LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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The lower-limit line has a clopen basis, is regular, and is Lindelöf under countable choice

Statement

The lower-limit line has a basis of clopen sets and is regular. Assuming the Axiom of Countable Choice, it is Lindelöf.

Facts & Assumptions

Given: The lower-limit line and, for the Lindelöf assertion, the Axiom of Countable Choice.

[F1]

Its basic open sets are the intervals $[a,b)$ (The lower-limit topology on $\mathbb{R}$ , with the half-open intervals $[a,b)$ as a basis).

[L1]

A clopen neighbourhood basis gives regularity through the closed-neighbourhood characterization (A space is regular if and only if every point has a neighbourhood base of closed neighbourhoods, if and only if $x \in U$ open gives an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$ ).

[L2]

$\mathbb Q$ is countably infinite and is dense in $\mathbb R$ ( $\mathbb{Q}$ is countably infinite, The rationals embed densely in the reals).

[A1]

The Axiom of Countable Choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[F2]

Lindelöf means that every open cover has an at most countable subcover (Countably compact, Lindel"of, sequentially compact, limit point compact and $\sigma$ -compact spaces, and relatively compact subsets).

Proof

technique · direct

1.1

Each $[a,b)$ is clopen: its complement is $(-\infty,a)\cup[b,\infty)$ , a union of lower-limit basic intervals. Hence the line is regular by [L1].

F1L1

1.2

Let $\mathcal U$ be an open cover and let $O$ be the union of the usual intervals $(a,b)$ for which $[a,b)$ lies in a member of $\mathcal U$ . The rational-endpoint intervals $[p,q)$ contained in members of $\mathcal U$ cover $O$ ; they form an at most countable family by [L2].

F1L2

1.3

Put $D=\mathbb R\setminus O$ . For $x\in D$ , some $[x,b_x)$ lies in a member of $\mathcal U$ , and $[x,b_x)\cap D=\{x\}$ . The first rational $q_x\in(x,b_x)$ in a fixed enumeration exists, and the intervals $(x,q_x)$ are pairwise disjoint; their first rationals $r_x$ are therefore distinct. Thus $x\mapsto r_x$ injects $D$ into $\mathbb Q$ , so $D$ is at most countable.

F1L2

2.1

By [A1], choose one member of $\mathcal U$ covering each point of the at most countable set $D$ . Together with one covering member for each rational-endpoint interval used in step 1.2, these form an at most countable subcover of $\mathcal U$ .

A1L2step 1.2step 1.3

3.1

Therefore the lower-limit line is Lindelöf under countable choice.

F2step 2.1∎

Depends on

Used by

Assuming countable choice, the lower-limit line is normal Corollary
Assuming choice, two paracompact lower-limit lines can have a nonparacompact product Counterexample
For the lower-limit line, χ=d=L=c=ℵ₀ and w=2^ℵ₀ under choice Example
Under choice, the lower-limit line is regular and separable but not second countable and therefore not metrizable Example
Assuming choice, refuted: paracompactness is productive False statement
FALSE: every regular space is metrizable False statement

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 123 results over 33 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, Sorgenfrey line (standard reference, not scraped)
Sorgenfrey topology (Encyclopedia of Mathematics) (standard reference, not scraped)