ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$ω + 1$ as a convergent sequence together with its limit, and, assuming countable choice, $[0, ω_{1})$ , in which every sequence lies inside an at most countable initial segment

Example

Give every ordinal its order topology (The order topology on an ordinal, with the half-open intervals $(α, β]$ and the initial segments $[0, β]$ as a basis), under which it is $T_{3}$ — that is $T_{1}$ , Hausdorff and regular (Every ordinal with its order topology has a basis of clopen sets, and is $T_{1}$ , Hausdorff and regular). Two ordinals are worked here.

The space $ω + 1$ . By the successor clause of ordinal addition (Ordinal addition $α + β$ ), $ω + 1 = ω^{+} = ω \cup {ω}$ , so the space is the set of natural numbers together with one extra point on top ( $ω$ is the least limit ordinal). Then:

Every $n \in ω$ is isolated: ${0} = [0, 0]$ and ${m^{+}} = (m, m^{+}]$ are basic open sets.
The sequence $x_{k} : = k$ ( $k \in N$ ) converges to $ω$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), and it converges to no other point of $ω + 1$ .

So $ω + 1$ is, as a topological space, exactly a convergent sequence together with its limit, and $ω$ is its unique non-isolated point.

The space $[0, ω_{1}) = ω_{1}$ . Let $ω_{1}$ be the first uncountable ordinal (The first uncountable ordinal $ω_{1} : = ℵ (ω)$ ), so that $ω_{1}$ is a limit ordinal, every ordinal below it is at most countable, and $ω_{1}$ itself is uncountable ( $ω_{1}$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF). As a set, $[0, ω_{1})$ is $ω_{1}$ , an ordinal being the set of ordinals below it (Ordinal (von Neumann)). Assuming the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )):

Every sequence $(x_{k})$ in $ω_{1}$ has an at most countable range, so there is $α < ω_{1}$ with $x_{k} \leq α$ for all $k$ ; hence the whole sequence lies inside the initial segment $[0, α] = α^{+}$ , which is an ordinal below $ω_{1}$ and is at most countable.
Consequently no sequence in $ω_{1}$ has a range cofinal in $ω_{1}$ (Cofinal subset of an ordinal).

Clause 3 is the fact the deleted Tychonoff plank consumes, and it is the reason $[0, ω_{1})$ behaves unlike any metrizable space: a sequence can never approach the "top" of $ω_{1}$ , because there is no top to approach along a sequence.

Facts & Assumptions

Given: Ordinals with their order topologies; the natural numbers $ω$ ; the first uncountable ordinal $ω_{1}$ ; a sequence $(x_{k})_{k \in N}$ in $ω_{1}$ ; and the Axiom of Countable Choice where stated.

[A1]

The basic open sets of an ordinal $γ$ are $[0, β]$ for $β \in γ$ and $(α, β]$ for $α < β$ in $γ$ ; they form a basis (The order topology on an ordinal, with the half-open intervals $(α, β]$ and the initial segments $[0, β]$ as a basis, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A2]

$α + 1 = α^{+}$ , by the clauses of ordinal addition at $0$ and at a successor (Ordinal addition $α + β$ ).

[L1]

$ω$ is an ordinal and a limit ordinal, every element of $ω$ is $0$ or a successor, and $m \in n$ is $m < n$ for naturals ( $ω$ is the least limit ordinal, Successor and limit ordinals, Basic closure properties of ordinals).

[L2]

$x_{k} \to p$ means: for every neighbourhood $N$ of $p$ there is $K$ with $x_{k} \in N$ for all $k \geq K$ ; an open set containing $p$ is such a neighbourhood (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L3]

$ω_{1}$ is uncountable, is a limit ordinal, and every ordinal below it is at most countable ( $ω_{1}$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF, The first uncountable ordinal $ω_{1} : = ℵ (ω)$ , Finite, countably infinite, countable, uncountable).

[L4]

Assuming ${A C}_{ω}$ , every at most countable subset of $ω_{1}$ is bounded below $ω_{1}$ : there is $α \in ω_{1}$ with $ξ \leq α$ for every $ξ$ in the subset (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, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L5]

The range of a sequence is nonempty and at most countable (A nonempty set is at most countable iff it is a surjective image of $N$ , Finite, countably infinite, countable, uncountable).

[L6]

For ordinals exactly one of $ξ < η$ , $ξ = η$ , $η < ξ$ holds; $α^{+}$ is an ordinal, and $α^{+} = {ξ : ξ \leq α}$ (Trichotomy and well-ordering of the ordinals, Basic closure properties of ordinals, Ordinal (von Neumann)).

