Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^X$ and on $C(X,Y)$

Let $X$ be a nonempty set 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).

Uniform convergence. A sequence $(f_k)$ in $Y^X$ (The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$) converges uniformly to $f\in Y^X$ if for every real $\epsilon >0$ there is $K\in \mathbb{N}$ such that

$d(f_k(x),f(x))<\epsilon \text{for every }x\in X\text{ and every }k\ge K\mathrm{.}$

The whole content is the quantifier order: one index $K$ must serve every point of $X$ at once, whereas pointwise convergence allows $K$ to depend on the point as well as on $\epsilon$. As everywhere in this library $\mathbb{N}$ contains $0$ and a sequence is indexed from $0$ (The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$).

The topology. The topology of uniform convergence (the uniform topology) on $Y^X$ is the metric topology ${\mathcal{T}}_{\bar{\rho }}$ (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) of the uniform metric

$\bar{\rho }(f,g)={\sup }_{x\in X}\min \{d(f(x),g(x)),1\}$

of 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 $Y^X$. Its basic open sets are the balls $B_{\bar{\rho }}(f,\epsilon )$ (Open ball, closed ball and sphere in a metric space), and $Y^X$ with this topology is a metrizable space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

On $C(X,Y)$. If $X$ carries a topology, the topology of uniform convergence on $C(X,Y)$ (Continuity of a map of topological spaces at a point and globally) is the subspace topology inherited from $Y^X$ (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). It is the metric topology of the restriction of $\bar{\rho }$ to $C(X,Y)\times C(X,Y)$: the subspace topology of a metric topology is the metric topology of the subspace metric (Isometry, isometric embedding, and the subspace metric on a subset, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). So the two readings of the phrase agree, and $C(X,Y)$ carrying it is again metrizable.

The name is justified by the next item. That convergence in ${\mathcal{T}}_{\bar{\rho }}$ is exactly uniform convergence in the sense defined above is not part of the definition; it is Convergence in the uniform metric is exactly uniform convergence: one $N$ serving every point ↗, and it is what entitles the topology to the name.

$X$ is nonempty throughout. The uniform metric is defined only for nonempty $X$ (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 $Y^X$), so the topology of uniform convergence is defined only there. The notion of uniform convergence itself makes sense for $X=\mathrm{\varnothing }$ and is vacuous, every sequence converging uniformly to the unique element of $Y^{\mathrm{\varnothing }}$; nothing below uses that case.

• Uniform convergence is a property of the metric $d$, not of the topology of $Y$. Both quantifiers above are about distances. Two metrics inducing the same topology on $Y$ can disagree about which sequences of functions converge uniformly, exactly as they can disagree about which sequences are Cauchy (Topologically, uniformly and Lipschitz equivalent metrics on a set). Read uniformly convergent as an abbreviation for uniformly convergent with respect to this metric, always.

• The truncation at $1$ does not affect the notion. The uniform metric truncates distances at $1$ so that a supremum exists without a boundedness hypothesis, and the next item shows that the truncation is invisible to convergence: below the threshold $1$ the truncated and untruncated distances agree, and convergence is a statement about arbitrarily small distances.

• Uniform convergence is strictly stronger than pointwise convergence. Taking $K$ from the uniform condition serves at each individual point, so a uniformly convergent sequence converges pointwise; the converse fails, and the companion page exhibits a sequence of continuous functions on $[0,1]$ converging pointwise to $0$ with $\bar{\rho }(f_k,0)=1$ for every $k$.