Closed Ordinal Space is Complete Order Space

Theorem
Let $\Gamma$ be a limit ordinal.

Let $\left[{0 \,.\,.\, \Gamma}\right]$ denote the closed ordinal space on $\Gamma$.

Then $\left[{0 \,.\,.\, \Gamma}\right]$ is a complete order space.

Proof
Let $H$ be a subset of an ordinal space.

Then $H$ has an infimum: its first element.

Let $H$ be a subset of $\left[{0 \,.\,.\, \Gamma}\right]$.

Then $H$ has a supremum.

Therefore $\left[{0 \,.\,.\, \Gamma}\right]$ is a complete order space.