac{1}{n+1},\frac{1}{n})$, where each of the open sets $(\frac{1}{n+1},\frac{1}{n})$ is in the basis for the standard topology.  Also, since the basis elements of the standard topology are a subset of the K-topology's basis elements, any open set the standard topology generates can also be generated by the K-topology.  I don't see how one is strictly finer than the other but according to Munkres, the K-topology is strictly finer than the standard topology.

Let $U=\Bbb R\setminus K$. By definition $U$ is open in the $K$-topology on $\Bbb R$, but it is not open in the usual topology: $0\in U$, but $U$ does not contain any open interval around $0$.
