CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$R$ covered by its closed singletons: every restriction of the indicator of ${0}$ is continuous and the map is not, so the closed pasting lemma needs finiteness

Statement refuted

Refuted: that continuity may be checked on an arbitrary closed cover. Claim 3 of Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous allows only finitely many closed pieces, and the restriction is not removable.

Witness. Give $R$ its usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $R$ 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) and let

$F : = {{t} : t \in R}$

be the family of its singletons, a cover of $R$ by closed sets. Let $f : R \to R$ be the indicator of ${0}$ , that is $f (0) = 1$ and $f (t) = 0$ for $t \neq 0$ . Then every restriction $f ∣_{{t}}$ is continuous for the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), and $f$ is not continuous (Continuity of a map of topological spaces at a point and globally).

Facts & Assumptions

Given: $R$ with its usual topology, the cover $F$ by singletons, and the function $f$ above.

[A1]

A map 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)).

[A2]

The subspace topology on $S \subseteq R$ has as open sets the traces $U \cap S$ with $U$ open in $R$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace); a topology on $S$ always contains $\emptyset$ and $S$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L1]

In the usual topology $B (x, r) = (x - r, x + r)$ , balls are open, and $U$ is open exactly when every $x \in U$ has some $r > 0$ with $(x - r, x + r) \subseteq U$ (The absolute value makes $R$ 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, Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ , and $1 / n > 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L3]

$0 < 1$ and hence $1 < 1 + 1$ (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities); consequently $1 \in (0, 1 + 1)$ and $0 \notin (0, 1 + 1)$ .

[L4]

Every singleton ${t} \subseteq R$ is closed in the usual topology, its complement $(- \infty, t) \cup (t, \infty)$ being a union of two open sets (The absolute value makes $R$ 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, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

Counterexample

technique · direct

1.1givenL4

$F$ covers $R$ , each $t \in R$ lying in ${t}$ , and each member is closed by [L4].

1.2A2A1

For every $t \in R$ the subspace ${t}$ carries only the two subsets $\emptyset$ and ${t}$ , both of which are open in it by [A2]; hence every function out of ${t}$ has open preimages and is continuous, and in particular $f ∣_{{t}}$ is.

1.3L1L3

$V : = B (1, 1) = (0, 1 + 1)$ is a ball, hence open in $R$ by [L1], and $f^{- 1} [V] = {0}$ : indeed $f (0) = 1 \in V$ by [L3], while $f (t) = 0 \notin V$ for $t \neq 0$ , again by [L3].

1.4L1L2

${0}$ is not open in the usual topology: for any $r > 0$ the ball $(- r, r)$ contains the point $1 / n$ for a natural $n \geq 1$ with $1 / n < r$ given by [L2], and $1 / n > 0$ , so $1 / n \in (- r, r) ∖ {0}$ ; hence no ball around $0$ lies inside ${0}$ .

2.1step 1.1step 1.2step 1.3step 1.4A1∎

By step 1.3 and step 1.4 the preimage under $f$ of the open set $V$ is not open, so $f$ is not continuous by [A1]; by steps 1.1 and 1.2 the family $F$ is a closed cover of $R$ every restriction to which is continuous. So claim 3 of Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous fails without the hypothesis that the cover be finite.

Remarks

Why an infinite closed cover is useless and an infinite open cover is not. The proof of claim 3 of Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous writes $f^{- 1} [F]$ as a union of sets closed in $R$ and concludes that it is closed; only finite unions of closed sets are closed (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and here $f^{- 1} [R ∖ V]$ is the union of the uncountably many closed sets ${t}$ , $t \neq 0$ , which is $R ∖ {0}$ , not closed. The open-cover version has no such restriction because arbitrary unions of open sets are open.
The singleton cover trivialises every function. For any spaces $X$ and $Y$ and any $f : X \to Y$ , the restriction of $f$ to a one-point subspace is continuous, so the singleton cover certifies nothing whatever. The witness is therefore the sharpest form of the failure rather than a delicate example, and the map $f$ could be replaced by any discontinuous function.
A two-piece closed cover of $R$ would have detected the discontinuity. For instance $(- \infty, 0]$ and $[0, \infty)$ are closed and cover $R$ , and $f$ restricted to $(- \infty, 0]$ is already discontinuous at $0$ by the argument of step 1.4 carried out inside that subspace.

$R$ covered by its closed singletons: every restriction of the indicator of ${0}$ is continuous and the map is not, so the closed pasting lemma needs finiteness

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

RR covered by its closed singletons: every restriction of the indicator of {0}{0} is continuous and the map is not, so the closed pasting lemma needs finiteness

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

$R$ covered by its closed singletons: every restriction of the indicator of ${0}$ is continuous and the map is not, so the closed pasting lemma needs finiteness