{"history":[],"article":{"id":"413c4e70-9c5d-483f-bc99-f133da79fd05","title":"Totally bounded space","slug":"totally-bounded-space","content":":::summary\n:::\n\n# \nIn [topology](https://bliptext.com/articles/topology) and related branches of [mathematics](https://bliptext.com/articles/mathematics), **total-boundedness** is a generalization of [compactness](https://bliptext.com/articles/compact-space) for circumstances in which a set is not necessarily [closed](https://bliptext.com/articles/closed-topology).\nA totally bounded set can be [covered](https://bliptext.com/articles/cover-topology) by [finitely](https://bliptext.com/articles/finite-set) many [subsets](https://bliptext.com/articles/subset) of every fixed “size” (where the meaning of “size” depends on the structure of the [ambient space](https://bliptext.com/articles/ambient-space)).\nThe term **precompact** (or **pre-compact**) is sometimes used with the same meaning, but **precompact** is also used to mean [relatively compact](https://bliptext.com/articles/relatively-compact).\nThese definitions coincide for subsets of a [complete metric space](https://bliptext.com/articles/complete-metric-space), but not in general.\n\n# In metric spaces\nA [metric space](https://bliptext.com/articles/metric-space) (M,d) is **totally bounded** if and only if for every real number, there exists a finite collection of [open balls](https://bliptext.com/articles/open-ball) of radius \\varepsilon whose centers lie in M and whose union contains M. Equivalently, the [metric space](https://bliptext.com/articles/metric-space) M is **totally bounded** if and only if for every , there exists a [finite cover](https://bliptext.com/articles/finite-cover) such that the radius of each element of the cover is at most \\varepsilon.\nThis is equivalent to the existence of a finite [ε-net](https://bliptext.com/articles/net-metric-spaces).\nA 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.\nEach totally [bounded](https://bliptext.com/articles/bounded-set) space is [bounded](https://bliptext.com/articles/bounded-set) (as the union of finitely many [bounded](https://bliptext.com/articles/bounded-set) sets is [bounded](https://bliptext.com/articles/bounded-set)).\nThe reverse is true for subsets of [Euclidean space](https://bliptext.com/articles/euclidean-space) (with the [subspace topology](https://bliptext.com/articles/subspace-topology)), but not in general.\nFor example, an infinite set equipped with the [discrete metric](https://bliptext.com/articles/discrete-metric) is bounded but not totally bounded: every discrete ball of radius or less is a singleton, and no finite union of singletons can cover an infinite set.\n# [[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. | upload.wikimedia.org/wikipedia/commons/e/e7/Unit///square///totally///bounded///space.png]\n\n## Uniform (topological) spaces\nA 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 a [uniform structure](https://bliptext.com/articles/uniform-structure).\nA subset S of a [uniform space](https://bliptext.com/articles/uniform-space) X is totally bounded if and only if, for any [entourage](https://bliptext.com/articles/entourage-topology) E, there exists a finite cover of S by subsets of X each of whose [Cartesian squares](https://bliptext.com/articles/cartesian-square) is a subset of E. (In other words, E replaces the \"size\"\nε\n, and a subset is of size E if its Cartesian square is a subset of E.)\nThe definition can be extended still further, to any category of spaces with a notion of [compactness](https://bliptext.com/articles/compactness-topology) and [Cauchy completion](https://bliptext.com/articles/cauchy-completion): a space is totally bounded if and only if its (Cauchy) completion is compact.\n\n# Examples and elementary properties\n\n## Comparison with compact sets\nIn metric spaces, a set is compact if and only if it is complete and totally bounded; without the [axiom of choice](https://bliptext.com/articles/axiom-of-choice) only the forward direction holds.\nPrecompact sets share a number of properties with compact sets.\n\n# In topological groups\nAlthough 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](https://bliptext.com/articles/separation-axiom).\nFor example, in metric spaces, a set is compact if and only if complete and totally bounded.\nUnder the definition below, the same holds for any topological vector space (not necessarily Hausdorff nor complete).\nThe general [logical](https://bliptext.com/articles/logic-symbolic) form of the [definition](https://bliptext.com/articles/definition) is: a subset S of a space X is totally bounded if and only if, [given any](https://bliptext.com/articles/given-any) size E, [there exists](https://bliptext.com/articles/there-exist) a finite cover \\mathcal{O} of S such that each element of \\mathcal{O} has size at most E. X is then totally bounded if and only if it is totally bounded when considered as a subset of itself.\nWe adopt the convention that, for any neighborhood of the identity, a subset is called (****') ****' if and only if A subset S of a [topological group](https://bliptext.com/articles/topological-group) X is (****') ****' if it satisfies any of the following equivalent conditions:\n

