TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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For a metric domain and a metric target the compact-open topology on $C (X, Y)$ is the topology of compact convergence

Statement

Let $(X, d_{X})$ and $(Y, d)$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), each carrying its metric topology, and let $C (X, Y)$ be the set of continuous maps $X \to Y$ (Continuity of a map of topological spaces at a point and globally). Then the compact-open topology (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ ) and the topology of compact convergence (The topology of compact convergence on $C (X, Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ ) on $C (X, Y)$ are the same topology.

Both halves are proved by exhibiting, around each point of a generating set of one topology, a generating set of the other inside it. No choice principle is used: the only cover produced below is indexed by pairs, so the indexed form of compactness (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) returns everything that is needed.

The metric hypothesis on the target is not removable by anything on this page. The topology of compact convergence is defined only for a metric target, since its basic sets are written with a distance in $Y$ ; the compact-open topology needs only the open sets of $Y$ . The theorem is a statement about the case where both are defined.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ and $(Y, d)$ with their metric topologies, the set $C (X, Y)$ of continuous maps, the sets $S (K, V)$ of The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ , the sets $B_{K} (f, ε)$ of The topology of compact convergence on $C (X, Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ , and the topologies $T_{c o}$ and $T_{c c}$ they respectively generate.

[L1]

The sets $S (K, V)$ , for $K \subseteq X$ compact and $V \subseteq Y$ open, are a subbasis for $T_{c o}$ ; $S (\emptyset, V) = S (K, Y) = C (X, Y)$ ; and finite intersections of subbasic sets are open (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ , 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, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L2]

The sets $B_{K} (f, ε)$ are a basis for $T_{c c}$ , and $B_{\emptyset} (f, ε) = C (X, Y)$ ; facts (U1), (U2) and (U3) of The topology of compact convergence on $C (X, Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ are available, in particular the existence of $\max_{x \in K} d (f (x), g (x))$ for $f, g \in C (X, Y)$ and nonempty compact $K$ (The topology of compact convergence on $C (X, Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ ).

[L3]

A topology generated by a family $S$ is contained in every topology containing $S$ , and a set all of whose points lie in a basic set inside it is a union of basic sets, hence open (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).

[L4]

The continuous image of a compact subset is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset, claim 2).

[L5]

For nonempty $A \subseteq Y$ the distance $y \mapsto d (y, A)$ is defined, is $1$ -Lipschitz and hence continuous, and satisfies $d (y, A) \leq d (y, z)$ for every $z \in A$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, 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, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, Greatest lower bound (infimum)).

[L6]

A continuous real-valued function on a nonempty compact metric space attains a least and a greatest value, and the restriction of a continuous map to a metric subspace is continuous (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Isometry, isometric embedding, and the subspace metric on a subset, 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).

[L7]

$\overset{ˉ}{B} (a, r)$ is closed and $B (a, r)$ is open in $(X, d_{X})$ ; relative openness in a subspace is tracing, so a closed subset of $X$ traces to a closed subset of any metric subspace, and a closed subset of a compact metric space is compact (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, 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, A closed subset of a compact metric space is compact, Open ball, closed ball and sphere in a metric space, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L8]

$K$ compact and $(U_{i})_{i \in I}$ open in $X$ with $K \subseteq ⋃_{i} U_{i}$ give $n \in N$ and indices $i_{0}, \dots, i_{n} \in I$ with $K \subseteq U_{i_{0}} \cup \dots \cup U_{i_{n}}$ , unless $K = \emptyset$ (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, claim 3).

[L9]

The minimum of a two-element set of reals exists and is one of the two elements; balls are open and $y \in B (y, s)$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Open ball, closed ball and sphere in a metric space, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Proof

technique · direct

1.1L1L2L3suffices: each subbasic compact open set is a union of basic compact convergence sets

First half: let $K \subseteq X$ be compact, let $V \subseteq Y$ be open and let $f \in S (K, V)$ ; it suffices to produce a real $ε > 0$ with $B_{K} (f, ε) \subseteq S (K, V)$ , since then every point of $S (K, V)$ lies in a basic set of $T_{c c}$ inside it.

