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

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On $C(X,Y)$ with $X$ and $Y$ metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence

Statement

Let $(X,d_X)$ be a nonempty metric space and let $(Y,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric), each carrying its metric topology, and write $\mathcal{T}_{\mathrm{pt}}$ , $\mathcal{T}_{\mathrm{cc}}$ and $\mathcal{T}_{\mathrm{u}}$ for the topologies of pointwise convergence (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C(X,Y)$ ), of compact convergence (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ ) and of uniform convergence (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C(X,Y)$ ) on $C(X,Y)$ . Then

$\mathcal{T}_{\mathrm{pt}} \;\subseteq\; \mathcal{T}_{\mathrm{cc}} \;\subseteq\; \mathcal{T}_{\mathrm{u}} ,$

that is, uniform convergence is finer than compact convergence, which is finer than pointwise convergence (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison for finer). The middle topology is also the compact-open topology (For a metric domain and a metric target the compact-open topology on $C(X,Y)$ is the topology of compact convergence, The compact-open topology on $C(X,Y)$ for a metric domain $X$ , with subbasis $S(K,V) = \{f : f[K] \subseteq V\}$ ).

No strictness is claimed. The theorem asserts the two inclusions and nothing more; that neither reverses in general is witnessed on the companion page, by a sequence converging pointwise but not on compact sets and by a sequence converging on compact sets but not uniformly. Those witnesses are not prerequisites of this theorem. Nonemptiness of $X$ is inherited from For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar\rho(f,g) = \sup_{x} \min\{d(f(x),g(x)), 1\}$ is a metric on $Y^{X}$ , which defines the uniform metric only there. No choice principle is used.

Facts & Assumptions

Given: A nonempty metric space $(X,d_X)$ , a metric space $(Y,d)$ , the set $C(X,Y)$ of continuous maps, and on it the three topologies named in the Statement; $\bar d = \min\{d,1\}$ and $\bar\rho$ are as in For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar\rho(f,g) = \sup_{x} \min\{d(f(x),g(x)), 1\}$ is a metric on $Y^{X}$ .

[L1]

$\mathcal{T}_{\mathrm{pt}}$ is generated by the sets $T(x,V) := \{\, g \in C(X,Y) : g(x) \in V \,\}$ for $x \in X$ and $V \subseteq Y$ open, these being the traces on $C(X,Y)$ of the subbasic sets of the product topology (The topology of pointwise convergence on $Y^{X}$ , which is the product topology, and its restriction to $C(X,Y)$ , 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, Basis and subbasis for a topology, and the topology generated by a family of sets).

[L2]

The sets $B_K(f,\varepsilon)$ are a basis for $\mathcal{T}_{\mathrm{cc}}$ , with $B_{\varnothing}(f,\varepsilon) = C(X,Y)$ , and $\max_{x \in K} d(f(x),g(x))$ exists for $f, g \in C(X,Y)$ and nonempty compact $K$ , by fact (U3) there (The topology of compact convergence on $C(X,Y)$ for metric $X$ and $Y$ : uniform convergence on each compact subset of $X$ ).

[L3]

$\mathcal{T}_{\mathrm{u}}$ is the metric topology on $C(X,Y)$ of the restriction of $\bar\rho$ , whose balls are the traces $B_{\bar\rho}(g,\delta) \cap C(X,Y)$ ; balls are open and $g \in B_{\bar\rho}(g,\delta)$ (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C(X,Y)$ , 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).

[L4]

A topology generated by a family is contained in every topology containing that family, and a set all of whose points lie in a member of a basis inside it is 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).

[L5]

A one-point subset of a metric space is compact, the one-point metric space being compact (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).

[L6]

A subset $V \subseteq Y$ is open exactly when each of its points has a ball around it inside $V$ (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, Open ball, closed ball and sphere in a metric space).

[L7]

$\bar d \le d$ and $\bar d \le 1$ ; if $\bar d(u,v) < 1$ then $\bar d(u,v) = d(u,v)$ ; and $\bar d(g(x),h(x)) \le \bar\rho(g,h)$ for every $x$ ( $\min(d,1)$ and $d/(1+d)$ are metrics uniformly equivalent to $d$ , so every metric space carries a bounded metric with the same topology, For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar\rho(f,g) = \sup_{x} \min\{d(f(x),g(x)), 1\}$ is a metric on $Y^{X}$ , Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct

1.1

For the first inclusion, let $x \in X$ , let $V \subseteq Y$ be open and let $f \in T(x,V)$ .

1.2

For the second inclusion, let $K \subseteq X$ be compact, let $f_0 \in C(X,Y)$ , let $\varepsilon > 0$ be real and let $g \in B_K(f_0,\varepsilon)$ .

