Countable Finite Complement Space is not Path-Connected

Theorem

Let $T = \struct {S, \tau}$ be a countable finite complement topology.

Then $T$ is not path-connected.

Proof

Aiming for a contradiction, suppose $f: \closedint 0 1 \to S$ is a path on $T$.

Then $f$ is by definition continuous.

By definition of the finite complement topology, for all $x \in S$, the set $\set x$ is closed.

Now consider the set:

$F = \set {\map {f^{-1} } x: x \in S}$

From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.

Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.

As $T$ is a countable finite complement topology, its underlying set $S$ is countably infinite by definition.

Hence $F$ is also countably infinite by nature of $f$ being a mapping.

Furthermore, we have $\ds \bigcup F = \closedint 0 1$.

Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\closedint 0 1$.

From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.

From this contradiction, $f$ cannot be continuous, and so cannot be a path on $T$.

So $T$ cannot be path-connected.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $18$. Finite Complement Topology on a Countable Space: $10$