[L7]

A subset $S$ of a limit ordinal $γ$ is cofinal in $γ$ when for every $ξ \in γ$ there is $σ \in S$ with $ξ \leq σ$ (Cofinal subset of an ordinal).

[L8]

Every ordinal with its order topology is $T_{1}$ , Hausdorff and regular (Every ordinal with its order topology has a basis of clopen sets, and is $T_{1}$ , Hausdorff and regular).

Verification

technique · direct

1.1

$ω + 1 = ω^{+} = ω \cup {ω}$ by [A2], so the points of the space are the natural numbers together with $ω$ .

A2L1

1.2

Let $N$ be a neighbourhood of $ω$ in $ω + 1$ ; by [A1] and [L2] there is a basic set $B$ with $ω \in B \subseteq N$ , and $B$ is $[0, β]$ with $ω \leq β$ or $(α, β]$ with $α < ω \leq β$ . In either case $(α, ω] \subseteq B$ for some $α \in ω$ , taking $α : = 0$ in the first case.

A1L2L6

1.3

Let $(x_{k})$ be a sequence in $ω_{1}$ ; its range $R : = {x_{k} : k \in N}$ is an at most countable subset of $ω_{1}$ by [L5].

2.1

Each $n \in ω$ is $0$ or a successor $m^{+}$ by [L1]; in the first case ${n} = [0, 0]$ and in the second ${n} = (m, m^{+}]$ , both basic open sets of $ω + 1$ by [A1], since $0 \in ω + 1$ and $m < m^{+} = n$ in $ω + 1$ . So every $n \in ω$ is isolated, which is claim 1.

step 1.1A1L1

2.2

Under step 1.2: $α \in ω$ , so for every $k > α$ one has $α < k \leq ω$ and hence $x_{k} = k \in (α, ω] \subseteq N$ ; so $x_{k} \to ω$ by [L2].

step 1.2L1L2L6

2.3

By [L4] there is $α \in ω_{1}$ with $ξ \leq α$ for every $ξ \in R$ , hence $x_{k} \leq α$ for every $k$ .

step 1.3L4

3.1

The sequence converges to no $n \in ω$ : by step 2.1 the set ${n}$ is an open neighbourhood of $n$ , and $x_{k} = k \neq n$ for every $k > n$ , so the sequence is not eventually in ${n}$ .

step 2.1L1L2

3.2

By [L6] the set $α^{+} = {ξ : ξ \leq α}$ contains every $x_{k}$ , and $α^{+} \in ω_{1}$ because $ω_{1}$ is a limit ordinal and $α \in ω_{1}$ ; so $α^{+}$ is an ordinal below $ω_{1}$ and is at most countable by [L3]. This is claim 3.

step 2.3L3L6

4.1

Steps 2.2 and 3.1 are claim 2.

step 2.2step 3.1

4.2

If some sequence had range cofinal in $ω_{1}$ , then by [L7] every $ξ \in ω_{1}$ would satisfy $ξ \leq x_{k}$ for some $k$ ; taking $ξ : = α^{+}$ of step 3.2 gives $α^{+} \leq x_{k} \leq α$ for some $k$ , contradicting $α < α^{+}$ by [L6]. So claim 4 holds.

step 2.3step 3.2L6L7

5.1

Both spaces are $T_{3}$ by [L8], and steps 2.1, 4.1, 3.2 and 4.2 are claims 1 to 4.

step 2.1step 4.1step 3.2step 4.2L8∎

Remarks

$ω + 1$ is the smallest interesting ordinal space. Every ordinal $γ \leq ω$ is discrete (The order topology on an ordinal, with the half-open intervals $(α, β]$ and the initial segments $[0, β]$ as a basis), so $ω + 1$ is where a non-isolated point appears for the first time, and it appears as the limit of the obvious sequence.
Clause 3 is where the countable choice enters and where it stays. It is inherited from 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 and from nothing else on this page; clauses 1, 2 and the $T_{3}$ property of both spaces are theorems of ZF (Every ordinal with its order topology has a basis of clopen sets, and is $T_{1}$ , Hausdorff and regular).
What clause 4 rules out. No sequence in $[0, ω_{1})$ can be used to approximate the space from below, which is why arguments about $ω_{1}$ are written with arbitrary at most countable sets rather than with sequences, and why the plank argument on this page bounds a set of ordinals rather than taking a limit of a sequence.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

60 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Order topology (Wikipedia) (standard reference, not scraped)
First uncountable ordinal (Wikipedia) (standard reference, not scraped)
L. Steen and J. Seebach, Counterexamples in Topology, §39-43 (standard reference, not scraped)

ω+1ω+1 as a convergent sequence together with its limit, and, assuming countable choice, [0,ω1)[0,ω1​), in which every sequence lies inside an at most countable initial segment