2.1

Under step 1.1: $f(x) \in V$ with $V$ open, so there is a real $\varepsilon > 0$ with $B(f(x),\varepsilon) \subseteq V$ ; and $\{x\}$ is a compact subset of $X$ .

step 1.1L5L6choose

2.2

Under step 1.2: if $K = \varnothing$ then $B_K(f_0,\varepsilon) = C(X,Y)$ , which is open in $\mathcal{T}_{\mathrm{u}}$ ; so assume $K \ne \varnothing$ , put $M := \max_{x \in K} d(f_0(x),g(x))$ , which exists and satisfies $M < \varepsilon$ , and put $\delta := \min\{\varepsilon - M,\ 1\}/2$ , a real with $0 < \delta \le 1/2$ and $2\delta \le \varepsilon - M$ .

step 1.2L2L3L7

3.1

Under step 1.1: $B_{\{x\}}(f,\varepsilon) \subseteq T(x,V)$ , since $g \in B_{\{x\}}(f,\varepsilon)$ means $d(f(x),g(x)) < \varepsilon$ , that is $g(x) \in B(f(x),\varepsilon) \subseteq V$ .

step 2.1L2

3.2

Under step 1.2 with $K \ne \varnothing$ : let $h \in B_{\bar\rho}(g,\delta) \cap C(X,Y)$ ; then for every $x \in X$ we have $\bar d(g(x),h(x)) \le \bar\rho(g,h) < \delta \le 1/2 < 1$ , hence $d(g(x),h(x)) = \bar d(g(x),h(x)) < \delta$ .

step 2.2L7

4.1

Under step 1.1: $f$ lies in the basic set $B_{\{x\}}(f,\varepsilon)$ of $\mathcal{T}_{\mathrm{cc}}$ , which by step 3.1 lies inside $T(x,V)$ ; as $f$ was an arbitrary point of $T(x,V)$ , the set $T(x,V)$ is open in $\mathcal{T}_{\mathrm{cc}}$ .

step 2.1step 3.1L2L4

4.2

Under step 1.2 with $K \ne \varnothing$ : for $x \in K$ we get $d(f_0(x),h(x)) \le d(f_0(x),g(x)) + d(g(x),h(x)) < M + \delta \le M + (\varepsilon - M) = \varepsilon$ , so $h \in B_K(f_0,\varepsilon)$ ; hence $B_{\bar\rho}(g,\delta) \cap C(X,Y) \subseteq B_K(f_0,\varepsilon)$ .

step 2.2step 3.2

5.1

By step 4.1 every generating set of $\mathcal{T}_{\mathrm{pt}}$ lies in $\mathcal{T}_{\mathrm{cc}}$ , so $\mathcal{T}_{\mathrm{pt}} \subseteq \mathcal{T}_{\mathrm{cc}}$ .

step 4.1L1L4

5.2

By steps 2.2 and 4.2 every point of every basic set of $\mathcal{T}_{\mathrm{cc}}$ has a ball of the uniform metric around it inside that set, so every such basic set is open in $\mathcal{T}_{\mathrm{u}}$ and hence $\mathcal{T}_{\mathrm{cc}} \subseteq \mathcal{T}_{\mathrm{u}}$ .

step 2.2step 4.2L2L3L4

6.1

Steps 5.1 and 5.2 are the two asserted inclusions, and the middle topology is the compact-open topology by For a metric domain and a metric target the compact-open topology on $C(X,Y)$ is the topology of compact convergence.

step 5.1step 5.2∎

Remarks

Where each inclusion comes from. The first is the observation that a one-point set is compact, so every constraint of the pointwise topology is already a constraint of the compact-convergence topology. The second is that $X$ itself need not be compact: a uniform bound over all of $X$ is at least as strong as a uniform bound over one compact set.
The truncation threshold appears once, in step 3.2, where the uniform distance has to be pushed below $1$ before it can be read as an untruncated distance. That costs nothing, since $\delta$ is being made small in any case.
Nonemptiness of $X$ is a hypothesis about the uniform metric only. The inclusion $\mathcal{T}_{\mathrm{pt}} \subseteq \mathcal{T}_{\mathrm{cc}}$ needs no such hypothesis; it is stated with it only because the theorem names all three topologies at once.
When the outer two coincide. If $X$ is itself compact, then $K = X$ is an admissible compact set and the chain collapses at its right end: compact convergence and uniform convergence agree on $C(X,Y)$ . The companion page works that case on $[0,1]$ and separates the two on $\mathbb{R}$ .

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 127 results over 22 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 convergence (Wikipedia) (standard reference, not scraped)
Uniform convergence (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §46 (standard reference, not scraped)

On C(X,Y)C(X,Y)C(X,Y) with XXX and YYY metric, uniform convergence is finer than compact convergence, which is finer than pointwise convergence