Countable Finite Complement Space is Not Path-Connected
Theorem
Let $T = \left({S, \tau}\right)$ be a finite complement topology on a countable set $S$.
Then $T$ is not path-connected.
Proof
Let $f: \left[{0 .. 1}\right] \to S$ be a path on $T$.
Then $f$ is by definition continuous.
From the finite complement topology, it follows that, for all $x \in S$, the set $\left\{{x}\right\}$ is closed.
Now consider the set:
- $F = \left\{{f^{-1} \left({x}\right): x \in S}\right\}$
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.
Furthermore, we have $\displaystyle \bigcup F = \left[{0 .. 1}\right]$.
Hence $F$ is a countable set of pairwise disjoint closed sets whose union is $\left[{0 .. 1}\right]$.
Major hole in proof: $F$ needn't be countably infinite, which is necessary. Compare when $f$ is constant. It surely will be a path
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From Unit Interval Not Countably Infinite Union of Disjoint Closed Sets, this is impossible.
From this contradiction, $f$ can not be continuous, and so cannot be a path on $T$.
So $T$ cannot be path-connected.
$\blacksquare$
Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{II}: \ 18: \ 10$
