Intopology and related branches ofmathematics,total-boundedness is a generalization ofcompactness for circumstances in which a set is not necessarilyclosed. A totally bounded set can becovered byfinitely manysubsets of every fixed “size” (where the meaning of “size” depends on the structure of theambient space).

The termprecompact (orpre-compact) is sometimes used with the same meaning, but precompact is also used to meanrelatively compact. These definitions coincide for subsets of acomplete metric space, but not in general.

Ametric space${\displaystyle (M,d)}$ istotally bounded if and only if for every real number${\displaystyle \epsilon >0}$, there exists a finite collection ofopen balls of radius${\displaystyle \epsilon }$ whose centers lie inM and whose union contains M. Equivalently, the metric spaceM is totally bounded if and only if for every${\displaystyle \epsilon >0}$, there exists afinite cover such that the radius of each element of the cover is at most${\displaystyle \epsilon }$. This is equivalent to the existence of a finiteε-net.^[1] A metric space is said to be totally bounded if every sequence admits a Cauchy subsequence; in complete metric spaces, a set is compact if and only if it is closed and totally bounded.^[2]

Each totally bounded space isbounded (as the union of finitely many bounded sets is bounded). The reverse is true for subsets ofEuclidean space (with thesubspace topology), but not in general. For example, an infinite set equipped with thediscrete metric is bounded but not totally bounded:^[3] every discrete ball of radius${\displaystyle \epsilon =1/2}$ or less is a singleton, and no finite union of singletons can cover an infinite set.

A metric appears in the definition of total boundedness only to ensure that each element of the finite cover is of comparable size, and can be weakened to that of auniform structure. A subsetS of auniform spaceX is totally bounded if and only if, for anyentourageE, there exists a finite cover ofS by subsets ofX each of whoseCartesian squares is a subset ofE. (In other words,E replaces the "size"ε, and a subset is of sizeE if its Cartesian square is a subset ofE.)^[4]

The definition can be extended still further, to any category of spaces with a notion ofcompactness andCauchy completion: a space is totally bounded if and only if its (Cauchy) completion is compact.

• Everycompact set is totally bounded, whenever the concept is defined.^{[clarification needed]}
• Every totally bounded set is bounded.
• A subset of thereal line, or more generally of finite-dimensionalEuclidean space, is totally bounded if and only if it isbounded.^[5][3]
• Theunit ball in aHilbert space, or more generally in aBanach space, is totally bounded (in the norm topology) if and only if the space has finitedimension.
• Equicontinuous bounded functions on a compact set are precompact in theuniform topology; this is theArzelà–Ascoli theorem.
• Ametric space isseparable if and only if it ishomeomorphic to a totally bounded metric space.^[3]
• The closure of a totally bounded subset is again totally bounded.^[6]

In metric spaces, a set is compact if and only if it is complete and totally bounded;^[5] without theaxiom of choice only the forward direction holds. Precompact sets share a number of properties with compact sets.

• Like compact sets, a finite union of totally bounded sets is totally bounded.
• Unlike compact sets, every subset of a totally bounded set is again totally bounded.
• The continuous image of a compact set is compact. Theuniformly continuous image of a precompact set is precompact.

Although the notion of total boundedness is closely tied to metric spaces, the greater algebraic structure of topological groups allows one to trade away someseparation properties. For example, in metric spaces, a set is compact if and only if complete and totally bounded. Under the definition below, the same holds for any topological vector space (not necessarily Hausdorff nor complete).^[6][7][8]

The generallogical form of thedefinition is: a subset${\displaystyle S}$ of a space${\displaystyle X}$ is totally bounded if and only if,given any size${\displaystyle E,}$there exists a finite cover${\displaystyle \mathcal{O}}$ of${\displaystyle S}$ such that each element of${\displaystyle \mathcal{O}}$ has size at most${\displaystyle E.}$${\displaystyle X}$ is then totally bounded if and only if it is totally bounded when considered as a subset of itself.

We adopt the convention that, for any neighborhood${\displaystyle U\subseteq X}$ of the identity, a subset${\displaystyle S\subseteq X}$ is called (left)${\displaystyle U}$-small if and only if${\displaystyle (-S)+S\subseteq U.}$^[6] A subset${\displaystyle S}$ of atopological group${\displaystyle X}$ is (left)totally bounded if it satisfies any of the following equivalent conditions:

