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In functional analysis and related areas of mathematics, a **complete topological vector space** is a topological vector space (TVS) 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 *Cauchy nets* or *Cauchy filters*, which are generalizations of *Cauchy sequences*, 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 completeness for metric spaces.
But unlike metric-completeness, TVS-completeness does not depend on any metric and is defined for *all* 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^{[note 1]} 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 (typically) *not* 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 (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 even if it is not metrizable or not Hausdorff, has a *completion*, which by definition is a complete TVS into which can be TVS-embedded as a dense vector subspace. Moreover, every Hausdorff TVS has a *Hausdorff* completion, which is necessarily unique up to TVS-isomorphism. However, as discussed below, all TVSs have infinitely many non-Hausdorff completions that are *not* TVS-isomorphic to one another.

This section summarizes the definition of a complete topological vector space (TVS) 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 (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 set

and for any the

where if then contains the diagonal

If is a symmetric set (that is, if ), then is ** symmetric**, which by definition means that holds where and in addition, this symmetric set's

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.

where denotes the

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 (

and similarly their

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**^{[4]} 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

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 *complete* if the converse is also always true.
That is, is complete if and only if the following holds:

- whenever is a net in then converges (to some point) in if and only if in

A similar characterization of completeness holds if filters and prefilters are used instead of nets.

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).^{[5]} Every convergent series is necessarily a Cauchy series. In a complete TVS, every Cauchy series is necessarily a convergent series.

A prefilter on a topological vector space is called a **Cauchy prefilter**^{[6]} if it satisfies any of the following equivalent conditions:

- in
- The family is a prefilter.
- Explicitly, means that for every neighborhood of the origin in there exist such that

- in
- The family is a prefilter equivalent to (
*equivalence*means these prefilters generate the same filter on ). - Explicitly, means that for every neighborhood of the origin in there exists some such that

- The family is a prefilter equivalent to (
- For every neighborhood of the origin in contains some -small set (that is, there exists some such that ).
^{[6]}- A subset is called
**-small**or**small of order**^{[6]}if

- A subset is called
- For every neighborhood of the origin in there exists some and some such that
^{[6]}- This statement remains true if "" is replaced with ""

- 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.^{[3]}

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

A topological vector space is called a **complete topological vector space** if any of the following equivalent conditions are satisfied:

- is a complete uniform space when it is endowed with its canonical uniformity.
- In the general theory of uniform spaces, a uniform space is called a complete uniform space if each Cauchy filter on converges to some point of in the topology induced by the uniformity. When is a TVS, the topology induced by the canonical uniformity is equal to 's given topology (so convergence in this induced topology is just the usual convergence in ).

- is a complete subset of itself.
- There exists a neighborhood of the origin in that is also a complete subset of
^{[6]}- This implies that every locally compact TVS is complete (even if the TVS is not Hausdorff).

- 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.

- Every Cauchy filter on converges in to at least one point of
- Every Cauchy net in converges in to at least one point of

where if in addition is pseudometrizable or metrizable (for example, a normed space) then this list can be extended to include:

- is sequentially complete.

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

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

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 ^{[7]}** (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.

For any subsets let^{[1]}

and let

A non-empty family is called a

- 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**^{[7]} is a fundamental system of entourages such that for every if and only if for all A uniformity is called a **translation-invariant uniformity**^{[7]} if it has a base of entourages that is translation-invariant.
The canonical uniformity on any TVS is translation-invariant.^{[7]}

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

**Relatives**

Let be arbitrary and let be the canonical projections onto the first and second coordinates, respectively.

For any define

where (respectively, ) is called the set of

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 **neighborhood prefilter** at a point and, respectively, on a subset are the families of sets:

and the filters on that each generates is known as the

and use the neighborhood definition of "open set" to obtain a topology on called the

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 **complete uniform space** (respectively, a **sequentially complete uniform space**) if every Cauchy prefilter (respectively, every elementary 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 (that is, 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 Cauchy filter on then although will be a Cauchy *pre*filter, it will be a Cauchy filter in if and only if is surjective.

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

- for every there is some such that if with and then

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.^{[note 3]}

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
- If is a metric on then any limit point is necessarily unique and the same is true for limits of Cauchy prefilters on

- 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.

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.
^{[8]}

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.^{[9]}

A pseudometric on a vector space is said to be a **translation invariant pseudometric** if for all vectors

Suppose is pseudometrizable TVS (for example, 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.^{[10]}
If is *not* translation-invariant, then may be possible for to be a complete TVS but to *not* be a complete pseudometric space^{[10]} (see this footnote^{[note 4]} for an example).^{[10]}

**Theorem ^{[11]}^{[12]}** (Klee) — Let be

Two norms on a vector space are called **equivalent** if and only if they induce the same topology.^{[13]} 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* equivalent to that Banach space's given norm.^{[note 6]}^{[13]}
All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.^{[14]} 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**^{[15]} 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 *Hausdorff* completion, which is necessarily unique up to TVS-isomorphism. However, all TVSs, even those that are Hausdorff, (already) complete, and/or metrizable have infinitely many non-Hausdorff completions that are *not* TVS-isomorphic to one another.