1.2L1L2L3suffices: each basic compact convergence set is a union of compact open sets

Second half: let $K \subseteq X$ be compact, let $f_{0} \in C (X, Y)$ , let $ε_{0} > 0$ be real and let $g \in B_{K} (f_{0}, ε_{0})$ ; it suffices to produce a finite intersection of sets $S (K^{'}, V^{'})$ containing $g$ and contained in $B_{K} (f_{0}, ε_{0})$ .

2.1step 1.1L1L2

In step 1.1, if $K = \emptyset$ or $V = Y$ then $S (K, V) = C (X, Y)$ , which is open in $T_{c c}$ , and $ε : = 1$ serves since $B_{K} (f, 1) \subseteq C (X, Y)$ ; so assume $K \neq \emptyset$ and $V \neq Y$ , whence $Y ∖ V$ is nonempty.

2.2step 1.2L1L2

In step 1.2, if $K = \emptyset$ then $B_{K} (f_{0}, ε_{0}) = C (X, Y) = S (\emptyset, Y)$ , which is subbasic and hence open in $T_{c o}$ ; so assume $K \neq \emptyset$ .

3.1step 2.1L4L5

Under step 2.1: $f [K]$ is a nonempty compact subset of $Y$ , and the function $ψ (y) : = d (y, Y ∖ V)$ is defined and continuous on $Y$ .

3.2step 2.2L2

Under step 2.2: $M : = \max_{x \in K} d (f_{0} (x), g (x))$ exists and satisfies $M < ε_{0}$ , because $g \in B_{K} (f_{0}, ε_{0})$ makes $ε_{0}$ a strict upper bound of the values; put $δ : = ε_{0} - M > 0$ .

4.1step 3.1L6choose

Under step 2.1: the restriction of $ψ$ to the nonempty compact metric subspace $f [K]$ is continuous, so it attains a least value $ε : = ψ (y_{0})$ at some $y_{0} \in f [K]$ , and $ε \leq ψ (y)$ for every $y \in f [K]$ .

4.2step 3.2constructL6L9

Under step 2.2: let $P$ be the set of pairs $(a, r)$ with $a \in K$ , $r > 0$ real and $g [\overset{ˉ}{B} (a, r) \cap K] \subseteq B (g (a), δ / 4)$ ; the family $(B (a, r))_{(a, r) \in P}$ consists of open subsets of $X$ and covers $K$ , since continuity of $g$ at $a \in K$ gives $s > 0$ with $g [B (a, s)] \subseteq B (g (a), δ / 4)$ and then $r : = s / 2$ satisfies $\overset{ˉ}{B} (a, r) \subseteq B (a, s)$ , so $(a, r) \in P$ and $a \in B (a, r)$ .

5.1step 4.1L5L9

Under step 2.1: $ε > 0$ , because $y_{0} \in f [K] \subseteq V$ with $V$ open gives a real $s > 0$ with $B (y_{0}, s) \subseteq V$ , so every $z \in Y ∖ V$ satisfies $d (y_{0}, z) \geq s$ , making $s$ a lower bound of the distances from $y_{0}$ to $Y ∖ V$ and hence $ψ (y_{0}) \geq s > 0$ .

5.2step 4.2L8

Under step 2.2: since $K \neq \emptyset$ is compact, there are $n \in N$ and pairs $(a_{0}, r_{0}), \dots, (a_{n}, r_{n}) \in P$ with $K \subseteq B (a_{0}, r_{0}) \cup \dots \cup B (a_{n}, r_{n})$ ; each index is a pair, so the centres and radii come back with the indices and nothing is selected.

6.1step 4.1step 5.1L5

Under step 2.1: for $u \in B_{K} (f, ε)$ and $x \in K$ we have $d (f (x), u (x)) < ε \leq ψ (f (x))$ by step 4.1, since $f (x) \in f [K]$ ; were $u (x) \in Y ∖ V$ , the distance $d (f (x), u (x))$ would be one of the distances from $f (x)$ to $Y ∖ V$ and hence at least $ψ (f (x))$ , which it is not; so $u (x) \in V$ .

