Compact Complement Topology is Irreducible

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Theorem

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


Then $T$ is an irreducible 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:

\(\displaystyle \complement_\R \left({V_1 \cup V_2}\right)\) \(\ne\) \(\displaystyle \varnothing\)
\(\displaystyle \implies \ \ \) \(\displaystyle \complement_\R \left({V_1}\right) \cap \complement_\R \left({V_2}\right)\) \(\ne\) \(\displaystyle \varnothing\) De Morgan's Laws: Complement of Union
\(\displaystyle \implies \ \ \) \(\displaystyle U_1 \cap U_2\) \(\ne\) \(\displaystyle \varnothing\)

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

$\blacksquare$


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