Explicitly, a topological vector spaces (TVS) is complete if every net, or equivalently, every filter, that is Cauchy with respect to the space's canonical uniformity necessarily converges to some point. Said differently, a TVS is complete if its canonical uniformity is a complete uniformity. The canonical uniformity on a TVS is the unique[note 2] translation-invariant uniformity that induces on the topology This notion of "TVS-completeness" depends only on vector subtraction and the topology of the TVS; consequently, it can be applied to all TVSs, including those whose topologies can not be defined in terms metrics or pseudometrics. A first-countable TVS is complete if and only if every Cauchy sequence (or equivalently, every elementary Cauchy filter) converges to some point.
Every topological vector space (TVS) is a commutative topological group with identity under addition and the canonical uniformity of a TVS is defined entirely in terms of subtraction (and thus addition); scalar multiplication is not involved and no additional structure is needed.
The diagonal of is the family For any the canonical entourage/canonical vicinity around is the set
where if then contains the diagonal
If is a symmetric set (that is, if ), then is symmetric, which by definition means that and in addition, this symmetric set's composition with itself is:
If is any neighborhood basis at the origin in then the family of subsets of :
or equivalently, if for every neighborhood of in there exists some such that for all with A Cauchy sequence is a Cauchy net that is a sequence. It suffices to check any of these defining conditions for any given neighborhood basis of in
Let denote the vector subtraction map If then in and so the continuity of guarantees that in (i.e. that is Cauchy). As explained below, is called complete if the converse of this statement is also always true. That is, is complete if and only if whenever is a net in then converges in if and only if in A similar characterization holds for filters and prefilters.
A series is called a Cauchy series (respectively, a convergent series) if the sequence of partial sums is a Cauchy sequence (respectively, a convergent sequence). Every convergent series is necessarily a Cauchy series. In a complete TVS, every Cauchy series is necessarily a convergent series.
For any a prefilter on is necessarily a subset of ; that is,
A subset of a TVS is called a complete subset if it satisfies any of the following equivalent conditions:
Every Cauchy prefilter on converges to at least one point of
If is Hausdorff then every prefilter on will converge to at most one point of But if is not Hausdorff then a prefilter may converge to multiple points in The same is true for nets.
Every Cauchy net in converges to at least one point of
is a complete uniform space (under the point-set topology definition of "complete uniform space") when is endowed with the uniformity induced on it by the canonical uniformity of
The subset is called a sequentially complete subset if every Cauchy sequence in (or equivalently, every elementary Cauchy filter/prefilter on ) converges to at least one point of
Importantly, convergence to points outside of does not prevent a set from being complete: If is not Hausdorff and if every Cauchy prefilter on converges to some point of then will be complete even if some or all Cauchy prefilters on also converge to points(s) in In short, there is no requirement that these Cauchy prefilters on converge only to points in The same can be said of the convergence of Cauchy nets in
As a consequence, if a TVS is not Hausdorff then every subset of the closure of in is complete because it is compact and every compact set is necessarily complete. In particular, if is a proper subset, such as for example, then would be complete even though every Cauchy net in (and also every Cauchy prefilter on ) converges to every point in including those points in that do not belong to This example also shows that complete subsets (and indeed, even compact subsets) of a non-Hausdorff TVS may fail to be closed. For example, if then if and only if is closed in
Complete topological vector space
A topological vector space is called a complete topological vector space if any of the following equivalent conditions are satisfied:
The existence of the canonical uniformity was demonstrated above by defining it. The theorem below establishes that the canonical uniformity of any TVS is the only uniformity on that is both (1) translation invariant, and (2) generates on the topology
Theorem(Existence and uniqueness of the canonical uniformity)—The topology of any TVS can be derived from a unique translation-invariant uniformity. If is any neighborhood base of the origin, then the family is a base for this uniformity.
This section is dedicated to explaining the precise meanings of the terms involved in this uniqueness statement.
Uniform spaces and translation-invariant uniformities
where (resp. ) is called the set of left (resp. right) -relatives of (points in) The maps and are the canonical projections onto the first and second coordinates, respectively.