: For any neighborhood U of the identity 0, there exist finitely many such that
For any neighborhood U of 0, there exists a finite subset such that (where the right hand side is the [Minkowski sum](https://bliptext.com/articles/minkowski-sum) ).
For any neighborhood U of 0, there exist finitely many subsets of X such that and each B_j is U-small.
For any given [filter subbase](https://bliptext.com/articles/filters-in-topology) \\mathcal{B} of the identity element's [neighborhood filter](https://bliptext.com/articles/neighborhood-filter) \\mathcal{N} (which consists of all neighborhoods of 0 in X) and for every there exists a cover of S by finitely many B-small subsets of X.
Any of the following three sets satisfies (any of the above definitions of) being (left) totally bounded:\n
The [closure](https://bliptext.com/articles/closure-topology) of S in X.
The image of S under the [canonical quotient](https://bliptext.com/articles/first-isomorphism-theorem) which is defined by (where 0 is the identity element).
The sum

\n\n****T****h****e**** ****t****e****r****m**** **** ****u****s****u****a****l****l****y**** ****a****p****p****e****a****r****s**** ****i****n**** ****t****h****e**** ****c****o****n****t****e****x****t**** ****o****f**** ****H****a****u****s****d****o****r****f****f**** ****t****o****p****o****l****o****g****i****c****a****l**** ****v****e****c****t****o****r**** ****s****p****a****c****e****s****.****\nIn that case, the following conditions are also all equivalent to S being (left) totally bounded:\n

In the [completion](https://bliptext.com/articles/complete-topological-vector-space) \\widehat{X} of X, the closure of S is compact.

Every ultrafilter on S is a [Cauchy filter](https://bliptext.com/articles/cauchy-filter).

\n\n****T****h****e**** ****d****e****f****i****n****i****t****i****o****n**** ****o****f**** **** ****i****s**** ****a****n****a****l****o****g****o****u****s****:**** ****s****i****m****p****l****y**** ****s****w****a****p**** ****t****h****e**** ****o****r****d****e****r**** ****o****f**** ****t****h****e**** ****p****r****o****d****u****c****t****s****.****\nCondition 4 implies any subset of is totally bounded (in fact, compact; see above).\nIf X is not Hausdorff then, for example, \\{ 0 \\} is a compact complete set that is not closed.\n\n## Topological vector spaces\nAny topological vector space is an abelian topological group under addition, so the above conditions apply.\nHistorically, statement 6(a) was the first reformulation of total boundedness for [topological vector spaces](https://bliptext.com/articles/topological-vector-space); it dates to a 1935 paper of John von Neumann.\nThis definition has the appealing property that, in a [locally convex space](https://bliptext.com/articles/locally-convex-space) endowed with the [weak topology](https://bliptext.com/articles/weak-topology-polar-topology), the precompact sets are exactly the [bounded sets](https://bliptext.com/articles/bounded-set-topological-vector-space).\nFor separable Banach spaces, there is a nice characterization of the precompact sets (in the norm topology) in terms of [weakly convergent](https://bliptext.com/articles/weak-topology) sequences of functionals: if X is a separable Banach space, then is precompact if and only if every [weakly convergent](https://bliptext.com/articles/weak-topology) sequence of functionals converges [uniformly](https://bliptext.com/articles/uniform-convergence) on S.\n\n### Interaction with convexity\n

The [balanced hull](https://bliptext.com/articles/balanced-set) of a totally bounded subset of a topological vector space is again totally bounded.
The [Minkowski sum](https://bliptext.com/articles/minkowski-sum) of two compact (totally bounded) sets is compact (resp. totally bounded).
In a locally convex (Hausdorff) space, the [convex hull](https://bliptext.com/articles/convex-hull) and the [disked hull](https://bliptext.com/articles/disked-hull) of a totally bounded set K is totally bounded if and only if K is complete.