For example, the vector space consisting of scalar-valued simple functions for which (where this seminorm is defined in the usual way in terms of Lebesgue integration) 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 (so as to obtain a Hausdorff TVS) results in (a space linearly isometrically-isomorphic to) the usual complete Hausdorff -space (endowed with the usual complete norm).

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 (where this "generalized" TVS is defined analogously to original space of scalar-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.

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.^{[16]}

However, every Hausdorff TVS has a *Hausdorff* completion that is unique up to TVS-isomorphism.^{[16]} But nevertheless, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.

**Example** (**Non-uniqueness of completions**):^{[15]}
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.^{[16]} But nevertheless, as shown above, every Hausdorff TVS still has infinitely many non-isomorphic non-Hausdorff completions.

**Properties of Hausdorff completions ^{[17]}** — 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 ^{[17]}** — 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

**Existence of Hausdorff completions**

A Cauchy filter on a TVS is called a **minimal Cauchy filter**^{[17]} 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

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 (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 ^{[17]}

If is a metrizable TVS then a Hausdorff completion of can be constructed using equivalence classes of Cauchy sequences instead of minimal Cauchy filters.

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 (without proof) 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 (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 ^{[18]}^{[19]}
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.

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,^{[19]} given any algebraic (and thus topological) complement of in and given any completion of the Hausdorff TVS such that then the natural inclusion^{[20]}

is a well-defined TVS-embedding of onto a dense vector subspace of the complete TVS where moreover,

**Theorem ^{[7]}^{[21]}** (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

is a neighborhood basis at the origin in

**Theorem ^{[22]}** (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.

**Grothendieck's Completeness Theorem**

Let denote the *equicontinuous compactology* on the continuous dual space which by definition consists of all equicontinuous weak-* closed and weak-* bounded absolutely convex subsets of ^{[23]} (which are necessarily weak-* compact subsets of ). Assume that every is endowed with the weak-* topology.
A filter on is said to *converge continuously* to if there exists some containing (that is, ) such that the trace of on which is the family converges to in (that is, if in the given weak-* topology).^{[24]}
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 (which is or ) where denotes any neighborhood basis at the origin in denotes the duality pairing, and denotes the filter generated by ^{[24]}
A map into a topological space (such as or ) is said to be *-continuous* if whenever a filter on converges continuously to then ^{[24]}

**Grothendieck's Completeness Theorem ^{[24]}** — If is a Hausdorff topological vector space then its completion is linearly isomorphic to the set of all -continuous linear functions on

If a TVS has any of the following properties then so does its completion:

- Hausdorff
- Locally convex
- Pseudometrizable
^{[16]} - Metrizable
^{[16]} - Seminormable
- Normable
- Moreover, if is a normed space, then the completion can be chosen to be a Banach space such that the TVS-embedding of into is an isometry.

- Hausdorff pre-Hilbert. That is, a TVS induced by an inner product.
^{[25]} - Nuclear
^{[26]} - Barrelled
^{[27]} - Mackey
^{[28]} - DF-space
^{[29]}

**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 ^{[25]}^{[21]}

**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 ^{[30]} The completion of a locally convex bornological space is a barrelled space.^{[27]} If and are DF-spaces then the projective tensor product, as well as its completion, of these spaces is a DF-space.^{[31]}

The completion of the projective tensor product of two nuclear spaces is nuclear.^{[26]} The completion of a nuclear space is TVS-isomorphic with a projective limit of Hilbert spaces.^{[26]}

If (meaning that the addition map is a TVS-isomorphism) has a Hausdorff completion then If in addition is an inner product space and and are orthogonal complements of each other in (that is, ), then and are orthogonal complements in the Hilbert space

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 ^{[26]}

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.^{[26]}

**Theorem** — ^{[11]}
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.
^{[32]}A product of Hausdorff completions of a family of (Hausdorff) TVSs is a Hausdorff completion of their product TVS.^{[32]}More generally, an arbitrary product of complete subsets of a family of TVSs is a complete subset of the product TVS.^{[33]} - The projective limit of a projective system of Hausdorff complete (resp. sequentially complete, quasi-complete) TVSs has that same property.
^{[32]}A projective limit of Hausdorff completions of an inverse system of (Hausdorff) TVSs is a Hausdorff completion of their projective limit.^{[32]} - If is a closed vector subspace of a complete pseudometrizable TVS then the quotient space is complete.
^{[3]} - Suppose is a
*complete*vector subspace of a metrizable TVS If the quotient space is complete then so is^{[3]}^{[34]}However, there exists a complete TVS having a closed vector subspace such that the quotient TVS is*not*complete.^{[17]} - Every F-space, Fréchet space, Banach space, and Hilbert space is a complete TVS.
- Strict LF-spaces and strict LB-spaces are complete.
^{[35]} - Suppose that is a dense subset of a TVS If every Cauchy filter on converges to some point in then is complete.
^{[34]} - 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 topology of uniform convergence on bounded subsets of If is a bornological space and if is complete then is a complete TVS.
^{[35]}In particular, the strong dual of a bornological space is complete.^{[35]}However, it need not be bornological. - Every quasi-complete DF-space is complete.
^{[29]} - 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.
^{[6]}

Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.^{[36]}
Every complete TVS is quasi-complete space and sequentially complete.^{[37]}
However, the converses of the above implications are generally false.^{[37]}
There exists a sequentially complete locally convex TVS that is not quasi-complete.^{[29]}

If a TVS has a complete neighborhood of the origin then it is complete.^{[38]}
Every complete *pseudo*metrizable TVS is a barrelled space and a Baire space (and thus non-meager).^{[39]}
The dimension of a complete metrizable TVS is either finite or uncountable.^{[19]}

Any neighborhood basis of any point in a TVS is a Cauchy prefilter.

Every convergent net (respectively, prefilter) in a TVS is necessarily a Cauchy net (respectively, a Cauchy prefilter).^{[6]}
Any prefilter that is subordinate to (that is, finer than) a Cauchy prefilter is necessarily also a Cauchy prefilter^{[6]} 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 (respectively, Cauchy prefilter), then that Cauchy net (respectively, that Cauchy prefilter) converges to in ^{[3]}
If a Cauchy filter in a TVS has an accumulation point then it converges to

Uniformly continuous maps send Cauchy nets to Cauchy nets.^{[3]}
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 ^{[17]}

If is an injective topological homomorphism from a complete TVS into a Hausdorff TVS then the image of (that is, ) is a closed subspace of ^{[34]}
If is a topological homomorphism from a complete metrizable TVS into a Hausdorff TVS then the range of is a closed subspace of ^{[34]}
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 ^{[40]}

**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 ^{[3]}
If in addition is a homomorphism then its unique uniformly continuous extension is also a homomorphism.^{[3]}
This remains true if "TVS" is replaced by "commutative topological group."^{[3]}
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.^{[41]} 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.)^{[41]}

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).^{[42]}

**Complete subsets**

Every complete subset of a TVS is sequentially complete.
A complete subset of a Hausdorff TVS is a closed subset of ^{[3]}^{[38]}

Every compact subset of a TVS is complete (even if the TVS is not Hausdorff or not complete).^{[3]}^{[38]}
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.^{[38]}

**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.^{[29]}
If is a complete TVS and is a closed vector subspace of such that is not complete, then does *not* have a topological complement in ^{[29]}

**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 ^{[29]}

**Relation to compact subsets**

A subset of a TVS (*not* assumed to be Hausdorff or complete) is compact if and only if it is complete and totally bounded.^{[43]}^{[proof 2]}
Thus a closed and totally bounded subset of a complete TVS is compact.^{[44]}^{[3]}

In a Hausdorff locally convex TVS, the convex hull of a precompact set is again precompact.^{[45]} Consequently, in a complete locally convex Hausdorff TVS, the closed convex hull of a compact subset is again compact.^{[46]}

The convex hull of compact subset of a Hilbert space is *not* necessarily closed and so also *not* necessarily compact. For example, let be the separable Hilbert space of square-summable sequences with the usual norm and let be the standard orthonormal basis (that is at the -coordinate). The closed set is compact but its convex hull is *not* a closed set because belongs to the closure of in but (since every sequence is a finite convex combination of elements of and so is necessarily in all but finitely many coordinates, which is not true of ).^{[47]} However, like in all complete Hausdorff locally convex spaces, the *closed* convex hull of this compact subset is compact.^{[46]} 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 (since ). The closed convex hull of in (here, "closed" means with respect to and not to as before) is equal to which is not compact (because it is not a complete subset). This shows that in a Hausdorff locally convex space that is not complete, the closed convex hull of compact subset might *fail* to be compact (although it will be precompact/totally bounded).

Every complete totally bounded set is relatively compact.^{[3]}
If is any TVS then the quotient map is a closed map^{[48]} and thus A subset of a TVS is totally bounded if and only if its image under the canonical quotient map is totally bounded.^{[19]} 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.^{[3]}
In a locally convex space, the convex hull and the disked hull of a totally bounded set is totally bounded.^{[36]} If is a subset of a TVS such that every sequence in has a cluster point in then is totally bounded.^{[19]} 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 (that is, its closure in the completion of is compact).^{[40]}

If is compact, then