Compact Complement Topology is Coarser than Euclidean Topology

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Theorem

Let $T = \left({\R, \tau}\right)$ 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.

$\blacksquare$


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