# Properties of this topology on $\Bbb X$.

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

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.

• 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. – Pranita Gupta Feb 1 at 17:49
• @PranitaGupta no we can have a finite subcover. Try to find a cover of size 3 without a smaller subcover. – Henno Brandsma Feb 1 at 17:52
• 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 . – Pranita Gupta Feb 1 at 17:57
• @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. – Henno Brandsma Feb 1 at 17:59
• @PranitaGupta it has several meanings. An irreducible cover is one where we cannot leave one out and still have a cover. – Henno Brandsma Feb 1 at 18:05