Extremally Disconnected Metric Space is Discrete

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Theorem

Let $M = \struct {A, d}$ be a metric space which is extremally disconnected.

Then $M$ is the discrete topology.


Proof

Let $M = \struct {A, d}$ be extremally disconnected.

Let $p \in A$.

As $M$ is a metric space, $\set p$ can be expressed as:

$\set p = \displaystyle \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$

where:

$\map {B_{1 / n} } p$ denotes the open $1 / n$-ball of $p$
$\paren {\map {B_{1 / n} } p}^-$ denotes the closure of $\map {B_{1 / n} } p$

That is, as the intersection of the closures of the open $1 / n$-ball of $p$ for all non-zero natural numbers.

Now let:

$\displaystyle U = \bigcup_{n \mathop \in \N_{>0} } \map {B_{1 / 2 n} } p \setminus \paren {\map {B_{1 / \paren {2 n + 1} } } p}^-$

Then either $U$ or the complementary set of annuli is an open set which has $p$ as a non-interior limit point provided $\set p$ is not open.



So if $M$ is not the discrete topology, it cannot be extremally disconnected.

$\blacksquare$


Sources