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

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If $X$ is a locally compact metric space then the evaluation map is continuous for the compact-open topology

Statement

Let $(X, d)$ be a locally compact metric space (Locally compact metric space: every point has a compact neighbourhood) carrying its metric topology, let $(Y, T_{Y})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), and give $C (X, Y)$ the compact-open topology (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ ). Then 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, the product carrying the product topology (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).

No hypothesis whatever is placed on $Y$ , which is an arbitrary topological space: the argument uses only that a point of $Y$ lies in an open set. No choice principle is used.

Local compactness is not removable. This page records as a false statement that the evaluation map is continuous for every metric domain, and its witness is $X = Q$ , a metric space that is locally compact at no point.

Facts & Assumptions

Given: A locally compact metric space $(X, d)$ with its metric topology, a topological space $(Y, T_{Y})$ , the set $C (X, Y)$ with the compact-open topology, and the evaluation map $e$ .

[L1]

A map $h$ into $Y$ is continuous exactly when for every point $p$ of its domain and every open $V \subseteq Y$ with $h (p) \in V$ there is an open $U$ of the domain with $p \in U$ and $h [U] \subseteq V$ (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 $f (\overline{A}) \subseteq \overline{f (A)}$ ).

[L2]

For a two-element index set the basic product-open sets are the boxes: $O \times U$ with $O$ open in $C (X, Y)$ and $U$ open in $X$ is open in $C (X, Y) \times X$ (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, 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).

[L3]

$S (K, V) = {g \in C (X, Y) : g [K] \subseteq V}$ is open in the compact-open topology for every compact $K \subseteq X$ and open $V \subseteq Y$ (The compact-open topology on $C (X, Y)$ for a metric domain $X$ , with subbasis $S (K, V) = {f : f [K] \subseteq V}$ ).

[L4]

Local compactness at $x$ gives a real $r_{0} > 0$ such that $\overset{ˉ}{B} (x, r)$ is compact for every real $r$ with $0 < r < r_{0}$ (In a locally compact metric space every point has arbitrarily small compact closed balls, hence a neighbourhood base of compact sets, Locally compact metric space: every point has a compact neighbourhood, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L5]

A subset $W \subseteq X$ is open exactly when each of its points has a ball around it inside $W$ ; balls are open; $B (x, r) \subseteq \overset{ˉ}{B} (x, r)$ ; and $\overset{ˉ}{B} (x, r) \subseteq B (x, s)$ whenever $0 < r < s$ (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, 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).

[L6]

The minimum of a two-element set of reals exists, is one of the two elements 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).

[L7]

$f \in C (X, Y)$ is continuous, so $f^{- 1} [V]$ is open in $X$ for every open $V \subseteq Y$ (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 $f (\overline{A}) \subseteq \overline{f (A)}$ ).

Proof

technique · direct

1.1

Let $(f, x) \in C (X, Y) \times X$ and let $V \subseteq Y$ be open with $e (f, x) = f (x) \in V$ .

1.2

Local compactness gives a real $r_{0} > 0$ such that $\overset{ˉ}{B} (x, r)$ is compact for every real $r$ with $0 < r < r_{0}$ .

L4choose

2.1

$f^{- 1} [V]$ is open in $X$ and contains $x$ , so there is a real $s > 0$ with $B (x, s) \subseteq f^{- 1} [V]$ .

step 1.1L5L7choose

3.1

Put $r : = \min {s, r_{0}} / 2$ , a real with $0 < r$ , $r < r_{0}$ and $r < s$ ; then $K : = \overset{ˉ}{B} (x, r)$ is compact and $K \subseteq B (x, s) \subseteq f^{- 1} [V]$ , that is $f [K] \subseteq V$ .

step 2.1step 1.2L5L6

4.1

Hence $f \in S (K, V)$ , which is open in the compact-open topology, and $x \in B (x, r)$ , which is open in $X$ ; so $S (K, V) \times B (x, r)$ is an open subset of the product containing $(f, x)$ .

step 3.1L2L3L5

5.1

For every $(g, y) \in S (K, V) \times B (x, r)$ : $y \in B (x, r) \subseteq \overset{ˉ}{B} (x, r) = K$ and $g [K] \subseteq V$ , so $e (g, y) = g (y) \in V$ ; that is, $e [S (K, V) \times B (x, r)] \subseteq V$ .

step 3.1step 4.1L3L5

6.1

Steps 4.1 and 5.1 exhibit, for the arbitrary point $(f, x)$ and the arbitrary open $V$ containing its image, an open set of the product around $(f, x)$ mapped into $V$ ; so $e$ is continuous.

step 1.1step 4.1step 5.1L1∎

Remarks

What the compact-open topology is doing. The whole proof is the single observation that $S (K, V)$ constrains a function on the whole of the compact set $K$ , so once $K$ is a neighbourhood of $x$ the constraint survives moving the point as well as moving the function. A topology whose basic sets constrain a function at finitely many points only, such as the topology of pointwise convergence, cannot do this, and the evaluation map is in general not continuous for it.
Where local compactness is spent. Once, at step 1.2, to produce a compact neighbourhood of $x$ inside the open set $f^{- 1} [V]$ . Every metric space has arbitrarily small closed balls inside such an open set; what local compactness adds is that they may be taken compact.
The converse is not asserted. Nothing here says that continuity of the evaluation map forces $X$ to be locally compact. That direction is true for Hausdorff spaces in the general theory and is not proved in this library.

Depends on

Used by

Dependency tree · two levels

60 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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

If XX is a locally compact metric space then the evaluation map is continuous for the compact-open topology