Arc-Connectedness in Uncountable Finite Complement Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \struct {S, \tau}$ 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 $\closedint 0 1$.

So by definition of cardinality we can set up a bijection $f: \closedint 0 1 \leftrightarrow X$ such that $\map f 0 = a$ and $\map f 1 = b$.

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 $\BB$ be a basis for $T$.

Let $B \in \BB$ 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.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Arc-Connectedness_in_Uncountable_Finite_Complement_Space&oldid=470521"