1. Definition: For any neighborhood${\displaystyle U}$ of the identity${\displaystyle 0,}$ there exist finitely many${\displaystyle x_1,\dots ,x_n\in X}$ such that$S\subseteq \bigcup _{j=1}^n\left(x_j+U\right):=\left(x_1+U\right)+\cdots +\left(x_n+U\right).$
2. For any neighborhood${\displaystyle U}$ of${\displaystyle 0,}$ there exists a finite subset${\displaystyle F\subseteq X}$ such that${\displaystyle S\subseteq F+U}$ (where the right hand side is theMinkowski sum${\displaystyle F+U:=\{f+u:f\in F,u\in U\}}$).
3. For any neighborhood${\displaystyle U}$ of${\displaystyle 0,}$ there exist finitely many subsets${\displaystyle B_1,\dots ,B_n}$ of${\displaystyle X}$ such that${\displaystyle S\subseteq B_1\cup \cdots \cup B_n}$ and each${\displaystyle B_j}$ is${\displaystyle U}$-small.^[6]
4. For any givenfilter subbase${\displaystyle \mathcal{B}}$ of the identity element'sneighborhood filter${\displaystyle \mathcal{N}}$ (which consists of all neighborhoods of${\displaystyle 0}$ in${\displaystyle X}$) and for every${\displaystyle B\in \mathcal{B},}$ there exists a cover of${\displaystyle S}$ by finitely many${\displaystyle B}$-small subsets of${\displaystyle X.}$^[6]
5. ${\displaystyle S}$ isCauchy bounded: for every neighborhood${\displaystyle U}$ of the identity and everycountably infinite subset${\displaystyle I}$ of${\displaystyle S,}$ there exist distinct${\displaystyle x,y\in I}$ such that${\displaystyle x-y\in U.}$^[6] (If${\displaystyle S}$ is finite then this condition issatisfied vacuously).
6. Any of the following three sets satisfies (any of the above definitions of) being (left) totally bounded:
   1. Theclosure${\displaystyle \overline{S}={\mathrm{cl}}_{X}S}$ of${\displaystyle S}$ in${\displaystyle X.}$^[6]
      • This set being in the list means that the following characterization holds:${\displaystyle S}$ is (left) totally bounded if and only if${\displaystyle {\mathrm{cl}}_{X}S}$ is (left) totally bounded (according to any of the defining conditions mentioned above). The same characterization holds for the other sets listed below.
   2. The image of${\displaystyle S}$ under thecanonical quotient${\displaystyle X\to X/\overline{\{0\}},}$ which is defined by${\displaystyle x\mapsto x+\overline{\{0\}}}$ (where${\displaystyle 0}$ is the identity element).
   3. The sum${\displaystyle S+{\mathrm{cl}}_X\{0\}.}$^[9]

The termpre-compact usually appears in the context of Hausdorff topological vector spaces.^[10][11] In that case, the following conditions are also all equivalent to${\displaystyle S}$ being (left) totally bounded:

7. In thecompletion${\displaystyle \hat{X}}$ of${\displaystyle X,}$ the closure${\displaystyle {\mathrm{cl}}_{\hat{X}}S}$ of${\displaystyle S}$ is compact.^[10][12]
8. Every ultrafilter on${\displaystyle S}$ is aCauchy filter.

The definition ofright totally bounded is analogous: simply swap the order of the products.

Condition 4 implies any subset of${\displaystyle {\mathrm{cl}}_X\{0\}}$ is totally bounded (in fact, compact; see§ Comparison with compact sets above). If${\displaystyle X}$ is not Hausdorff then, for example,${\displaystyle \{0\}}$ is a compact complete set that is not closed.^[6]

Any topological vector space is an abelian topological group under addition, so the above conditions apply. Historically, statement 6(a) was the first reformulation of total boundedness fortopological vector spaces; it dates to a 1935 paper of John von Neumann.^[13]

This definition has the appealing property that, in alocally convex space endowed with theweak topology, the precompact sets are exactly thebounded sets.

For separable Banach spaces, there is a nice characterization of the precompact sets (in the norm topology) in terms of weakly convergent sequences of functionals: if${\displaystyle X}$ is a separable Banach space, then${\displaystyle S\subseteq X}$ is precompact if and only if everyweakly convergent sequence of functionals convergesuniformly on${\displaystyle S.}$^[14]

• Thebalanced hull of a totally bounded subset of a topological vector space is again totally bounded.^[6][15]
• TheMinkowski sum of two compact (totally bounded) sets is compact (resp. totally bounded).
• In a locally convex (Hausdorff) space, theconvex hull and thedisked hull of a totally bounded set${\displaystyle K}$ is totally bounded if and only if${\displaystyle K}$ is complete.^[16]

• Compact space
• Locally compact space
• Measure of non-compactness
• Orthocompact space
• Paracompact space
• Relatively compact subspace

• Uniform spaces
• Metric geometry
• Topology
• Functional analysis
• Compactness (mathematics)

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