Compact Complement Topology is Coarser than Euclidean Topology

Theorem

Let $T = \struct {\R, \tau}$ be the compact complement topology on $\R$.

Then $\tau$ is coarser than the usual (Euclidean) topology in $\R$.

Proof

Let $U \in \tau$.

Then $V = \R \setminus U$ is compact.

So by definition $V$ is closed in the Euclidean topology.

That is, $U = \R \setminus V$ is open in the Euclidean topology.

That is, every open set in the compact complement topology is also open in the Euclidean topology.

Hence the result, by definition of coarser topology.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $22$. Compact Complement Topology: $6$