For given the usual topology $\tau$ on $\Bbb{R}$, define the compact complement topology on $\mathbb{R}$ to be $$\tau'=\{A\subseteq \Bbb{R}:A^C\text{ is compact in }(\Bbb{R},\tau)\} \cup \{\emptyset \}.$$

Hausdorff : let $G_1$ and $G_2$ be disjoint open sets containing $x$ and $y$ for $x \ne y$ then $$G_1 \cap G_2 = \varnothing $$ $$G_1^C \cup G_2^C = \Bbb R$$ but $\Bbb R$ is not compact. So space is not Hausdorff.

Connectedness: Will the above also work for connectedness? Showing it's connected?

And are there any easy proof on compactness? I already saw this link but can't understand.


1 Answer 1


Yes, it works for connectedness. You have essentially shown that any two non-empty open subsets (so those sets with compact complement) always intersect. A space where this holds is called hyperconnected (see Wikipedia) and such spaces are connected, by the definition of connectedness.

Simple fact: as compact subsets of the standard topology are closed in that topology, all open subsets of the compact complement topology are standard open too.

Compactness is not too hard: suppose $\{U_i: i \in I\}$ is an open cover of $\mathbb{R}$ by non-empty open sets. Take any $U_{i_0}$ in this cover. Then $U_{i_0}^c$ is compact (in the usual topology) and the other sets of the cover are an (also standard) open cover of the complement so there is a finite subset $F \subseteq I\setminus \{i_0\}$ that covers $U_{i_0}^c$ and then $\{U_i: i \in F \cup \{i_0\}\}$ is a finite subcover of our original cover. Hence the space is compact.

  • $\begingroup$ Doesn't that mean just two open sets are enough for cover . If I could I could have given two upvotes one for compactness and other for its simplicity. $\endgroup$ Feb 1, 2019 at 17:49
  • $\begingroup$ @PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover. $\endgroup$ Feb 1, 2019 at 17:52
  • $\begingroup$ I Don't understand . Why cover of size three? If we take two open sets that cover X then we can take same subcover and cover ,and that will be finite . $\endgroup$ Feb 1, 2019 at 17:57
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    $\begingroup$ @PranitaGupta you ask whether two subsets are enough for all subcovers and I claim this is not the case. Sure, there are two element covers, but also irreducible three element open covers etc. My argument just needs to show finiteness which I did. $\endgroup$ Feb 1, 2019 at 17:59
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    $\begingroup$ @PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover. $\endgroup$ Feb 1, 2019 at 18:05

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