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 $T$?

  • $\begingroup$ This is sometimes called the $K$-topology on $\Bbb R$; the linked article should give you a good start on answering your own question. $\endgroup$ – Brian M. Scott Jan 26 '13 at 0:14
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    $\begingroup$ It's quite different actually. Check a couple of the basic concepts at $0$. $\endgroup$ – gnometorule Jan 26 '13 at 0:49
  • $\begingroup$ $\frac{1}{n}$ in the usual topology converges to $0$, but in the new topology, $\mathbb{R}\setminus A$ is a neighbourhood of $0$ that misses the sequence, so this is no longer true in the new topology. $\endgroup$ – Henno Brandsma Jan 26 '13 at 7:53

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.


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