# 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.

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Just a question: What does K-Topology mean? I've never heart of it. – Stefan H. Jan 15 at 12:01
K- topology is the topology on R generated by the basis that consists of the open intervals $(a,b)$, as well as the sets $(a,b)-K$, where $K=\{\frac{1}{n}\, n\in \mathbb{N}\}$ – Akt904 Jan 23 at 5:16

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 $0$.