Given is symmetric if while a subset is called -small if Two points and are -close if If and then:
if and only if Moreover, and is symmetric.
If and then
A non-empty family is called a base of entiyrages or a fundamental system of entourages if is a prefilter on satisfying all of the following conditions:
Every set in contains the diagonal of as a subset; that is, for every Said differently, the prefilter is fixed on
For every there exists some such that
For every there exists some such that
A uniformity or uniform structure on is a filter on that is generated by some base of entourages in which case we say that is a base of entourages for
For a commutative additive group a translation-invariant fundamental system of entourages is a fundamental system of entourages such that for every if and only if for all A uniformity is called a translation-invariant uniformity if it has a base of entourages that is translation-invariant. The same canonical uniformity would result by using a neighborhood basis of the origin rather the filter of all neighborhoods of the origin. The canonical uniformity on any TVS is translation-invariant.
Let be a base of entourages on For every and , the neighborhood prefilter on (resp. at ) is the set
and the filter on that it generates is known as the neighborhood filter of (resp. of ). Assigning to every the neighborhood prefilter
generates a topology on called the topology induced by or the induced topology. A subset is open in this topology if and only if any of the following equivalent conditions hold:
For every there exists some such that
For every there exists some such that
The closure of a subset in this topology is:
Cauchy prefilters and complete uniformities
A prefilter on a uniform space with uniformity is called a Cauchy prefilter if for every entourages there exists some such that
A uniform space is called a complete uniform space (resp. a sequentially complete uniform space) if every Cauchy prefilter (resp. every elementary Cauchy prefilter) on converges to at least one point of when is endowed with the topology induced by
If is a TVS then for any and
The topology induced on by the canonical uniformity is the same as the topology that started with (i.e. it is ).
Let and be TVSs, and be a map. Then is uniformly continuous if for every neighborhood of the origin in there exists a neighborhood of the origin in such that for all if then
Suppose that is uniformly continuous. If is a Cauchy net in then is a Cauchy net in If is a Cauchy prefilter in (meaning that is a family of subsets of that is Cauchy in ) then is a Cauchy prefilter in However, if is a Cuachy filter on then although will be a Cauchy prefilter, it will be a Cauchy filter in if and only if is surjective.
TVS completeness vs completeness of (pseudo)metrics
We review the basic notions related to the general theory of complete pseudometric spaces. Recall that every metric is a pseudometric and that a pseudometric is a metric if and only if implies Thus every metric space is a pseudometric space and a pseudometric space is a metric space if and only if is a metric.
Every pseudometric on a set induces the usual canonical topology on which we'll denote by ; it also induces a canonical uniformity on which we'll denote by The topology on induced by the uniformity is equal to A net in is Cauchy with respect to if and only if it is Cauchy with respect to the uniformity The pseudometric space is a complete (resp. a sequentially complete) pseudometric space if and only if is a complete (resp. a sequentially complete) uniform space. Moreover, the pseudometric space (resp. the uniform space ) is complete if and only if it is sequentially complete.
A pseudometric space (for example, a metric space) is called complete and is called a complete pseudometric if any of the following equivalent conditions hold:
Every Cauchy prefilter on converges to at least one point of
The above statement but with the word "prefilter" replaced by "filter."
Every Cauchy net in converges to at least one point of
Every Cauchy sequence in converges to at least one point of
Thus to prove that is complete, it suffices to only consider Cauchy sequences in (and it is not necessary to consider the more general Cauchy nets).
The canonical uniformity on induced by the pseudometric is a complete uniformity.
If is a metric on then any such limit point is necessarily unique and the same is true for limits of Cauchy prefilters on
And if addition is a metric then we may add to this list:
Every decreasing sequence of closed balls whose diameters shrink to has non-empty intersection.
