Uncountable Closed Ordinal Space is not Separable
Jump to navigation
Jump to search
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 separable space.
Proof
Let $H \subseteq \closedint 0 \Omega$ be a countable subset of $\closedint 0 \Omega$.
From Uncountable Open Ordinal Space is not Separable, there exists an open interval $\openint \sigma \Omega$ in the complement of $H^-$ in $\hointr 0 \Omega$, and so also in $\closedint 0 \Omega$.
Thus the closure of $H$ in $\closedint 0 \Omega$ does not equal $\closedint 0 \Omega$.
Thus $H$ is not everywhere dense in $\closedint 0 \Omega$.
Hence, by definition, $\closedint 0 \Omega$ is not separable.
$\blacksquare$
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 $[0, \Omega]$: $3$