Complete topological vector space


In functional analysis and related areas of mathematics, a complete topological vector space is a topological vector space with the property that whenever points get progressively closer to each other, then there exists some point towards which they all get closer.
The notion of "points that get progressively closer" is made rigorous by or, which are generalizations of, while "point towards which they all get closer" means that this Cauchy net or filter converges to
The notion of completeness for TVSs uses the theory of uniform spaces as a framework to generalize the notion of metric space|completeness for metric spaces].
But unlike metric-completeness, TVS-completeness does not depend on any metric and is defined for TVSs, including those that are not metrizable or Hausdorff.
Completeness is an extremely important property for a topological vector space to possess.
The notions of completeness for normed spaces and metrizable TVSs, which are commonly defined in terms of completeness of a particular norm or metric, can both be reduced down to this notion of TVS-completeness - a notion that is independent of any particular norm or metric.
A metrizable topological vector space with a translation invariant metric is complete as a TVS if and only if is a complete metric space, which by definition means that every -Cauchy sequence converges to some point in
Prominent examples of complete TVSs that are also metrizable include all F-spaces and consequently also all Fréchet spaces, Banach spaces, and Hilbert spaces.
Prominent examples of complete TVS that are metrizable include strict LF-spaces such as the space of test functions with it canonical LF-topology, the strong dual space of any non-normable Fréchet space, as well as many other polar topologies on continuous dual space or other topologies on spaces of linear maps.
Explicitly, a topological vector spaces is complete if every net, or equivalently, every filter, that is Cauchy with respect to the space's 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 translation-invariant uniformity that induces on the topology
This notion of "TVS-completeness" depends 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 converges to some point.
Every topological vector space even if it is not metrizable or not Hausdorff, has a, which by definition is a complete TVS into which can be TVS-embedded as a dense vector subspace. Moreover, every Hausdorff TVS has a completion, which is necessarily unique up to TVS-isomorphism. However, as discussed below, all TVSs have infinitely many non-Hausdorff completions that are TVS-isomorphic to one another.

Definitions

This section summarizes the definition of a complete topological vector space in terms of both nets and prefilters.
Information about convergence of nets and filters, such as definitions and properties, can be found in the article about filters in topology.
Every topological vector space is a commutative topological group with identity under addition and the canonical uniformity of a TVS is defined in terms of subtraction ; scalar multiplication is not involved and no additional structure is needed.

Canonical uniformity

The ' of is the set
and for any the '/ is the set
where if then contains the diagonal
If is a symmetric set, then is , which by definition means that holds where and in addition, this symmetric set's with itself is:
If is any neighborhood basis at the origin in then the family of subsets of
is a prefilter on
If is the neighborhood filter at the origin in then forms a base of entourages for a uniform structure on that is considered canonical.
Explicitly, by definition, is the filter on generated by the above prefilter:
where denotes the of in
The same canonical uniformity would result by using a neighborhood basis of the origin rather the filter of all neighborhoods of the origin. If is any neighborhood basis at the origin in then the filter on generated by the prefilter is equal to the canonical uniformity induced by

Cauchy net

The general theory of uniform spaces has its own definition of a "Cauchy prefilter" and "Cauchy net". For the canonical uniformity on these definitions reduce down to those given below.
Suppose is a net in and is a net in
The product becomes a directed set by declaring if and only if and
Then
denotes the ', where in particular If then the image of this net under the vector addition map denotes the ' of these two nets:
and similarly their ' is defined to be the image of the product net under the vector subtraction map :
In particular, the notation denotes the -indexed net and not the -indexed net since using the latter as the definition would make the notation useless.
A net in a TVS is called a Cauchy net if
Explicitly, this means that for every neighborhood of in there exists some index such that for all indices that satisfy and
It suffices to check any of these defining conditions for any given neighborhood basis of in
A Cauchy sequence is a sequence that is also a Cauchy net.
If then in and so the continuity of the vector subtraction map which is defined by guarantees that in where and
This proves that every convergent net is a Cauchy net.
By definition, a space is called if the converse is also always true.
That is, is complete if and only if the following holds:
A similar characterization of completeness holds if filters and prefilters are used instead of nets.
A series is called a ' if the sequence of partial sums is a Cauchy sequence. Every convergent series is necessarily a Cauchy series. In a complete TVS, every Cauchy series is necessarily a convergent series.