A pseudometric on a vector space is said to be a translation invariant pseudometric if for all vectors
Suppose is pseudometrizable TVS (e.g. a metrizable TVS) and that is any pseudometric on such that the topology on induced by is equal to If is translation-invariant, then is a complete TVS if and only if is a complete pseudometric space. If is not translation-invariant, then may be possible for to be a complete TVS but to not be a complete pseudometric space (see this footnote[note 4] for an example).
Theorem(Klee)—Let be any[note 5] metric on a vector space such that the topology induced by on makes into a topological vector space. If is a complete metric space then is a complete-TVS.
Complete norms and equivalent norms
Two norms on a vector space are called equivalent if and only if they induce the same topology. If and are two equivalent norms on a vector space then the normed space is a Banach space if and only if is a Banach space. See this footnote for an example of a continuous norm on a Banach space that is not is not equivalent to that Banach space's given norm.[note 6] All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space. Every Banach space is a complete TVS. A normed space is a Banach space (that is, its canonical norm-induced metric is complete) if and only if it is complete as a topological vector space.
A completion of a TVS is a complete TVS that contains a dense vector subspace that is TVS-isomorphic to
Non-uniqueness of all completions
As the example below shows, regardless of whether or not a space is Hausdorff or already complete, every topological vector space (TVS) has infinitely many non-isomorphic completions.
However, every Hausdorff TVS has a Hausdorff completion that is unique up to TVS-isomorphism. But nevertheless, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.
Example (Non-uniqueness of completions): Let denote any complete TVS and let denote any TVS endowed with the indiscrete topology, which recall makes into a complete TVS. Since both and are complete TVSs, so is their product If and are non-empty open subsets of and respectively, then and which shows that is a dense subspace of Thus by definition of "completion," is a completion of (it doesn't matter that is already complete). So by identifying with if is a dense vector subspace of then has both and as completions.
Every Hausdorff TVS has a Hausdorff completion that is unique up to TVS-isomorphism. But nevertheless, as shown above, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.
Properties of Hausdorff completions—Suppose that and are Hausdorff TVSs with complete. Suppose that is a TVS-embedding onto a dense vector subspace of Then
Universal property: for every continuous linear map into a complete Hausdorff TVS there exists a unique continuous linear map such that
If is a TVS embedding onto a dense vector subspace of a complete Hausdorff TVS having the above universal property, then there exists a unique (bijective) TVS-isomorphism such that
Corollary—Suppose is a complete Hausdorff TVS and is a dense vector subspace of Then every continuous linear map into a complete Hausdorff TVS has a unique continuous linear extension to a map
A Cauchy filter on a TVS is called a minimal Cauchy filter if there does not exist a Cauchy filter on that is strictly coarser than (that is, "strictly coarser than " means contained as a proper subset of ).
If is a Cauchy filter on then the filter generated by the following prefilter:
is the unique minimal Cauchy filter on that is contained as a subset of  In particular, for any the neighborhood filter at is a minimal Cauchy filter.
Let be the set of all minimal Cauchy filters on and let be the map defined by sending to the neighborhood filter of in We endow with the following vector space structure. Given and a scalar let (resp. denote the unique minimal Cauchy filter contained in the filter generated by (resp. ).
For every balanced neighborhood of the origin in let
If is Hausdorff then the collection of all sets as ranges over all balanced neighborhoods of the origin in forms a vector topology on making into a complete Hausdorff TVS. Moreover, the map is a TVS-embedding onto a dense vector subspace of 
If is a metrizable TVS then a Hausdorff completion of can be constructed using equivalence classes of Cauchy sequences instead of minimal Cauchy filters.
We now show how every non-Hausdorff TVS can be TVS-embedded onto a dense vector subspace of a complete TVS. The proof that every Hausdorff TVS has a Hausdorff completion is widely available so we use its conclusion to prove that every non-Hausdorff TVS also has a completion. These details are sometimes useful for extending results from Hausdorff TVSs to non-Hausdorff TVSs.
