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 $Y^X$

Let $X$ be a nonempty set, let $(Y,d)$ be a metric space (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and write

$\bar{d}(u,v):=\min \{d(u,v),1\}(u,v\in Y),$

which is a metric on $Y$ with $\bar{d}\le 1$ everywhere ($\min (d,1)$ and $d\mathrm{/}(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology, claims 1 and 2). For $f,g\in Y^X$ (The topology of pointwise convergence on $Y^X$, which is the product topology, and its restriction to $C(X,Y)$) put

$R(f,g):=\{\bar{d}(f(x),g(x)):x\in X\}\subseteq \mathbb{R},\bar{\rho }(f,g):=\sup R(f,g)\mathrm{.}$

This is well defined: $R(f,g)$ is nonempty because $X$ is, and $1$ is an upper bound of it, so the least upper bound exists (Complete ordered field (least-upper-bound property)) and is unique (Suprema and infima are unique).

Then $\bar{\rho }$ is a metric on $Y^X$ (Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric), the uniform metric, and $\bar{\rho }(f,g)\le 1$ for all $f,g$.

Both hypotheses are used and neither is decoration. Nonemptiness of $X$ is what makes $R(f,g)$ nonempty; for $X=\mathrm{\varnothing }$ the set $Y^X$ has a single element, but $\sup \mathrm{\varnothing }$ is undefined under the real-valued supremum convention used here (Conventions: $\sup \mathrm{\varnothing }$, unbounded sets, and the extended reals). The extended real line is introduced later and is not the codomain of this metric. Truncating $d$ at $1$ is what makes $R(f,g)$ bounded above with no boundedness hypothesis on $f$ and $g$; that is the whole reason the truncation is there.

• This replaces the published supremum metric here; it does not generalise it. The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set is the metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ on the bounded real-valued functions on a nonempty set, and its Statement is about that set of functions and that target. It cannot carry $C(X,Y)$ for a metric target $Y$, and it cannot carry unbounded functions at all. The metric above is defined on all of $Y^X$, for an arbitrary metric target, at the cost of truncating distances at $1$. Where both are defined the two are different functions: they disagree at every pair whose distance somewhere exceeds $1$. The companion page verifies, on $C([0,1],\mathbb{R})$, that they are nevertheless uniformly equivalent there and so induce the same topology; no wider claim than that is made here.

• The supremum need not be attained, so $\bar{\rho }(f,g)$ is a supremum and not a maximum. It is attained when $X$ is a nonempty compact metric space and $f,g$ are continuous, by the extreme value theorem (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value); nothing below assumes it in general.

• Truncation is what removes the boundedness hypothesis, and it is topologically free. $\min \{d,1\}$ is uniformly equivalent to $d$, hence topologically equivalent to it ($\min (d,1)$ and $d\mathrm{/}(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology, claim 3), so nothing about the topology of $Y$ is changed by the truncation. What is changed is the numerical value of the distance, and every statement below that compares $\bar{\rho }$ with an untruncated distance says at which threshold the two agree.