Cauchy filter and Cauchy prefilter

A prefilter on a topological vector space is called a Cauchy prefilter if it satisfies any of the following equivalent conditions:

  1. in
    • The family is a prefilter.
    • Explicitly, means that for every neighborhood of the origin in there exist such that
  2. in
    • The family is a prefilter equivalent to .
    • Explicitly, means that for every neighborhood of the origin in there exists some such that
  3. For every neighborhood of the origin in contains some -small set.
    • A subset is called -small or if
  4. For every neighborhood of the origin in there exists some and some such that
    • This statement remains true if "" is replaced with ""
  5. Every neighborhood of the origin in contains some subset of the form where and
It suffices to check any of the above conditions for any given neighborhood basis of in
A Cauchy filter is a Cauchy prefilter that is also a filter on
If is a prefilter on a topological vector space and if then in if and only if and is Cauchy.

Complete subset

For any a prefilter ' is necessarily a subset of ; that is,
A subset of a TVS is called a ' if it satisfies any of the following equivalent conditions:

  1. 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.
  2. Every Cauchy net in converges to at least one point of
  3. is a complete uniform space when is endowed with the uniformity induced on it by the canonical uniformity of
The subset is called a if every Cauchy sequence in converges to at least one point of
Importantly, : 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 converge to points in In short, there is no requirement that these Cauchy prefilters on converge to points in The same can be said of the convergence of Cauchy nets in
As a consequence, if a TVS is 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 Cauchy net in converges to point in including those points in that do not belong to
This example also shows that complete 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 ' if any of the following equivalent conditions are satisfied:

  1. is a complete uniform space when it is endowed with its canonical uniformity.
  2. is a complete subset of itself.
  3. There exists a neighborhood of the origin in that is also a complete subset of
  4. Every Cauchy prefilter on converges in 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.
  5. Every Cauchy filter on converges in to at least one point of
  6. Every Cauchy net in converges in to at least one point of
where if in addition is pseudometrizable or metrizable then this list can be extended to include:

  1. is sequentially complete.
A topological vector space is ' if any of the following equivalent conditions are satisfied:

  1. is a sequentially complete subset of itself.
  2. Every Cauchy sequence in converges in to at least one point of
  3. Every elementary Cauchy prefilter on converges in to at least one point of
  4. Every elementary Cauchy filter on converges in to at least one point of

Uniqueness of the canonical uniformity

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 translation invariant, and generates on the topology
This section is dedicated to explaining the precise meanings of the terms involved in this uniqueness statement.

Uniform spaces and translation-invariant uniformities

For any subsets let
and let
A non-empty family is called a ' or a ' if is a prefilter on satisfying all of the following conditions:

  1. Every set in contains the diagonal of as a subset; that is, for every Said differently, the prefilter is on
  2. For every there exists some such that
  3. For every there exists some such that
A ' or ' 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 a commutative additive group a ' is a fundamental system of entourages such that for every if and only if for all A uniformity is called a ' if it has a base of entourages that is translation-invariant.
The canonical uniformity on any TVS is translation-invariant.
The binary operator satisfies all of the following:

  • If and then
  • Associativity:
  • Identity:
  • Zero:
Symmetric entourages
Call a subset symmetric if which is equivalent to
This equivalence follows from the identity and the fact that if then if and only if
For example, the set is always symmetric for every
And because if and are symmetric then so is

Topology generated by a uniformity

Relatives
Let be arbitrary and let be the canonical projections onto the first and second coordinates, respectively.
For any define
where is called the set of left '-relatives of
Denote the special case where is a singleton set for some by:
If then
Moreover, right distributes over both unions and intersections, meaning that if then and
Neighborhoods and open sets
Two points and are -close if and a subset is called -small if
Let be a base of entourages on The ' at a point and, respectively, on a subset are the families of sets:
and the filters on that each generates is known as the ' of .
Assign to every the neighborhood prefilter
and use the neighborhood definition of "open set" to obtain a topology on called the topology induced by or the '.
Explicitly, a subset is open in this topology if and only if for every there exists some such that that is, is open if and only if 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 entourage there exists some such that
A uniform space is called a ' if every Cauchy prefilter on converges to at least one point of when is endowed with the topology induced by
Case of a topological vector space'
If is a topological vector space then for any and
and the topology induced on by the canonical uniformity is the same as the topology that started with.

Uniform continuity

Let and be TVSs, and be a map. Then is 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 then is a Cauchy prefilter in However, if is a Cauchy filter on then although will be a Cauchy filter, it will be a Cauchy filter in if and only if is surjective.

