Countable Finite Complement Space is not Locally Path-Connected

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ be a countable finite complement topology.


Then $T$ is not locally path-connected.


Proof

Let $\mathcal B$ be a basis for $T$.

Let $B \in \mathcal B$.


Aiming for a contradiction, suppose $f: \left[{0 \,.\,.\, 1}\right] \to B$ is a path on $T$.

Then $f$ is by definition continuous.


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.

From Basis of Countable Finite Complement Topology consists of Countably Infinite Sets, $B$ is a countably infinite set.

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

Furthermore, we have $\displaystyle \bigcup F = \left[{0 \,.\,.\, 1}\right]$.

Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\left[{0 \,.\,.\, 1}\right]$.

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 $B$.

So $B$ cannot be path-connected.

So, by definition, $T$ is not locally path-connected.

$\blacksquare$


Sources