","current_revision":null,"created_by":"kACnuQmH5k9Nr5VNVEpVeg7XpvUmMctD","created_at":"2025-01-25T21:25:58.398Z","updated_at":"2025-01-25T21:25:58.398Z","search_vector":"'bound':2 'space':3 'total':1","search_content":" \n \n\n# \nIn [topology](https://bliptext.com/articles/topology) and related branches of [mathematics](https://bliptext.com/articles/mathematics), **total-boundedness** is a generalization of [compactness](https://bliptext.com/articles/compact-space) for circumstances in which a set is not necessarily [closed](https://bliptext.com/articles/closed-topology).\nA totally bounded set can be [covered](https://bliptext.com/articles/cover-topology) by [finitely](https://bliptext.com/articles/finite-set) many [subsets](https://bliptext.com/articles/subset) of every fixed “size” (where the meaning of “size” depends on the structure of the [ambient space](https://bliptext.com/articles/ambient-space)).\nThe term **precompact** (or **pre-compact**) is sometimes used with the same meaning, but **precompact** is also used to mean [relatively compact](https://bliptext.com/articles/relatively-compact).\nThese definitions coincide for subsets of a [complete metric space](https://bliptext.com/articles/complete-metric-space), but not in general.\n\n# In metric spaces\nA [metric space](https://bliptext.com/articles/metric-space) (M,d) is **totally bounded** if and only if for every real number, there exists a finite collection of [open balls](https://bliptext.com/articles/open-ball) of radius \\varepsilon whose centers lie in M and whose union contains M. Equivalently, the [metric space](https://bliptext.com/articles/metric-space) M is **totally bounded** if and only if for every , there exists a [finite cover](https://bliptext.com/articles/finite-cover) such that the radius of each element of the cover is at most \\varepsilon.\nThis is equivalent to the existence of a finite [ε-net](https://bliptext.com/articles/net-metric-spaces).\nA 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.\nEach totally [bounded](https://bliptext.com/articles/bounded-set) space is [bounded](https://bliptext.com/articles/bounded-set) (as the union of finitely many [bounded](https://bliptext.com/articles/bounded-set) sets is [bounded](https://bliptext.com/articles/bounded-set)).\nThe reverse is true for subsets of [Euclidean space](https://bliptext.com/articles/euclidean-space) (with the [subspace topology](https://bliptext.com/articles/subspace-topology)), but not in general.\nFor example, an infinite set equipped with the [discrete metric](https://bliptext.com/articles/discrete-metric) is bounded but not totally bounded: every discrete ball of radius or less is a singleton, and no finite union of singletons can cover an infinite set.\n# [[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. | upload.wikimedia.org/wikipedia/commons/e/e7/Unit///square///totally///bounded///space.png]\n\n## Uniform (topological) spaces\nA 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 a [uniform structure](https://bliptext.com/articles/uniform-structure).\nA subset S of a [uniform space](https://bliptext.com/articles/uniform-space) X is totally bounded if and only if, for any [entourage](https://bliptext.com/articles/entourage-topology) E, there exists a finite cover of S by subsets of X each of whose [Cartesian squares](https://bliptext.com/articles/cartesian-square) is a subset of E. (In other words, E replaces the \"size\"\nε\n, and a subset is of size E if its Cartesian square is a subset of E.)\nThe definition can be extended still further, to any category of spaces with a notion of [compactness](https://bliptext.com/articles/compactness-topology) and [Cauchy completion](https://bliptext.com/articles/cauchy-completion): a space is totally bounded if and only if its (Cauchy) completion is compact.\n\n# Examples and elementary properties\n\n## Comparison with compact sets\nIn metric spaces, a set is compact if and only if it is complete and totally bounded; without the [axiom of choice](https://bliptext.com/articles/axiom-of-choice) only the forward direction holds.\nPrecompact sets share a number of properties with compact sets.\n\n# In topological groups\nAlthough 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](https://bliptext.com/articles/separation-axiom).\nFor example, in metric spaces, a set is compact if and only if complete and totally bounded.\nUnder the definition below, the same holds for any topological vector space (not necessarily Hausdorff nor complete).\nThe general [logical](https://bliptext.com/articles/logic-symbolic) form of the [definition](https://bliptext.com/articles/definition) is: a subset S of a space X is totally bounded if and only if, [given any](https://bliptext.com/articles/given-any) size E, [there exists](https://bliptext.com/articles/there-exist) a finite cover \\mathcal{O} of S such that each element of \\mathcal{O} has size at most E. X is then totally bounded if and only if it is totally bounded when considered as a subset of itself.