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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

1.1

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

givenL4

1.2

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.

A2A1

1.3

$V : = B (1, 1) = (0, 1 + 1)$ 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].

L1L3

1.4

${0}$ 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 ${0}$ .

L1L2

2.1

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.

step 1.1step 1.2step 1.3step 1.4A1∎

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.

Depends on

Used by

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Sources

Pasting lemma (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)

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