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 $\varepsilon > 0$ there is $K \in \mathbb{N}$ such that

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

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 $\varepsilon$. 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 $B_{\bar\rho}(f,\varepsilon)$ (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 $X = \varnothing$ and is vacuous, every sequence converging uniformly to the unique element of $Y^{\varnothing}$; 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 $k$.