Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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.

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The lower-limit plane has a countable dense set and a closed discrete antidiagonal of size ∣R∣

Statement

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

Facts & Assumptions

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

[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, 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 R is uncountable (A product of two at most countable sets is at most countable, R is uncountable (Cantor's nested intervals, 1874)).

Proof

technique · direct
1.1

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

F1F2L1
1.2

The map x↦(x,−x) is a bijection from R onto D, so D has cardinality ∣R∣ and is uncountable.

L1
1.3

For (x,−x)∈D, the rectangle [x,x+1)×[−x,−x+1) meets D only at (x,−x), so D is discrete in its subspace topology.

F1
1.4

If (u,v)∉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.

F1
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 · two levels

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Sources