Compact Complement Topology is Hyperconnected

Theorem

Let $T = \left({\R, \tau}\right)$ be the compact complement topology on $\R$.

Then $T$ is a hyperconnected space.

Proof

Let $U_1, U_2 \in \tau$ be open in $T$.

Let $\complement_\R \left({U_1}\right) = V_1$ and $\complement_\R \left({U_2}\right) = V_2$.

By definition of compact complement topology, $V_1, V_2 \subseteq \R$ such that $V_1$ and $V_2$ are both compact.

$V_1$ and $V_2$ are both bounded by definition of compact in $\R$, so their union is likewise bounded: above by $\max \left\{{\sup V_1, \sup V_2}\right\}$ and below by $\min \left\{{\inf V_1, \inf V_2}\right\}$.

So $V_1 \cup V_2$ can not equal $\R$ as $\R$ is not bounded.

So:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \complement_\R \left({V_1 \cup V_2}\right)\)	\(\ne\)	\(\displaystyle \varnothing\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \complement_\R \left({V_1}\right) \cap \complement_\R \left({V_2}\right)\)	\(\ne\)	\(\displaystyle \varnothing\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	De Morgan's laws
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle U_1 \cap U_2\)	\(\ne\)	\(\displaystyle \varnothing\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So any two open sets in $T$ are not disjoint and so $T$ is hyperconnected by definition.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{II}: \ 22: \ 3$