Compact Complement Topology is Irreducible

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Theorem

Let $T = \struct {\R, \tau}$ 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 $\relcomp \R {U_1} = V_1$ and $\relcomp \R {U_2} = V_2$.

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 $\max \set {\sup V_1, \sup V_2}$ and below by $\min \set {\inf V_1, \inf V_2}$.

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

So:

\(\ds \relcomp \R {V_1 \cup V_2}\) \(\ne\) \(\ds \O\)
\(\ds \leadsto \ \ \) \(\ds \relcomp \R {V_1} \cap \relcomp \R {V_2}\) \(\ne\) \(\ds \O\) De Morgan's Laws: Complement of Union
\(\ds \leadsto \ \ \) \(\ds U_1 \cap U_2\) \(\ne\) \(\ds \O\)

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

$\blacksquare$


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