Finite $\varepsilon$-net and totally bounded 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 $\varepsilon$ be a real with $\varepsilon > 0$.

• A finite $\varepsilon$-net for $(X,d)$ is a finite subset $F \subseteq X$ with $X = \bigcup_{y \in F} B(y, \varepsilon) ,$ the balls being those of $(X,d)$ (Open ball, closed ball and sphere in a metric space). Finite is the listing form fixed in Open cover, subcover, compact metric space, and compact subset of a metric space: $F = \emptyset$, or $F = \{y_0, \dots, y_m\}$ for some $m \in \mathbb{N}$ and points $y_0, \dots, y_m \in X$.
• $(X,d)$ is totally bounded when it has a finite $\varepsilon$-net for every real $\varepsilon > 0$.
• A subset $A \subseteq X$ is totally bounded when the metric subspace $(A,d_A)$ is (Isometry, isometric embedding, and the subspace metric on a subset); its nets are then finite subsets of $A$ and its balls are the balls $B_A(a,\varepsilon)$ of the subspace.

The empty space is totally bounded, the empty net serving for every $\varepsilon$, since a union over no indices is empty. Every space listed as $\{x_0, \dots, x_n\}$ is totally bounded too, itself being an $\varepsilon$-net for every $\varepsilon$.

The centres are required to lie in the space. Writing the condition with centres in $X$ and balls of $X$ is what makes total boundedness a property of the metric space $(X,d)$ alone, matching the treatment of compactness in Open cover, subcover, compact metric space, and compact subset of a metric space. For a subset $A$ this matters: the nets of $(A,d_A)$ consist of points of $A$, not of nearby points of the ambient space.

Total boundedness is stronger than boundedness and is not the same thing. A totally bounded space is bounded in the sense of Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space — that is claim 1 of A totally bounded metric space is bounded, every subspace of a totally bounded space is totally bounded, and the closure of a totally bounded subset is totally bounded — and the converse fails, as FALSE: a bounded metric space is totally bounded records. Boundedness asks for one ball containing the space; total boundedness asks for finitely many balls of every prescribed radius, and it is the second condition that controls how spread out the space is at small scales.

Why $\varepsilon$ ranges over the reals here. Convergence and the Cauchy condition are tested against rational $\varepsilon$ in this library (Convergence of a sequence in a metric space: $x_k \to x$ iff $d(x_k, x) \to 0$ in $\mathbb{R}$, Cauchy sequence in a metric space), because that is how Limits and Cauchy sequences of reals is written; total boundedness is not a limit condition and is stated for real $\varepsilon > 0$ directly. Nothing turns on the difference: a net for a rational $\varepsilon' \le \varepsilon$ is a net for $\varepsilon$, since $B(y,\varepsilon') \subseteq B(y,\varepsilon)$.

A net is not unique and is not part of the data. Total boundedness asserts that nets exist; it names none. Producing one net for each $\varepsilon$ simultaneously, as a function of $\varepsilon$, is a further act of selection, and where a proof needs that function it says so and pays for it — see A complete, totally bounded metric space is compact, proved from countable choice used exactly once and A compact metric space has a countable dense subset, by countable choice, each of which spends the Axiom of Countable Choice (The Axiom of Countable Choice ($\mathrm{AC}_\omega$)) exactly once and at exactly that point.