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 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.

The lower-limit plane has a countable dense set and a closed discrete antidiagonal of size $|\mathbb{R}|$

Statement

In the square of the lower-limit line, $\mathbb Q\times\mathbb Q$ is a countable dense subset, while $D=\{(x,-x):x\in\mathbb R\}$ is closed and discrete and has the same cardinality as $\mathbb R$ .

Facts & Assumptions

Given: The lower-limit plane and its basic rectangles $[a,b)\times[c,d)$ .

[F1]

Basic product-open sets restrict finitely many coordinates; for this binary product they are the basic rectangles (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The lower-limit topology on $\mathbb{R}$ , with the half-open intervals $[a,b)$ as a basis).

[F2]

A subset is dense iff it meets every nonempty basic open set, the rational numbers are countably infinite, and a rational lies strictly between any two distinct reals (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, $\mathbb{Q}$ is countably infinite, The rationals embed densely in the reals).

[L1]

A product of two at most countable sets is at most countable, and $\mathbb R$ is uncountable (A product of two at most countable sets is at most countable, $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)).

Proof

technique · direct

1.1

Every nonempty $[a,b)\times[c,d)$ contains a point of $\mathbb Q\times\mathbb Q$ : choose rationals $p\in[a,b)$ and $q\in[c,d)$ . Hence $\mathbb Q\times\mathbb Q$ is dense, and it is at most countable by [L1].

F1F2L1

1.2

The map $x\mapsto(x,-x)$ is a bijection from $\mathbb R$ onto $D$ , so $D$ has cardinality $|\mathbb R|$ and is uncountable.

1.3

For $(x,-x)\in D$ , the rectangle $[x,x+1)\times[-x,-x+1)$ meets $D$ only at $(x,-x)$ , so $D$ is discrete in its subspace topology.

1.4

If $(u,v)\notin D$ and $u+v>0$ , every sufficiently small lower-limit rectangle at $(u,v)$ has positive coordinate sum; if $u+v<0$ , choose its two right endpoints so that their total increment is less than $-(u+v)$ . In either case the rectangle misses $D$ , so the complement of $D$ is open.

2.1

Therefore $D$ is closed discrete, with the stated cardinality, and the plane has the stated countable dense subset.

step 1.1step 1.2step 1.3step 1.4∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 111 results over 25 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

G. Gruenhage, General Topology Course Notes, Sorgenfrey plane (standard reference, not scraped)
Sorgenfrey plane (Wikipedia) (standard reference, not scraped)
Sorgenfrey topology (Encyclopedia of Mathematics) (standard reference, not scraped)

The lower-limit plane has a countable dense set and a closed discrete antidiagonal of size ∣R∣|\mathbb{R}|∣R∣