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I am having troubles understanding the difference between Standard Topology generated by $(a,b)=\{x|a<x<b\}$ and K-Topology which is the union of all open intervals along with all sets of the form $(a,b) - K$ where $K=\{1/n|n \in \mathbb{N}\}$. A similar question was asked; however, I am not sure if I understand it correctly. Isn't the set $(a,b) \cup ((a,b)-K)$ just $(a,b)$? Which makes me think Standard Topology and K-Topology are the same but I know that is not the case. So I am applying the definitions wrong and I would appreciate clarification on this.

Rob
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  • Along with does not mean union with. The set $(-1,1)\setminus K$ is by definition open in the $K$-topology, but it is not open in the usual topology, because it contains no open interval centred at $0$. – Brian M. Scott Sep 26 '20 at 20:16
  • So let's say if $K$ ends at 2 then we are interested in sets $(-1,1)$ and $(-1,1)-{1/2}$ but not their union. Correct? – Rob Sep 26 '20 at 20:20
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    No, that’s not right. $K$ is the set of all fractions $\frac1n$ for $n\in\Bbb Z^+$. The definition of the $K$-topology says that $(-1,1)$ and $(-1,1)\setminus K$ are both basic open sets. – Brian M. Scott Sep 26 '20 at 20:22
  • I meant to say assume $K$ ends at $n=2$ so $K={1,1/2}$. I think I am getting your point though. For me it's easier to build intuition with finite sets first before moving on to infinite sets. – Rob Sep 26 '20 at 20:25
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    $K$ doesn’t end at any $n$: $K$ is the set of all fractions $\frac1n$. Thus, $$(-1,1)\setminus K=(-1,0]\cup\left(\frac12,1\right)\cup\left(\frac13,\frac12\right)\cup\left(\frac14,\frac13\right)\cup\ldots;.$$ – Brian M. Scott Sep 26 '20 at 20:27
  • I understand that part no problem. I just wanted to clarify that it is not their unions, that is $(a,b) \cup (a,b) - K$ is not what the definition is saying but $(a,b)$ and $(a,b)-K$ – Rob Sep 26 '20 at 20:30
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    Yes, this base for the $K$-topology contains each ordinary open interval $(a,b)$ and each set of the form $(a,b)\setminus K$. – Brian M. Scott Sep 26 '20 at 20:32

1 Answers1

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In the standard topology the set $K=\{\frac1n\mid n =1,2,\ldots\}$ is not closed. This is because $0 \in \overline{K}$ but $0 \notin K$.

The space $\Bbb R_K$ has the smallest topology that has all the open sets of the standard topology but in which $K$ is closed. This means that $K^\complement$ (the complement of $K$ in the reals) is also open (besides the standard open sets), so also any intersection of an open interval $(a,b)$ with $K^\complement$ is open and this is exactly $(a,b) - K$. So we have to add these as open sets too. But that is all we need to do, and we get $\Bbb R_K$ as described by Munkres. All points still have the same neighbourhood base as they had (in the standard topology), only $0$ gets a different base (so that it is no longer in the closure of $K$), namely all sets of the form $(-\varepsilon, \varepsilon) - K$, which all miss $K$, ensuring that $0 \notin \overline{K}$ in the new topology.

The effect of this addition of a single closed set is that we cannot separate $0$ from the new closed set $K$ by disjoint open sets, so $\Bbb R_K$ is not regular (the raison d'être of this example, probably: to be a simple pedagogical example in a text book to show that not all Hausdorff spaces need to be regular).

We also destroy local compactness ($\Bbb R$ is locally compact but $\Bbb R_K$ is not), and (local) path-connectedness as well, so it's a useful example in a text book: it's easily explained but can serve as illustrative example. It's also used in the book as an easy way to construct a quotient map whose square is not quotient etc. It's called Smirnov's deleted sequence topology here as that is its name in Counterexamples in Topology.

Steven Clontz
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Henno Brandsma
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