# 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

- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*(2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces