False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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FALSE: the evaluation map on $C(X,Y)$ with the compact-open topology is continuous for every metric $X$

Statement

False claim: for every metric space $X$ and every topological space $Y$ the evaluation map $e : C(X,Y) \times X \to Y$ , $e(f,x) = f(x)$ (The evaluation map $e : C(X,Y) \times X \to Y$ , $e(f,x) = f(x)$ ), is continuous when $C(X,Y)$ carries the compact-open topology (The compact-open topology on $C(X,Y)$ for a metric domain $X$ , with subbasis $S(K,V) = \{f : f[K] \subseteq V\}$ ).

The witness is $X = \mathbb{Q}$ , the rationals inside $\mathbb{R}$ with the metric $d(s,t) = |s-t|$ , and $Y = \mathbb{R}$ with the same metric. The load-bearing fact is that a compact subset of $\mathbb{Q}$ has empty interior in $\mathbb{Q}$ : it is closed in $\mathbb{R}$ , and a subset of $\mathbb{Q}$ closed in $\mathbb{R}$ that contained a $\mathbb{Q}$ -ball would contain a whole real interval, which is uncountable while $\mathbb{Q}$ is not.

What the true theorem on this page requires is therefore not decoration. Continuity of the evaluation map is proved here under the hypothesis that $X$ is locally compact (Locally compact metric space: every point has a compact neighbourhood, If $X$ is a locally compact metric space then the evaluation map is continuous for the compact-open topology), and $\mathbb{Q}$ is a metric space that is locally compact at no point, exactly because of the fact just named.

No choice principle is used.

Facts & Assumptions

Given: The rationals $\mathbb{Q}$ inside $\mathbb{R}$ with the metric $d(s,t) = |s-t|$ (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, Isometry, isometric embedding, and the subspace metric on a subset), the target $\mathbb{R}$ with the same metric, the constant function $z : \mathbb{Q} \to \mathbb{R}$ with value $0$ , and the open interval $V := (-1,1)$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L1]

A compact subset of a metric space is closed in it and bounded; compactness of a subset is a property of the subspace metric alone, so a compact subset of $\mathbb{Q}$ is a compact subset of $\mathbb{R}$ (A compact subset of a metric space is closed and bounded, Open cover, subcover, compact metric space, and compact subset of a metric space, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, Isometry, isometric embedding, and the subspace metric on a subset).

[L2]

A union of two closed subsets of a metric space is closed, its complement being an intersection of two open sets; iterating covers any finite list, and $\varnothing$ is closed (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, 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).

[L3]

Strictly between any two reals lies a rational (The rationals embed densely in the reals).

[L4]

For nonempty $A$ the closure is $\overline{A} = \{\, x : d(x,A) = 0 \,\}$ , a closed set equals its closure and contains it, and $d(x,A) \ge 0$ with $d(x,A) \le d(x,a)$ for every $a \in A$ ; $d(x,A) = 0$ when $x \in A$ (The closure of a nonempty $A$ is $\{x : d(x,A) = 0\}$ , equals $A$ together with its limit points, and is the smallest closed superset, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Greatest lower bound (infimum)).

[L5]

Every nondegenerate open interval of $\mathbb{R}$ is uncountable, $\mathbb{Q}$ is at most countable, and a subset of an at most countable set is at most countable (Every nondegenerate interval of $\mathbb{R}$ is uncountable, $\mathbb{Q}$ is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L6]

For nonempty $A \subseteq \mathbb{Q}$ the map $s \mapsto d(s,A)$ is $1$ -Lipschitz, so $s \mapsto c\, d(s,A)$ is Lipschitz with constant $c$ for a real $c > 0$ , hence continuous ( $|d(x,A) - d(y,A)| \le d(x,y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent, Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Absolute value in an ordered field).

[L7]

Continuity of a map at a point, in the open-set form, and the fact that the boxes $O \times U$ with $O$ open in $C(\mathbb{Q},\mathbb{R})$ and $U$ open in $\mathbb{Q}$ form a basis for the product topology, while the finite intersections of the sets $S(K,W)$ form a basis for the compact-open topology (Continuity of a map of topological spaces at a point and globally, 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 $f(\overline{A}) \subseteq \overline{f(A)}$ , 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 compact-open topology on $C(X,Y)$ for a metric domain $X$ , with subbasis $S(K,V) = \{f : f[K] \subseteq V\}$ , Basis and subbasis for a topology, and the topology generated by a family of sets, 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, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L8]

Balls of the subspace $\mathbb{Q}$ are traces of balls of $\mathbb{R}$ : $B_{\mathbb{Q}}(q,r) = (q-r, q+r) \cap \mathbb{Q}$ ; and a subset of a metric space is open exactly when each of its points has a ball around it inside the set (Isometry, isometric embedding, and the subspace metric on a subset, 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, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L9]

The minimum of a two-element set of reals exists, is one of them, and is at most each of them (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Refutation

technique · contradiction

1.1

Suppose the claim holds; in particular, with $X := \mathbb{Q}$ and $Y := \mathbb{R}$ , the evaluation map $e : C(\mathbb{Q},\mathbb{R}) \times \mathbb{Q} \to \mathbb{R}$ is continuous.

assume-contra

1.2

The constant function $z$ is continuous, and $e(z,0) = 0 \in V$ with $V$ open in $\mathbb{R}$ .

givenL6L8

2.1

Continuity of $e$ at the point $(z,0)$ gives, by the two bases of [L7], a natural $n$ , compact sets $K_0, \dots, K_{n-1} \subseteq \mathbb{Q}$ , open sets $V_0, \dots, V_{n-1} \subseteq \mathbb{R}$ and an open $U \subseteq \mathbb{Q}$ with $0 \in U$ , such that $z \in O := S(K_0,V_0) \cap \dots \cap S(K_{n-1},V_{n-1})$ and $e[O \times U] \subseteq V$ .

step 1.1step 1.2L7choose

3.1

Fix a real $\varepsilon > 0$ with $B_{\mathbb{Q}}(0,\varepsilon) \subseteq U$ .

step 2.1L8choose

3.2

Put $K := K_0 \cup \dots \cup K_{n-1}$ , with $K := \varnothing$ when $n = 0$ ; each $K_j$ is a compact subset of $\mathbb{R}$ and hence closed in $\mathbb{R}$ , so $K$ is a subset of $\mathbb{Q}$ closed in $\mathbb{R}$ .

step 2.1L1L2

4.1

$B_{\mathbb{Q}}(0,\varepsilon) \not\subseteq K$ : otherwise $(-\varepsilon,\varepsilon) \cap \mathbb{Q} \subseteq K$ , and this set is nonempty since it contains $0$ ; every $t \in (-\varepsilon,\varepsilon)$ then satisfies $d(t, (-\varepsilon,\varepsilon) \cap \mathbb{Q}) = 0$ , because for a real $s > 0$ the set $(\max\{t-s,-\varepsilon\},\ \min\{t+s,\varepsilon\})$ is a nondegenerate open interval and so contains a rational within $s$ of $t$ ; hence $t$ lies in the closure of that set, which the closed $K$ contains, so $(-\varepsilon,\varepsilon) \subseteq K \subseteq \mathbb{Q}$ , making the uncountable interval $(-\varepsilon,\varepsilon)$ at most countable, which is false.

step 3.2L3L4L5L8L9

5.1

Fix $q \in B_{\mathbb{Q}}(0,\varepsilon)$ with $q \notin K$ .

step 4.1choose

6.1

Put $\eta := \min\{d(q,K),\ 1\}$ when $K \ne \varnothing$ and $\eta := 1$ when $K = \varnothing$ ; then $0 < \eta \le 1$ , because $K$ closed in $\mathbb{R}$ and $q \notin K$ give $q \notin \overline{K}$ and hence $d(q,K) \ne 0$ , while $d(q,K) \ge 0$ ; and $|q - t| \ge d(q,K) \ge \eta$ for every $t \in K$ .

step 3.2step 5.1L4L9

6.2

$q \in U$ , since $|q - 0| < \varepsilon$ puts $q$ in $B_{\mathbb{Q}}(0,\varepsilon) \subseteq U$ .

step 3.1step 5.1

7.1

Put $A := \{\, s \in \mathbb{Q} : |s - q| \ge \eta \,\}$ , which is nonempty because $q + 1 \in \mathbb{Q}$ and $|(q+1) - q| = 1 \ge \eta$ , and define $g : \mathbb{Q} \to \mathbb{R}$ by $g(s) := (2/\eta)\, d(s,A)$ .

step 6.1constructL4

8.1

$g$ is Lipschitz with constant $2/\eta$ , hence continuous, so $g \in C(\mathbb{Q},\mathbb{R})$ .

step 7.1L6

8.2

$g$ vanishes on $K$ : every $t \in K$ lies in $\mathbb{Q}$ and satisfies $|t - q| \ge \eta$ by step 6.1, so $t \in A$ and $d(t,A) = 0$ .

step 6.1step 7.1L4

8.3

$g(q) \ge 2$ : every $s \in A$ satisfies $|q - s| \ge \eta$ , so $\eta$ is a lower bound of the distances from $q$ to the members of $A$ and $d(q,A) \ge \eta$ , whence $g(q) = (2/\eta)\,d(q,A) \ge 2$ .

step 7.1L4

9.1

$g \in O$ : for each $j < n$ with $K_j \ne \varnothing$ we have $z \in S(K_j,V_j)$ , so $0 \in V_j$ , and $g[K_j] \subseteq \{0\} \subseteq V_j$ by step 8.2 since $K_j \subseteq K$ ; for $K_j = \varnothing$ the condition is vacuous; and for $n = 0$ the set $O$ is the whole of $C(\mathbb{Q},\mathbb{R})$ .

step 2.1step 3.2step 8.1step 8.2

10.1

Hence $(g,q) \in O \times U$ while $e(g,q) = g(q) \ge 2$ , so $e(g,q) \notin (-1,1) = V$ , contradicting $e[O \times U] \subseteq V$ of step 2.1; the assumption of step 1.1 is therefore false, and the claim fails for $X = \mathbb{Q}$ and $Y = \mathbb{R}$ .

step 2.1step 8.3step 9.1step 6.2discharge-contradiction∎

Remarks

The obstruction is exactly the absence of compact neighbourhoods. A basic compact-open neighbourhood of a function constrains it on finitely many compact sets, and in $\mathbb{Q}$ those sets have empty interior, so they leave rational points arbitrarily close to $0$ completely unconstrained. The bump $g$ exploits one such point. In a locally compact domain no such point exists near $0$ : some compact set is a neighbourhood of $0$ , and the argument of If $X$ is a locally compact metric space then the evaluation map is continuous for the compact-open topology goes through.
Separate continuity is not at issue. For each fixed $f$ the map $x \mapsto f(x)$ is continuous, and for each fixed $x$ the map $f \mapsto f(x)$ is continuous for the compact-open topology (The evaluation map $e : C(X,Y) \times X \to Y$ , $e(f,x) = f(x)$ ). What fails above is joint continuity, and the witness is the standard warning that separate continuity in each variable does not give continuity on the product.
The bump function is built from a distance and needs no maximum or truncation. Taking $g(s) = (2/\eta)\,d(s,A)$ with $A$ the set of rationals at distance at least $\eta$ from $q$ makes $g$ Lipschitz by $|d(x,A) - d(y,A)| \le d(x,y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz alone, vanish on $A$ and hence on $K$ , and take a value at least $2$ at $q$ . Nothing about its exact shape matters.

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: 196 results over 38 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

Compact-open topology (Wikipedia) (standard reference, not scraped)
Locally compact space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §46 (standard reference, not scraped)

FALSE: the evaluation map on C(X,Y)C(X,Y)C(X,Y) with the compact-open topology is continuous for every metric XXX