Let denote the closure of the origin in where is endowed with its subspace topology induced by (so that has the indiscrete topology). Since has the trivial topology, it is easily shown that every vector subspace of that is an algebraic complement of in is necessarily a topological complement of in  Let denote any topological complement of in which is necessarily a Hausdorff TVS (since it is TVS-isomorphic to the quotient TVS [note 7]). Since is the topological direct sum of and (which means that in the category of TVSs), the canonical map
is a TVS-isomorphism. Let denote the inverse of this canonical map. (As a side note, it follows that every open and every closed subset of satisfies [proof 1])
The Hausdorff TVS can be TVS-embedded, say via the map onto a dense vector subspace of its completion Since and are complete, so is their product Let denote the identity map and observe that the product map is a TVS-embedding whose image is dense in Define the map[note 8]
which is a TVS-embedding of onto a dense vector subspace of the complete TVS Moreover, observe that the closure of the origin in is equal to and that and are topological complements in
To summarize, given any algebraic (and thus topological) complement of in and given any completion of the Hausdorff TVS such that then the natural inclusion
is a well-defined TVS-embedding of onto a dense vector subspace of the complete TVS where moreover, we have
Topology of a completion
Theorem(Topology of a completion)—Let be a complete TVS and let be a dense vector subspace of If is any neighborhood base of the origin in then the set
is a neighborhood of the origin in the completion of
If is locally convex and is a family of continuous seminorms on that generate the topology of then the family of all continuous extensions to of all members of is a generating family of seminorms for
Said differently, if is a completion of a TVS with and if is a neighborhood base of the origin in then the family of sets
Theorem(Completions of quotients)—Let be a metrizable topological vector space and let be a closed vector subspace of Suppose that is a completion of Then the completion of is TVS-isomorphic to If in addition is a normed space, then this TVS-isomorphism is also an isometry.
Properties preserved by completions
If a TVS has any of the following properties then so does its completion:
Every inner product space has a completion that is a Hilbert space, where the inner product is the unique continuous extension to of the original inner product The norm induced by is also the unique continuous extension to of the norm induced by 
Properties of maps preserved by extensions to a completion
If is a nuclear linear operator between two locally convex spaces and if be a completion of then has a unique continuous linear extension to a nuclear linear operator 
Let and be two Hausdorff TVSs with complete. Let be a completion of Let denote the vector space of continuous linear operators and let denote the map that sends every to its unique continuous linear extension on Then is a (surjective) vector space isomorphism. Moreover, maps families of equicontinuous subsets onto each other. Suppose that is endowed with a -topology and that denotes the closures in of sets in Then the map is also a TVS-isomorphism.
Examples and sufficient conditions for a complete TVS
Theorem— Let be any (not assumed to be translation-invariant) metric on a vector space such that the topology induced by on makes into a topological vector space. If is a complete metric space then is a complete-TVS.
Any TVS endowed with the trivial topology is complete and every one of its subsets is complete. Moreover, every TVS with the trivial topology is compact and hence locally compact. Thus a complete seminormable locally convex and locally compact TVS need not be finite-dimensional if it is not Hausdorff.
An arbitrary product of complete (resp. sequentially complete, quasi-complete) TVSs has that same property. If all spaces are Hausdorff, then the converses are also true. A product of Hausdorff completions of a family of (Hausdorff) TVSs is a Hausdorff completion of their product TVS. More generally, an arbitrary product of complete subsets of a family of TVSs is a complete subset of the product TVS.
The projective limit of a projective system of Hausdorff complete (resp. sequentially complete, quasi-complete) TVSs has that same property. A projective limit of Hausdorff completions of an inverse system of (Hausdorff) TVSs is a Hausdorff completion of their projective limit.
Suppose is a complete vector subspace of a metrizable TVS If the quotient space is complete then so is  However, there exists a complete TVS having a closed vector subspace such that the quotient TVS is not complete.
Let and be Hausdorff TVS topologies on a vector space such that If there exists a prefilter such that is a neighborhood basis at the origin for and such that every is a complete subset of then is a complete TVS.
