Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and let $A, B \subseteq X$.

Bounded subset. $A$ is bounded if $A = \emptyset$ or there are $x_0 \in X$ and a real $r > 0$ with $A \subseteq B(x_0, r)$ (Open ball, closed ball and sphere in a metric space). The space $(X,d)$ is a bounded metric space if $X$ is a bounded subset of itself.

Diameter, for nonempty bounded $A$ only. Suppose $A$ is nonempty and bounded, and put

$D(A) := \{\, d(a,b) : a, b \in A \,\} \subseteq \mathbb{R}.$

Then $D(A)$ is nonempty, since $A$ is, and it is bounded above: fixing $x_0$ and $r$ with $A \subseteq B(x_0,r)$, every $a, b \in A$ satisfy $d(a,b) \le d(a,x_0) + d(x_0,b) < r + r$ by the triangle inequality, symmetry (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) and addition of inequalities (Order is preserved by adding a constant and by adding inequalities, Ordered field), so $r + r$ is an upper bound of $D(A)$ (Lower bound, bounded below, bounded set). Hence $D(A)$ has a least upper bound in $\mathbb{R}$ by the least-upper-bound property (Complete ordered field (least-upper-bound property)), and that bound is unique (Suprema and infima are unique). Define

$\operatorname{diam}(A) := \sup D(A).$

Distance from a point to a set, for nonempty $A$ only. Let $x \in X$ and let $A$ be nonempty, and put $E(x,A) := \{\, d(x,a) : a \in A \,\}$. Then $E(x,A)$ is nonempty and bounded below by $0$, since a metric is nonnegative (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Lower bound, bounded below, bounded set), so it has a greatest lower bound (Every nonempty set bounded below has an infimum, Greatest lower bound (infimum)), unique by Suprema and infima are unique. Define

$d(x,A) := \inf E(x,A).$

Distance between two sets, for nonempty $A$ and $B$ only. Put $E(A,B) := \{\, d(a,b) : a \in A,\ b \in B \,\}$, again nonempty and bounded below by $0$, and define

$d(A,B) := \inf E(A,B).$

Every one of the three scope restrictions is load bearing. In this library $\sup$ and $\inf$ denote real numbers and are written only after existence has been established; the extended real line is introduced on a later page and is not used for the suprema and infima taken here, and no convention $\sup \emptyset = -\infty$ is in force in this development (Conventions: $\sup \emptyset$, unbounded sets, and the extended reals). Accordingly:

• $\operatorname{diam}(A)$ is defined exactly when $A$ is nonempty and bounded. It is not defined for $A = \emptyset$, and it is not defined, not even as an infinite value, for an unbounded $A$.
• $d(x,A)$ is defined exactly when $A \ne \emptyset$, and $d(A,B)$ exactly when both $A$ and $B$ are nonempty. No boundedness is needed for these two, because $0$ is always a lower bound.

• Diameter and the distance functions are nonnegative. For nonempty bounded $A$ and any $a \in A$ we have $0 = d(a,a) \in D(A)$, so $\operatorname{diam}(A) \ge 0$; and $d(x,A) \ge 0$, $d(A,B) \ge 0$ because $0$ is a lower bound of the sets they are infima of (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Greatest lower bound (infimum)).
• $d(A,B)$ is not a metric on the nonempty subsets of $X$. It is symmetric and vanishes on $A = B$, but two distinct disjoint sets can be at distance $0$, so the separation axiom (M1) of Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric fails; the witness is on the companion page. The letter $d$ is reused for three different functions here, the metric, the point-to-set distance and the set-to-set distance, only because the arguments make the intended one unambiguous.
• $d(x,A)$ is the special case $d(\{x\},A)$, since $E(\{x\},A) = E(x,A)$, and the two infima therefore agree by uniqueness (Suprema and infima are unique).