K-topology satisfies the Hausdorff axiom?

Let R be the set of all real numbers and let K={1/n, n is a natural number}. Generate a topology on R by taking as basis all open intervals (a,b) and all sets of the form (a,b)-K (the set of all elements in (a,b) that are not in K). The topology generated is known as the K-topology on R.

K-topology satisfies the Hausdorff axiom. I don't know how to prove it at all. This is my HW. Please help me.

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Would you please give the definition of "K-topology" so that we know what you're talking about? Most of us here are not sitting in your course, so even if we might be able to answer your question, we can't because we don't know what exactly you're asking. – t.b. Oct 23 '11 at 14:54
Hint: The K-topology is, by definition, finer than the standard topology on $\mathbb R$. – Rasmus Oct 23 '11 at 15:00
Let R be the set of all real numbers and let K={1/n, n is a natural number}. Generate a topology on R by taking as basis all open intervals (a,b) and all sets of the form (a,b)-K (the set of all elements in (a,b) that are not in K). The topology generated is known as the K-topology on R. – Jihyun Kim Oct 23 '11 at 15:24
@Jihyun Kim: You should edit that into your question, so it's more visible. – Chris Eagle Oct 23 '11 at 15:31

Hint: We want to show that if $a$ and $b$ are distinct points of $\mathbb{R}$, they can be "separated" by disjoint sets that are open in the $K$-topology. Now use the fact that every set which is open in the ordinary topology on $\mathbb{R}$ is open in the $K$-topology.