Totally bounded space

[0, 1]² is a totally bounded space because for every ε > 0, the unit square can be covered by finitely many open discs of radius ε.

A metric space ${\displaystyle (M,d)}$ is totally bounded if and only if for every real number ${\displaystyle \varepsilon >0}$, there exists a finite collection of open balls of radius ${\displaystyle \varepsilon }$ whose centers lie in M and whose union contains M. Equivalently, the metric space M is totally bounded if and only if for every ${\displaystyle \varepsilon >0}$, there exists a finite cover such that the radius of each element of the cover is at most ${\displaystyle \varepsilon }$. This is equivalent to the existence of a finite ε-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 is bounded (as the union of finitely many bounded sets is bounded). The reverse is true for subsets of Euclidean space (with the subspace topology), but not in general. For example, an infinite set equipped with the discrete metric is bounded but not totally bounded:[3] every discrete ball of radius ${\displaystyle \varepsilon =1/2}$ or less is a singleton, and no finite union of singletons can cover an infinite set.

• Every compact set is totally bounded, whenever the concept is defined.
• Every totally bounded set is bounded.
• A subset of the real line, or more generally of finite-dimensional Euclidean space, is totally bounded if and only if it is bounded.[5][3]
• The unit ball in a Hilbert space, or more generally in a Banach space, is totally bounded (in the norm topology) if and only if the space has finite dimension.
• Equicontinuous bounded functions on a compact set are precompact in the uniform topology; this is the Arzelà–Ascoli theorem.
• A metric space is separable if and only if it is homeomorphic to a totally bounded metric space.[3]
• The closure of a totally bounded subset is again totally bounded.[6]

Although the notion of total boundedness is closely tied to metric spaces, the greater algebraic structure of topological groups allows one to trade away some separation 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 general logical form of the definition 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 a topological 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 ${\displaystyle x_{1},\ldots ,x_{n}\in X}$ such that ${\textstyle S\subseteq \bigcup _{j=1}^{n}\left(x_{j}+U\right):=\left(x_{1}+U\right)+\cdots +\left(x_{n}+U\right).}$
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 the Minkowski 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 ${\displaystyle B_{1},\ldots ,B_{n}}$ 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 given filter subbase ${\displaystyle {\mathcal {B}}}$ of the identity element's neighborhood 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. ${\displaystyle S}$ is Cauchy bounded: for every neighborhood ${\displaystyle U}$ of the identity and every countably 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 is satisfied vacuously).
6. Any of the following three sets satisfies (any of the above definitions of) being (left) totally bounded:
   1. The closure ${\displaystyle {\overline {S}}=\operatorname {cl} _{X}S}$ 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 ${\displaystyle \operatorname {cl} _{X}S}$ 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 the canonical 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 ${\displaystyle S+\operatorname {cl} _{X}\{0\}.}$[9]

The term pre-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 the completion ${\displaystyle {\widehat {X}}}$ of ${\displaystyle X,}$ the closure ${\displaystyle \operatorname {cl} _{\widehat {X}}S}$ of ${\displaystyle S}$ is compact.[10][12]
8. Every ultrafilter on ${\displaystyle S}$ is a Cauchy filter.

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

Condition 4 implies any subset of ${\displaystyle \operatorname {cl} _{X}\{0\}}$ 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]

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