\nWe adopt the convention that, for any neighborhood of the identity, a subset is called (****') ****' if and only if A subset S of a [topological group](https://bliptext.com/articles/topological-group) X is (****') ****' if it satisfies any of the following equivalent conditions:\n \n : For any neighborhood U of the identity 0, there exist finitely many such that \n For any neighborhood U of 0, there exists a finite subset such that (where the right hand side is the [Minkowski sum](https://bliptext.com/articles/minkowski-sum) ). \n For any neighborhood U of 0, there exist finitely many subsets of X such that and each B_j is U-small. \n For any given [filter subbase](https://bliptext.com/articles/filters-in-topology) \\mathcal{B} of the identity element's [neighborhood filter](https://bliptext.com/articles/neighborhood-filter) \\mathcal{N} (which consists of all neighborhoods of 0 in X) and for every there exists a cover of S by finitely many B-small subsets of X. \n**** ****S**** ****i****s**** ****:**** ****f****o****r**** ****e****v****e****r****y**** ****n****e****i****g****h****b****o****r****h****o****o****d**** ****U**** ****o****f**** ****t****h****e**** ****i****d****e****n****t****i****t****y**** ****a****n****d**** ****e****v****e****r****y**** ****c****o****u****n****t****a****b****l****y**** ****i****n****f****i****n****i****t****e**** ****s****u****b****s****e****t**** ****I**** ****o****f**** ****S****,**** ****t****h****e****r****e**** ****e****x****i****s****t**** ****d****i****s****t****i****n****c****t**** ****x****,**** ****y**** ****\\****i****n**** ****I**** ****s****u****c****h**** ****t****h****a****t**** ****(****I****f**** ****S**** ****i****s**** ****f****i****n****i****t****e**** ****t****h****e****n**** ****t****h****i****s**** ****c****o****n****d****i****t****i****o****n**** ****i****s**** ****s****a****t****i****s****f****i****e****d**** ****v****a****c****u****o****u****s****l****y****)****.**** ****\n Any of the following three sets satisfies (any of the above definitions of) being (left) totally bounded:\n The [closure](https://bliptext.com/articles/closure-topology) of S in X. The image of S under the [canonical quotient](https://bliptext.com/articles/first-isomorphism-theorem) which is defined by (where 0 is the identity element). \n The sum \n \n \n****T****h****e**** ****t****e****r****m**** **** ****u****s****u****a****l****l****y**** ****a****p****p****e****a****r****s**** ****i****n**** ****t****h****e**** ****c****o****n****t****e****x****t**** ****o****f**** ****H****a****u****s****d****o****r****f****f**** ****t****o****p****o****l****o****g****i****c****a****l**** ****v****e****c****t****o****r**** ****s****p****a****c****e****s****.****\nIn that case, the following conditions are also all equivalent to S being (left) totally bounded:\n In the [completion](https://bliptext.com/articles/complete-topological-vector-space) \\widehat{X} of X, the closure of S is compact. \n Every ultrafilter on S is a [Cauchy filter](https://bliptext.com/articles/cauchy-filter). \n \n****T****h****e**** ****d****e****f****i****n****i****t****i****o****n**** ****o****f**** **** ****i****s**** ****a****n****a****l****o****g****o****u****s****:**** ****s****i****m****p****l****y**** ****s****w****a****p**** ****t****h****e**** ****o****r****d****e****r**** ****o****f**** ****t****h****e**** ****p****r****o****d****u****c****t****s****.****\nCondition 4 implies any subset of is totally bounded (in fact, compact; see above).\nIf X is not Hausdorff then, for example, \\{ 0 \\} is a compact complete set that is not closed.\n\n## Topological vector spaces\nAny topological vector space is an abelian topological group under addition, so the above conditions apply.\nHistorically, statement 6(a) was the first reformulation of total boundedness for [topological vector spaces](https://bliptext.com/articles/topological-vector-space); it dates to a 1935 paper of John von Neumann.\nThis definition has the appealing property that, in a [locally convex space](https://bliptext.com/articles/locally-convex-space) endowed with the [weak topology](https://bliptext.com/articles/weak-topology-polar-topology), the precompact sets are exactly the [bounded sets](https://bliptext.com/articles/bounded-set-topological-vector-space).\nFor separable Banach spaces, there is a nice characterization of the precompact sets (in the norm topology) in terms of [weakly convergent](https://bliptext.com/articles/weak-topology) sequences of functionals: if X is a separable Banach space, then is precompact if and only if every [weakly convergent](https://bliptext.com/articles/weak-topology) sequence of functionals converges [uniformly](https://bliptext.com/articles/uniform-convergence) on S.\n\n### Interaction with convexity\n \n The [balanced hull](https://bliptext.com/articles/balanced-set) of a totally bounded subset of a topological vector space is again totally bounded. \n The [Minkowski sum](https://bliptext.com/articles/minkowski-sum) of two compact (totally bounded) sets is compact (resp. totally bounded). \n In a locally convex (Hausdorff) space, the [convex hull](https://bliptext.com/articles/convex-hull) and the [disked hull](https://bliptext.com/articles/disked-hull) of a totally bounded set K is totally bounded if and only if K is complete. \n ","revisionCount":0},"hasMore":false,"nextCursor":null}

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