Every convergent net (resp. prefilter) in a TVS is necessarily a Cauchy net (resp. a Cauchy prefilter). Any prefilter that is subordinate to (i.e. finer than) a Cauchy prefilter is necessarily also a Cauchy prefilter and any prefilter finer than a Cauchy prefilter is also a Cauchy prefilter. The filter associated with a sequence in a TVS is Cauchy if and only if the sequence is a Cauchy sequence. Every convergent prefilter is a Cauchy prefilter.
If is a TVS and if is a cluster point of a Cauchy net (resp. Cauchy prefilter), then that Cauchy net (resp. that Cauchy prefilter) converges to in  If a Cauchy filter in a TVS has an accumulation point then it converges to
Uniformly continuous maps send Cauchy nets to Cauchy nets. A Cauchy sequence in a Hausdorff TVS A Cauchy sequence in a Hausdorff TVS when considered as a set, is not necessarily relatively compact (that is, its closure in is not necessarily compact[note 9]) although it is precompact (that is, its closure in the completion of is compact).
Every Cauchy sequence is a bounded subset but this is not necessarily true of Cauchy net. For example, let have it usual order, let denote any preorder on the non-indiscrete TVS (that is, does not have the trivial topology; it is also assumed that ) and extend these two preorders to the union by declaring that holds for every and Let be defined by if and otherwise (that is, if ), which is a net in since the preordered set is directed (this preorder on is also partial order (respectively, a total order) if this is true of ). This net is a Cauchy net in because it converges to the origin, but the set is not a bounded subset of (because does not have the trivial topology).
Suppose that is a family of TVSs and that denotes the product of these TVSs. Suppose that for every index is a prefilter on Then the product of this family of prefilters is a Cauchy filter on if and only if each is a Cauchy filter on 
Suppose that is a uniformly continuous map from a dense subset of a TVS into a complete Hausdorff TVS Then has a unique uniformly continuous extension to all of  If in addition is a homomorphism then its unique uniformly continuous extension is also a homomorphism. This remains true if "TVS" is replaced by "commutative topological group." The map is not required to be a linear map and that is not required to be a vector subspace of
Uniformly continuous linear extensions
Suppose be a continuous linear operator between two Hausdorff TVSs. If is a dense vector subspace of and if the restriction to is a topological homomorphism then is also a topological homomorphism. So if and are Hausdorff completions of and respectively, and if is a topological homomorphism, then 's unique continuous linear extension is a topological homomorphism. (Note that it's possible for to be surjective but for to not be injective.)
Suppose and are Hausdorff TVSs, is a dense vector subspace of and is a dense vector subspaces of If are and are topologically isomorphic additive subgroups via a topological homomorphism then the same is true of and via the unique uniformly continuous extension of (which is also a homeomorphism).
Every compact subset of a TVS is complete (even if the TVS is not Hausdorff or not complete). Closed subsets of a complete TVS are complete; however, if a TVS is not complete then is a closed subset of that is not complete. The empty set is complete subset of every TVS. If is a complete subset of a TVS (the TVS is not necessarily Hausdorff or complete) then any subset of that is closed in is complete.
Every complete totally bounded set is relatively compact. If is any TVS then the quotient map is a closed map and thus A subset of a TVS is totally bounded if and only if its image under the canonical quotient map is totally bounded. Thus is totally bounded if and only if is totally bounded. In any TVS, the closure of a totally bounded subset is again totally bounded. In a locally convex space, the convex hull and the disked hull of a totally bounded set is totally bounded. If is a subset of a TVS such that every sequence in has a cluster point in then is totally bounded. A subset of a Hausdorff TVS is totally bounded if and only if every ultrafilter on is Cauchy, which happens if and only if it is pre-compact (i.e. its closure in the completion of is compact).
If is compact, then and this set is compact. Thus the closure of a compact set is compact[note 10] (that is, all compact sets are relatively compact). Thus the closure of a compact set is compact. Every relatively compact subset of a Hausdorff TVS is totally bounded.
In a complete locally convex space, the convex hull and the disked hull of a compact set are both compact. More generally, if is a compact subset of a locally convex space, then the convex hull (resp. the disked hull ) is compact if and only if it is complete. Every subset