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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 with the product of its ordinal order topologies, let , and let . Then is compact, Hausdorff, and normal, while is an open regular subspace that is not normal.
Facts & Assumptions
Given: The Axiom of Countable Choice and the ordinal product above.
The Axiom of Countable Choice, under which every at most countable subset of is bounded below (The Axiom of Countable Choice (), Assuming countable choice: every at most countable subset of is bounded below , so no at most countable subset of is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable).
Ordinal spaces have clopen bases and are ; every successor ordinal is compact (Every ordinal with its order topology has a basis of clopen sets, and is , Hausdorff and regular, Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, is countably compact and sequentially compact while is compact, Ordinal addition ).
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 and , , , , regularity, , complete regularity, and Tychonoffness are productive, , , , regularity, , complete regularity, and Tychonoffness are hereditary).
Normality separates disjoint closed subsets by disjoint open sets (Normal spaces and spaces, with the source disagreement over whether normality includes stated explicitly).
The ordinal order-topology basis gives neighbourhoods of , singleton neighbourhoods of , and neighbourhoods of (The order topology on an ordinal, with the half-open intervals and the initial segments as a basis, Every ordinal with its order topology has a basis of clopen sets, and is , Hausdorff and regular).
Proof
The factors and are compact and by [L1], so [L2] makes compact, Hausdorff, regular, and normal.
Since is , is closed; hence is open. Its regularity follows from the hereditary regularity conclusion in [L2].
Put and , regarded as subsets of . The clopen ordinal basis shows that they are disjoint closed subsets of .
Suppose, for a contradiction, that is normal. Choose disjoint open with and .
For , let . By [F2] each is nonempty; [A1] chooses simultaneously. The countable set is bounded by some .
Put . Since , [F2] gives and with .
The point lies in by step 3.2 and in by step 3.1, because . This contradicts , so is not normal; together with steps 1.1 and 1.2 this proves all the stated properties.
Depends on
- The order topology on an ordinal, with the half-open intervals $(\alpha, \beta]$ and the initial segments $[0, \beta]$ as a basis
- Every ordinal with its order topology has a basis of clopen sets, and is $T_1$, Hausdorff and regular
- $T_0$, $T_1$, $T_2$, regularity, $T_3$, complete regularity, and Tychonoffness are hereditary
- $T_0$, $T_1$, $T_2$, regularity, $T_3$, complete regularity, and Tychonoffness are productive
- 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 $T_4$
- Assuming countable choice: every at most countable subset of $\omega_1$ is bounded below $\omega_1$, so no at most countable subset of $\omega_1$ is cofinal in it, and a supremum of at most countably many at most countable ordinals is at most countable
- Every successor ordinal is compact in its order topology and every limit ordinal is not; and, assuming countable choice, $\omega_1$ is countably compact and sequentially compact while $\omega_1 + 1$ is compact
- Ordinal addition $\alpha + \beta$
- Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
Used by
- Assuming countable choice, normality is not open-hereditary: deleting one corner of the Tychonoff plank Counterexample
- Assuming countable choice, the deleted Tychonoff plank worked as T₃ but not normal inside its compact Hausdorff normal parent Example
- Assuming countable choice, refuted: every regular space is normal False statement
- Assuming countable choice, normality is not hereditary, even to open regular subspaces Theorem
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 144 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
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)