False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31

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Refuted: separability is hereditary

Statement

Separability is hereditary.

Facts & Assumptions

Given: The lower-limit plane $P$ and its antidiagonal $A=\{(x,-x):x\in\mathbb R\}$ .

[L1]

Products of at most countable sets are at most countable (A product of two at most countable sets is at most countable).

[L2]

The rational numbers are at most countable and dense in the real line, and the real line is uncountable ( $\mathbb{Q}$ is countably infinite, The rationals embed densely in the reals, $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874)).

[L3]

Separability is the existence of an at most countable dense subset, and a property is hereditary when every subspace has it (Separability: the existence of an at most countable dense subset, Hereditary, open-hereditary and closed-hereditary properties of topological spaces).

[F1]

The half-open intervals $[a,b)$ , $a<b$ , satisfy the basis criterion, and products of their members form a basis for the product topology (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis, 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).

Refutation

technique · direct

1.1

The half-open intervals cover $\mathbb R$ , and if two contain $x$ , then $[x,c)$ lies in their intersection for some $c>x$ ; hence [F1] makes them a basis. The rational grid $\mathbb Q\times\mathbb Q$ is at most countable by [L1] and [L2], and density of $\mathbb Q$ makes it meet every nonempty basic lower-limit rectangle, so it is dense in $P$ .

L1L2L3F1

1.2

For each $x\in\mathbb R$ , the basic rectangle $[x,x+1)\times[-x,-x+1)$ meets $A$ only in $(x,-x)$ ; hence $A$ is discrete in its subspace topology.

given

2.1

The map $x\mapsto(x,-x)$ is a bijection from the uncountable set $\mathbb R$ onto $A$ , so a dense subset of the discrete space $A$ must be all of $A$ and cannot be at most countable.

step 1.2L2L3

3.1

Thus $P$ is separable by step 1.1 but has the nonseparable subspace $A$ by step 2.1, refuting heredity of separability.

step 1.1step 2.1L3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 113 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

UCR General Topology Notes (standard reference, not scraped)
Sorgenfrey plane (Wikipedia) (standard reference, not scraped)