How does this topology differ from the usual topology on $\mathbb{R}$? Let $A:=\{\frac{1}{n}\,|\,n=1,2,3,...\}$ and let $T$ be the usual topology on $\mathbb{R}$.  Now consider the set $\mathcal{B}:=T \cup \{\mathbb{R}\setminus A\}$.  This set forms a subbasis for a topology (call it $T'$) on $\mathbb{R}$.
Take an open interval $(a,b)$.  If $(a,b) \cap A = \varnothing$, then we just obtain another open interval or union of open intervals.  If $(a,b) \cap A \ne \varnothing$ and we take the intersection with $\mathbb{R}\setminus A$, then $(a,b)$ gets split into pieces at each element of $A$, so for example, $(\frac{3}{10},\frac{3}{4}) \cap \mathbb{R}\setminus A = (\frac{3}{10},\frac{1}{3}) \cup (\frac{1}{3},\frac{1}{2}) \cup (\frac{1}{2},\frac{3}{4})$.
Even taking $(0,1) \cap \mathbb{R}\setminus A$, we obtain a countable union of intervals of the form $(\frac{1}{n},\frac{1}{m})$ with $n>m$ consecutive positive integers, and so this is an open set in $T$.
So my question is: how does $T'$ differ from $T$?
 A: As Brian M. Scott pointed out, K-topology on Wikipedia is a good resource for learning about the definition and properties of this topology.
The most significant difference between the K-topology and the standard topology on $\mathbb{R}$ is that $\mathbb{R}\setminus A=\{\frac{1}{n}\,|\,n=1,2,3,...\}$ is an open set containing no member of $A$, so $A$ is closed, and unlike in the standard topology does not possess $0$ as a limit point since $\mathbb{R}\setminus A$ is an open set containing $0$ but no point of $A$.
The result is that $A$ has no limit points, and so it is not compact (since it is not even limit point compact).  The article goes on to say that no subspace of $\mathbb{R}$ with the K-topology can be compact.  This is certainly a strong distinction from $\mathbb{R}$ with the usual topology, where every closed and bounded subset is compact by the Heine-Borel Theorem.
My personal confusion resulted from focusing too much on the sets of the form $(a,b)\setminus A$, and not enough on the open set $\mathbb{R}\setminus A$, which is really what makes the difference.  The sets $(a,b)\setminus A$ are a byproduct resulting from taking intersections to ensure that the new collection of open sets is indeed a topology.
