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Assuming countable choice, the deleted Tychonoff plank is a regular nonnormal open subspace of a compact Hausdorff normal space

Statement

Assume the Axiom of Countable Choice. Let $P = (ω_{1} + 1) \times (ω + 1)$ with the product of its ordinal order topologies, let $p = (ω_{1}, ω)$ , and let $T = P ∖ {p}$ . Then $P$ is compact, Hausdorff, and normal, while $T$ is an open regular subspace that is not normal.

Facts & Assumptions

Given: The Axiom of Countable Choice and the ordinal product $P$ above.

[A1]

The Axiom of Countable Choice, under which every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Assuming countable choice: every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ , so no at most countable subset of $ω_{1}$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).

[L1]

Ordinal spaces have clopen bases and are $T_{3}$ ; every successor ordinal is compact (Every ordinal with its order topology has a basis of clopen sets, and is $T_{1}$ , Hausdorff and regular, Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, $ω_{1}$ is countably compact and sequentially compact while $ω_{1} + 1$ is compact, Ordinal addition $α + β$ ).

[L2]

Finite products of compact spaces are compact, compact Hausdorff spaces are normal, and the positive preservation theorems preserve regularity, Hausdorffness, and regularity under subspaces (A product of finitely many compact spaces is compact in the product topology, A compact Hausdorff space is regular and normal, hence $T_{3}$ and $T_{4}$ , $T_{0}$ , $T_{1}$ , $T_{2}$ , regularity, $T_{3}$ , complete regularity, and Tychonoffness are productive, $T_{0}$ , $T_{1}$ , $T_{2}$ , regularity, $T_{3}$ , complete regularity, and Tychonoffness are hereditary).

[F1]

Normality separates disjoint closed subsets by disjoint open sets (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

[F2]

The ordinal order-topology basis gives neighbourhoods $(α, ω_{1}]$ of $ω_{1}$ , singleton neighbourhoods ${n}$ of $n < ω$ , and neighbourhoods $(m, ω]$ of $ω$ (The order topology on an ordinal, with the half-open intervals $(α, β]$ and the initial segments $[0, β]$ as a basis, Every ordinal with its order topology has a basis of clopen sets, and is $T_{1}$ , Hausdorff and regular).

Proof

technique · contradiction

1.1

The factors $ω_{1} + 1$ and $ω + 1$ are compact and $T_{3}$ by [L1], so [L2] makes $P$ compact, Hausdorff, regular, and normal.

L1L2

1.2

Since $P$ is $T_{1}$ , ${p}$ is closed; hence $T$ is open. Its regularity follows from the hereditary regularity conclusion in [L2].

1.3

Put $E = {ω_{1}} \times ω$ and $F = ω_{1} \times {ω}$ , regarded as subsets of $T$ . The clopen ordinal basis shows that they are disjoint closed subsets of $T$ .

2.1

Suppose, for a contradiction, that $T$ is normal. Choose disjoint open $U, V \subseteq T$ with $E \subseteq U$ and $F \subseteq V$ .

F1step 1.3assume-contra

3.1

For $n < ω$ , let $C_{n} = {ξ < ω_{1} : (ξ, ω_{1}] \times {n} \subseteq U}$ . By [F2] each $C_{n}$ is nonempty; [A1] chooses $α_{n} \in C_{n}$ simultaneously. The countable set ${α_{n} : n < ω}$ is bounded by some $α < ω_{1}$ .

A1F2step 2.1

3.2

Put $β = α + 1 < ω_{1}$ . Since $(β, ω) \in V$ , [F2] gives $γ < β$ and $m < ω$ with $(γ, β] \times (m, ω] \subseteq V$ .

F2step 2.1

4.1

The point $(β, m + 1)$ lies in $V$ by step 3.2 and in $U$ by step 3.1, because $β > α \geq α_{m + 1}$ . This contradicts $U \cap V = \emptyset$ , so $T$ is not normal; together with steps 1.1 and 1.2 this proves all the stated properties.

step 1.1step 1.2step 3.1step 3.2discharge-contradiction∎

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Sources

L. A. Steen and J. A. Seebach, Counterexamples in Topology, deleted Tychonoff plank (standard reference, not scraped)
Tychonoff plank (Wikipedia) (standard reference, not scraped)