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

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 ${\mathcal{T}}_{\mathrm{p}\mathrm{t}}$, ${\mathcal{T}}_{\mathrm{c}\mathrm{c}}$ 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

${\mathcal{T}}_{\mathrm{p}\mathrm{t}}\subseteq {\mathcal{T}}_{\mathrm{c}\mathrm{c}}\subseteq {\mathcal{T}}_{\mathrm{u}},$

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.

• 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 ${\mathcal{T}}_{\mathrm{p}\mathrm{t}}\subseteq {\mathcal{T}}_{\mathrm{c}\mathrm{c}}$ 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 $\mathbb{R}$.