TVS completeness vs completeness of (pseudo)metrics

Preliminaries: Complete pseudometric spaces

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.
If is a subset of a pseudometric space then the diameter of is defined to be
A prefilter on a pseudometric space is called a -Cauchy prefilter or simply a Cauchy prefilter if for each real there is some such that the diameter of is less than
Suppose is a pseudometric space. A net in is called a -Cauchy net or simply a Cauchy net if is a Cauchy prefilter, which happens if and only if
or equivalently, if and only if in This is analogous to the following characterization of the converge of to a point: if then in if and only if in
A Cauchy sequence is a sequence that is also a Cauchy net.
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 pseudometric space if and only if is a complete uniform space.
Moreover, the pseudometric space is complete if and only if it is sequentially complete.
A pseudometric space is called complete and is called a complete pseudometric if any of the following equivalent conditions hold:

  1. Every Cauchy prefilter on converges to at least one point of
  2. The above statement but with the word "prefilter" replaced by "filter."
  3. Every Cauchy net in converges to at least one point of
    • If is a metric on then any limit point is necessarily unique and the same is true for limits of Cauchy prefilters on
  4. 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 .
  5. The canonical uniformity on induced by the pseudometric is a complete uniformity.
And if addition is a metric then we may add to this list:

  1. Every decreasing sequence of closed balls whose diameters shrink to has non-empty intersection.

Complete pseudometrics and complete TVSs

Every F-space, and thus also every Fréchet space, Banach space, and Hilbert space is a complete TVS. Note that every F-space is a Baire space but there are normed spaces that are Baire but not Banach.
A pseudometric on a vector space is said to be a if for all vectors
Suppose is pseudometrizable TVS and that is 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 translation-invariant, then may be possible for to be a complete TVS but to be a complete pseudometric space.

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 equivalent to that Banach space's given norm.
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 if and only if it is complete as a topological vector space.

Completions

A completion of a TVS is a complete TVS that contains a dense vector subspace that is TVS-isomorphic to In other words, it is a complete TVS into which can be TVS-embedded as a dense vector subspace. Every TVS-embedding is a uniform embedding.
Every topological vector space has a completion. Moreover, every Hausdorff TVS has a completion, which is necessarily unique up to TVS-isomorphism. However, all TVSs, even those that are Hausdorff, complete, and/or metrizable have infinitely many non-Hausdorff completions that are TVS-isomorphic to one another.

Examples of completions

For example, the vector space consisting of scalar-valued simple functions for which becomes a seminormed space when endowed with this seminorm, which in turn makes it into both a pseudometric space and a non-Hausdorff non-complete TVS; any completion of this space is a non-Hausdorff complete seminormed space that when quotiented by the closure of its origin results in the usual complete Hausdorff -space.
As another example demonstrating the usefulness of completions, the completions of topological tensor products, such as projective tensor products or injective tensor products, of the Banach space with a complete Hausdorff locally convex TVS results in a complete TVS that is TVS-isomorphic to a "generalized" -space consisting -valued functions on . Similarly, the completion of the injective tensor product of the space of scalar-valued -test functions with such a TVS is TVS-isomorphic to the analogously defined TVS of -valued test functions.

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 has infinitely many non-isomorphic completions.
However, every Hausdorff TVS has a completion that is unique up to TVS-isomorphism. But nevertheless, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.
Example :
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 .
So by identifying with if is a dense vector subspace of then has both and as completions.

Hausdorff completions

Every Hausdorff TVS has a 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.
Existence of Hausdorff completions
A Cauchy filter on a TVS is called a if there does exist a Cauchy filter on that is strictly coarser than .
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
Endow with the following vector space structure:
Given and a scalar let denote the unique minimal Cauchy filter contained in the filter generated by .
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.

Non-Hausdorff completions

This subsection details 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 and so this fact will be used to show 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 .
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 is the topological direct sum of and, the canonical map
is a TVS-isomorphism.
Let denote the inverse of this canonical map.
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
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 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,

Topology of a completion

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
is a neighborhood basis at the origin in
Grothendieck's Completeness Theorem
Let denote the on the continuous dual space which by definition consists of all equicontinuous weak-* closed and weak-* bounded absolutely convex subsets of . Assume that every is endowed with the weak-* topology.
A filter on is said to to if there exists some containing such that the trace of on which is the family converges to in .
The filter [|converges continuously] to if and only if converges continuously to the origin, which happens if and only if for every the filter in the scalar field where denotes any neighborhood basis at the origin in denotes the duality pairing, and denotes the filter generated by
A map into a topological space is said to be if whenever a filter on converges continuously to then