6.2step 5.2L7L9

Under step 2.2: for $j \leq n$ put $K_{j} : = \overset{ˉ}{B} (a_{j}, r_{j}) \cap K$ and $V_{j} : = B (g (a_{j}), δ / 2)$ ; each $K_{j}$ is closed in the compact metric space $(K, d_{K})$ , being the trace on $K$ of the closed set $\overset{ˉ}{B} (a_{j}, r_{j})$ , hence is compact, and each $V_{j}$ is open in $Y$ .

7.1step 1.1step 2.1step 6.1L3

Under step 2.1: step 6.1 holds for every $x \in K$ , so $u [K] \subseteq V$ and $u \in S (K, V)$ ; hence $B_{K} (f, ε) \subseteq S (K, V)$ , which is what step 1.1 required, and every $S (K, V)$ is open in $T_{c c}$ , so $T_{c o} \subseteq T_{c c}$ .

7.2step 4.2step 5.2step 6.2L1

Under step 2.2: $g \in S (K_{j}, V_{j})$ for every $j \leq n$ , since $K_{j} \subseteq \overset{ˉ}{B} (a_{j}, r_{j}) \cap K$ and $(a_{j}, r_{j}) \in P$ give $g [K_{j}] \subseteq B (g (a_{j}), δ / 4) \subseteq V_{j}$ ; so $g \in O : = S (K_{0}, V_{0}) \cap \dots \cap S (K_{n}, V_{n})$ , a finite intersection of subbasic sets and hence open in $T_{c o}$ .

8.1step 5.2step 6.2step 7.2L9

Under step 2.2: let $h \in O$ and $x \in K$ ; step 5.2 gives $j \leq n$ with $x \in B (a_{j}, r_{j})$ , so $x \in K_{j}$ , whence $d (h (x), g (a_{j})) < δ / 2$ and $d (g (x), g (a_{j})) < δ / 4$ , so $d (h (x), g (x)) < δ / 2 + δ / 4 < δ$ by the triangle inequality.

9.1step 3.2step 8.1

Under step 2.2: therefore $d (f_{0} (x), h (x)) \leq d (f_{0} (x), g (x)) + d (g (x), h (x)) < M + δ = ε_{0}$ for every $x \in K$ , that is $h \in B_{K} (f_{0}, ε_{0})$ ; so $O \subseteq B_{K} (f_{0}, ε_{0})$ , which is what step 1.2 required.

10.1step 2.2step 9.1L2L3

By step 9.1 every point of every basic set of $T_{c c}$ is interior to it in $T_{c o}$ , so every basic set of $T_{c c}$ is open in $T_{c o}$ , and since those basic sets generate, $T_{c c} \subseteq T_{c o}$ .

11.1step 7.1step 10.1∎

With step 7.1 the two inclusions give $T_{c o} = T_{c c}$ .

Remarks

The first half is where the compactness of the image is used, through the extreme value theorem applied to the distance to the closed set $Y ∖ V$ . Without it the number $ε$ of step 4.1 would be an infimum that might be $0$ , and the conclusion would fail: the set $S (K, V)$ genuinely needs $f [K]$ to sit at a positive distance from the complement of $V$ , and that is a consequence of compactness, not of openness of $V$ .
The cases $K = \emptyset$ and $V = Y$ are disposed of first for a reason. In both, $S (K, V)$ is the whole space and the distance $d (f [K], Y ∖ V)$ is not defined — in the first because there is no point of $K$ to measure from, in the second because $Y ∖ V$ is empty and this library defines the distance to a set only for a nonempty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).
The second half is a covering argument and is where the compact-open topology earns its subbasis. A single set $S (K, V)$ cannot control $g$ uniformly on $K$ ; what does is a finite family of sets $S (K_{j}, V_{j})$ on which $g$ varies by less than a quarter of the slack. That the pieces $K_{j}$ are again compact is A closed subset of a compact metric space is compact applied inside $K$ .
This is the theorem that lets the rest of the page use whichever description is convenient. The comparison of the three topologies is proved against compact convergence, while the evaluation map and the exponential law are proved against the compact-open topology, and the two are the same topology whenever both are defined.

For a metric domain and a metric target the compact-open topology on $C (X, Y)$ is the topology of compact convergence

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a metric domain and a metric target the compact-open topology on C(X,Y)C(X,Y) is the topology of compact convergence

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

For a metric domain and a metric target the compact-open topology on $C (X, Y)$ is the topology of compact convergence