Indiscrete Space is Arc-Connected iff Uncountable

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.


Then $T$ is arc-connected if and only if $S$ is an uncountable set.


Proof

Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space such that $S$ is uncountable.

Let $a, b \in S$.


Consider an injection $f: \left[{0 \,.\,.\, 1}\right] \to S$ such that $f \left({0}\right) = a$ and $f \left({1}\right) = b$.

This can always be found because $S$ is itself uncountable.


From Mapping to Indiscrete Space is Continuous, we have that $f$ is ‎continuous.

Thus $T$ is arc-connected.


Now suppose $S$ is an indiscrete topological space which is arc-connected.

Then there exists an injection $f: \left[{0 \,.\,.\, 1}\right] \to S$ such that $f \left({0}\right) = a$ and $f \left({1}\right) = b$.


This can only exist if $S$ is uncountable.

$\blacksquare$


Sources