Arc-Connectedness in Uncountable Finite Complement Space

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

If the Continuum Hypothesis is accepted as true, then:
 * $T$ is arc-connected


 * $T$ is locally arc-connected.

Proof
Let $a, b \in S$ such that $a \ne b$.

Let us assume the truth of the Continuum Hypothesis.

Then $a$ and $b$ are contained in a subset $X \subseteq S$ whose cardinality is the same as that of $\left[{0 \,.\,.\, 1}\right]$.

So by definition of cardinality we can set up a bijection $f: \left[{0 \,.\,.\, 1}\right] \leftrightarrow X$ such that $f \left({0}\right) = a$ and $f \left({1}\right) = b$.

Each element of $S$ is closed from Finite Complement Space is $T_1$.

Hence $f$ is continuous and therefore an arc in $T$.

As $a$ and $b$ are arbitrary, $T$ is arc-connected.

Now let $\mathcal B$ be a basis for $T$.

Let $B \in \mathcal B$ and let $a, b \in B$.

By definition of a finite complement topology on a uncountable set, $B$ is uncountable.

So, by the same argument as above, we can set up an arc $f$ between any points of $B$.

So $T$ is locally arc-connected.