Uncountable Open Ordinal Space is not Lindelöf
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Theorem
Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a Lindelöf space.
Proof
Aiming for a contradiction, suppose $\hointr 0 \Omega$ is a Lindelöf space.
From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.
From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.
From Lindelöf $T_3$ Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompact space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Paracompact, $\hointr 0 \Omega$ is not a paracompact space.
Hence the result by Proof by Contradiction.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $42$. Open Ordinal Space $[0, \Omega)$: $10$