Closed Ordinal Space is Complete Order Space

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\Gamma$ be a limit ordinal.

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


Then $\closedint 0 \Gamma$ 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 $\closedint 0 \Gamma$.

Then $H$ has a supremum.

Therefore $\closedint 0 \Gamma$ is a complete order space.

$\blacksquare$


Sources