Uncountable Closed Ordinal Space is not First-Countable

Theorem

Let $\Omega$ denote the first uncountable ordinal.

Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.

Then $\closedint 0 \Omega$ is not a first-countable space.

Proof

From Omega is Closed in Uncountable Closed Ordinal Space but not G-Delta Set, $\set \Omega$ cannot be expressed as a countable intersection of open sets of $\closedint 0 \Omega$.

Thus, by definition, $\Omega$ does not have a countable local basis.

Hence the result by definition of first-countable space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $43$. Closed Ordinal Space $[0, \Omega]$: $2$