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

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$\mathbb{Q}^n$ is a countable dense subset of $\mathbb{R}^n$ , and rational open boxes form a countable basis

Statement

Let $n\ge1$ . The set $\mathbb Q^n$ is countable and dense in $\mathbb R^n$ . Moreover the rational open boxes

$\prod_{i<n}(a_i,b_i),\qquad a_i,b_i\in\mathbb Q,\quad a_i<b_i,$

form a countable basis for the product topology on $\mathbb R^n$ .

Facts & Assumptions

Given: $n\ge1$ , the product $\mathbb R^n=\prod_{i<n}\mathbb R$ , and the rationals embedded in $\mathbb R$ .

[L1]

$\mathbb Q$ is countably infinite, and every finite power of an at most countable set is at most countable ( $\mathbb{Q}$ is countably infinite, Every finite power of an at most countable set is at most countable).

[L2]

Every nonempty open subset of $\mathbb R$ contains a rational point (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$ , and every nonempty open subset of $\mathbb{R}$ is uncountable).

[L3]

The product topology has a basis of finite-coordinate boxes, which for the finite index set $n$ are products of open subsets of $\mathbb R$ (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).

[L4]

A family is a basis when each point of each open set lies in one of its members contained in that open set (Basis and subbasis for a topology, and the topology generated by a family of sets).

[L5]

Finite choices may be assembled into a tuple, and a subset of an at most countable set is at most countable (Every natural-number-indexed list of nonempty sets has a choice function on its family of values, Every subset of an at most countable set is at most countable).

[L6]

If $U\subseteq\mathbb R$ is open and $x\in U$ , then some open interval about $x$ is contained in $U$ (Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace).

Proof

technique · constructive

1.1

By [L1], $\mathbb Q^n$ is at most countable. It is infinite because the injection $q\mapsto(q,0,\ldots,0)$ embeds $\mathbb Q$ in it, hence it is countable.

1.2

Let $U=\prod_{i<n}U_i$ be a nonempty basic product-open set. Every $U_i$ is nonempty and open, so [L2] gives a rational $q_i\in U_i$ ; finite choice supplies the tuple $q=(q_i)_{i<n}\in\mathbb Q^n\cap U$ .

L2L5choose

1.3

Let $x\in U=\prod_{i<n}U_i$ be a basic product-open neighbourhood. For each $i<n$ , use [L6] to choose $r_i>0$ with $(x_i-r_i,x_i+r_i)\subseteq U_i$ , then use [L2] to choose rationals $x_i-r_i<a_i<x_i<b_i<x_i+r_i.$ Finite choice assembles these choices, and then $x\in\prod_{i<n}(a_i,b_i)\subseteq U$ .

L2L5L6choose

2.1

Every nonempty open subset contains a nonempty basic product-open set about each of its points by [L3], so step 1.2 shows that every nonempty open subset meets $\mathbb Q^n$ . Thus $\mathbb Q^n$ is dense.

L3step 1.2

3.1

Step 1.3 and [L4] show that rational open boxes form a basis. They are indexed by a subset of $\mathbb Q^{2n}$ , which is at most countable by [L1], so the basis is at most countable by [L5].

L1L4L5step 1.3∎

Depends on

Used by

Under choice, for the usual real line, w=d=χ=L=c=ℵ₀ under the raw convention Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 147 results over 27 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

Euclidean space (standard reference, not scraped)
Separable space (Wikipedia) (standard reference, not scraped)

Qn\mathbb{Q}^nQn is a countable dense subset of Rn\mathbb{R}^nRn, and rational open boxes form a countable basis