Properties preserved by completions

If a TVS has any of the following properties then so does its completion:
Completions of Hilbert spaces
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
Other preserved properties
If is a Hausdorff TVS, then the continuous dual space of is identical to the continuous dual space of the completion of The completion of a locally convex bornological space is a barrelled space. If and are DF-spaces then the projective tensor product, as well as its completion, of these spaces is a DF-space.
The completion of the projective tensor product of two nuclear spaces is nuclear. The completion of a nuclear space is TVS-isomorphic with a projective limit of Hilbert spaces.
If has a Hausdorff completion then
If in addition is an inner product space and and are orthogonal complements of each other in, then and are orthogonal complements in the Hilbert space

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 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


  • 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 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 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 TVSs has that same property. A projective limit of Hausdorff completions of an inverse system of TVSs is a Hausdorff completion of their projective limit.
  • If is a closed vector subspace of a complete pseudometrizable TVS then the quotient space is complete.
  • Suppose is a 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 complete.
  • Every F-space, Fréchet space, Banach space, and Hilbert space is a complete TVS.
  • Strict LF-spaces and strict LB-spaces are complete.
  • Suppose that is a dense subset of a TVS If every Cauchy filter on converges to some point in then is complete.
  • The Schwartz space of smooth functions is complete.
  • The spaces of distributions and test functions is complete.
  • Suppose that and are locally convex TVSs and that the space of continuous linear maps is endowed with the [Topologies on spaces of linear maps|topology of uniform convergence on bounded subsets] of If is a bornological space and if is complete then is a complete TVS. In particular, the strong dual of a bornological space is complete. However, it need not be bornological.
  • Every quasi-complete DF-space is 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.

Properties

Complete TVSs

Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.
Every complete TVS is quasi-complete space and sequentially complete.
However, the converses of the above implications are generally false.
There exists a sequentially complete locally convex TVS that is not quasi-complete.
If a TVS has a complete neighborhood of the origin then it is complete.
Every complete metrizable TVS is a barrelled space and a Baire space.
The dimension of a complete metrizable TVS is either finite or uncountable.

Cauchy nets and prefilters

Any neighborhood basis of any point in a TVS is a Cauchy prefilter.
Every convergent net in a TVS is necessarily a Cauchy net.
Any prefilter that is subordinate to 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, then that Cauchy net 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 when considered as a set, is not necessarily relatively compact although it is precompact.
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 and extend these two preorders to the union by declaring that holds for every and
Let be defined by if and otherwise, which is a net in since the preordered set is directed. This net is a Cauchy net in because it converges to the origin, but the set is not a bounded subset of .
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

Maps

If is an injective topological homomorphism from a complete TVS into a Hausdorff TVS then the image of is a closed subspace of
If is a topological homomorphism from a complete metrizable TVS into a Hausdorff TVS then the range of is a closed subspace of
If is a uniformly continuous map between two Hausdorff TVSs then the image under of a totally bounded subset of is a totally bounded subset of
Uniformly continuous extensions
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.
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 .

Subsets

Complete subsets
Every complete subset of a TVS is sequentially complete.
A complete subset of a Hausdorff TVS is a closed subset of
Every compact subset of a TVS is 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 then any subset of that is closed in is complete.
Topological complements
If is a non-normable Fréchet space on which there exists a continuous norm then contains a closed vector subspace that has no topological complement.
If is a complete TVS and is a closed vector subspace of such that is not complete, then does have a topological complement in
Subsets of completions
Let be a separable locally convex metrizable topological vector space and let be its completion. If is a bounded subset of then there exists a bounded subset of such that
Relation to compact subsets
A subset of a TVS is compact if and only if it is complete and totally bounded.
Thus a closed and totally bounded subset of a complete TVS is compact.
In a Hausdorff locally convex TVS, the convex hull of a precompact set is again precompact. Consequently, in a complete locally convex Hausdorff TVS, the closed convex hull of a compact subset is again compact.
The convex hull of compact subset of a Hilbert space is necessarily closed and so also necessarily compact. For example, let be the separable Hilbert space [Lp space|] of square-summable sequences with the usual norm and let be the standard orthonormal basis. The closed set is compact but its convex hull is a closed set because belongs to the closure of in but . However, like in all complete Hausdorff locally convex spaces, the convex hull of this compact subset is compact. The vector subspace is a pre-Hilbert space when endowed with the substructure that the Hilbert space induces on it but is not complete and . The closed convex hull of in is equal to which is not compact. This shows that in a Hausdorff locally convex space that is not complete, the closed convex hull of compact subset might to be compact.
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.
If is compact, then and this set is compact. Thus the closure of a compact set is 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 is compact if and only if it is complete.
Every subset of is compact and thus complete. In particular, if is not Hausdorff then there exist compact complete sets that are not closed.