CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The diagonal $x \mapsto (x,x,\dots)$ from $\mathbb{R}$ into $\mathbb{R}^{\mathbb{N}}$ is continuous for the product topology and not for the box topology

Statement refuted

Refuted: that the box topology has the characteristic property of a product, that is, that a map into $\prod_i X_i$ with all components continuous is continuous for the box topology. Equivalently, this exhibits again that the two topologies differ (FALSE: the product topology and the box topology agree on every product).

Witness. Let $P := \mathbb{R}^{\mathbb{N}} = \prod_{k \in \mathbb{N}}\mathbb{R}$ with every factor carrying the usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and let

$\Delta : \mathbb{R} \to P, \qquad \Delta(t)_k := t \quad (k \in \mathbb{N})$

be the diagonal map. Every component $\pi_k \circ \Delta$ is the identity of $\mathbb{R}$ , hence continuous. Then $\Delta$ is continuous for the product topology (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2) and is not continuous for the box topology: the box

$B \;:=\; \prod_{k \in \mathbb{N}} \Big(-\tfrac{1}{k+1},\ \tfrac{1}{k+1}\Big)$

is box-open and $\Delta^{-1}[B] = \{0\}$ , which is not open in $\mathbb{R}$ .

Facts & Assumptions

Given: $P = \prod_{k \in \mathbb{N}}\mathbb{R}$ , the diagonal $\Delta$ , and the box $B$ above; $1/(k+1)$ abbreviates $1/\iota(k+1)$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

[A1]

A basis for the box topology is the family of all boxes $\prod_k U_k$ with every $U_k$ open in $\mathbb{R}$ (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, The box topology is finer than the product topology, the two agree for a finite index set in ZF, and, assuming the Axiom of Choice for nonempty factors, the box topology is strictly finer whenever infinitely many factors have a nonempty proper open subset).

[A2]

A map into a product with the product topology is continuous exactly when all its components are (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 2).

[L1]

A map of spaces is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$ , clause (b); Continuity of a map of topological spaces at a point and globally).

[L2]

$(a,b) = \{s : a < s < b\}$ is open in the usual topology of $\mathbb{R}$ , and a subset of $\mathbb{R}$ is open there exactly when each of its points has a bounded open interval around it inside the set (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L3]

For every real $\eta > 0$ there is a natural $m \ge 1$ with $1/m < \eta$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ); $\iota(k+1) \ge 1 > 0$ and $0 < u \le v$ gives $0 < 1/v \le 1/u$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, The canonical natural $\iota(n) = n \cdot 1_F$ of a field).

Counterexample

technique · direct

1.1

Each component $\pi_k \circ \Delta$ is the identity map of $\mathbb{R}$ , since $(\pi_k \circ \Delta)(t) = \Delta(t)_k = t$ ; the identity is continuous, its preimages being the sets themselves.

givenL1

1.2

Each factor of $B$ is a bounded open interval and $1/(k+1) > 0$ by [L3], so $B$ is a box with open factors and hence open in the box topology.

A1L2L3

1.3

$\{0\}$ is not open in $\mathbb{R}$ : for every $r > 0$ the interval $(-r,r)$ contains $r/2$ , which is different from $0$ ; so no bounded open interval around $0$ lies inside $\{0\}$ .

2.1

$0 \in \Delta^{-1}[B]$ , since $\Delta(0)_k = 0 \in (-1/(k+1),\ 1/(k+1))$ for every $k$ by [L3].

step 1.2L2L3

2.2

$\Delta$ is continuous for the product topology, by step 1.1 and [A2].

step 1.1A2

3.1

$\Delta^{-1}[B] = \{0\}$ : a real $t$ lies in it exactly when $|t| < 1/(k+1)$ for every $k \in \mathbb{N}$ ; if $t \ne 0$ then $|t| > 0$ and [L3] gives a natural $m \ge 1$ with $1/m < |t|$ , and taking $k := m - 1$ contradicts that condition. With step 2.1 this gives the stated equality.

step 2.1L2L3

4.1

By steps 1.2, 3.1 and 1.3 the preimage under $\Delta$ of a box-open set is not open in $\mathbb{R}$ , so $\Delta$ is not continuous into $P$ with the box topology, by [L1]; by step 2.2 it is continuous into $P$ with the product topology, although its components are the same in both cases. That refutes the claim.

step 1.2step 2.2step 3.1step 1.3L1∎

Remarks

This is the practical reason the product topology is the default. The characteristic property of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice is what makes a map into a product easy to build, and the box topology has no such property: here every component is the identity, and continuity still fails. Any construction that assembles a map coordinate by coordinate would break in the box topology.
The two topologies are separated by this single map. If they agreed, the same map could not be continuous for one and not for the other; so this item reproves the strictness recorded in FALSE: the product topology and the box topology agree on every product, by a different route and with the same box.
Nothing here needs a choice principle, the box $B$ and the map $\Delta$ being written down by formulas, and the only existential step being the Archimedean one of For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 16 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

Box topology (Wikipedia) (standard reference, not scraped)
Product topology (Wikipedia) (standard reference, not scraped)

The diagonal x↦(x,x,… )x \mapsto (x,x,\dots)x↦(x,x,…) from R\mathbb{R}R into RN\mathbb{R}^{\mathbb{N}}RN is continuous for the product topology and not for the box topology