Complete topological vector space

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 ${\displaystyle (X,\tau )}$ is the only uniformity on ${\displaystyle X}$ that is both (1) translation invariant, and (2) generates on ${\displaystyle X}$ the topology ${\displaystyle \tau .}$

Theorem (Existence and uniqueness of the canonical uniformity)—The topology of any TVS can be derived from a unique translation-invariant uniformity. If ${\displaystyle \mathcal{N}(0)}$ is any neighborhood base of the origin, then the family ${\displaystyle \left\{\mathrm{\Delta }(N):N\in \mathcal{N}(0)\right\}}$ 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

For any subsets ${\displaystyle \mathrm{\Phi },\mathrm{\Psi }\subseteq X\times X,}$ let $${\displaystyle {\mathrm{\Phi }}^{\mathrm{op}}\stackrel{\text{def}}{=}\{(y,x):(x,y)\in \mathrm{\Phi }\}}$$ and let $${\displaystyle \begin{array}{cc}\mathrm{\Phi }\circ \mathrm{\Psi }& \stackrel{\text{def}}{=}\left\{(x,z):\text{ there exists }y\in X\text{ such that }(x,y)\in \mathrm{\Psi }\text{ and }(y,z)\in \mathrm{\Phi }\right\}\\ & =\bigcup _{y\in X}\{(x,z):(x,y)\in \mathrm{\Psi }\text{ and }(y,z)\in \mathrm{\Phi }\}\end{array}}$$ A non-empty family ${\displaystyle \mathcal{B}\subseteq \mathrm{\wp }(X\times X)}$ is called a base of entourages or a fundamental system of entourages if ${\displaystyle \mathcal{B}}$ is a prefilter on ${\displaystyle X\times X}$ satisfying all of the following conditions:

1. Every set in ${\displaystyle \mathcal{B}}$ contains the diagonal of ${\displaystyle X}$ as a subset; that is, ${\displaystyle {\mathrm{\Delta }}_X\stackrel{\text{def}}{=}\{(x,x):x\in X\}\subseteq \mathrm{\Phi }}$ for every ${\displaystyle \mathrm{\Phi }\in \mathcal{B}.}$ Said differently, the prefilter ${\displaystyle \mathcal{B}}$ is fixed on ${\displaystyle {\mathrm{\Delta }}_X.}$
2. For every ${\displaystyle \mathrm{\Omega }\in \mathcal{B}}$ there exists some ${\displaystyle \mathrm{\Phi }\in \mathcal{B}}$ such that ${\displaystyle \mathrm{\Phi }\circ \mathrm{\Phi }\subseteq \mathrm{\Omega }.}$
3. For every ${\displaystyle \mathrm{\Omega }\in \mathcal{B}}$ there exists some ${\displaystyle \mathrm{\Phi }\in \mathcal{B}}$ such that ${\displaystyle \mathrm{\Phi }\subseteq {\mathrm{\Omega }}^{\mathrm{op}}\stackrel{\text{def}}{=}\{(y,x):(x,y)\in \mathrm{\Omega }\}.}$

A uniformity or uniform structure on ${\displaystyle X}$ is a filter ${\displaystyle \mathcal{U}}$ on ${\displaystyle X\times X}$ that is generated by some base of entourages ${\displaystyle \mathcal{B},}$ in which case we say that ${\displaystyle \mathcal{B}}$ is a base of entourages for ${\displaystyle \mathcal{U}.}$

For a commutative additive group ${\displaystyle X,}$ a translation-invariant fundamental system of entourages is a fundamental system of entourages ${\displaystyle \mathcal{B}}$ such that for every ${\displaystyle \mathrm{\Phi }\in \mathcal{B},}$ ${\displaystyle (x,y)\in \mathrm{\Phi }}$ if and only if ${\displaystyle (x+z,y+z)\in \mathrm{\Phi }}$ for all ${\displaystyle x,y,z\in X.}$ A uniformity ${\displaystyle \mathcal{B}}$ is called a translation-invariant uniformity if it has a base of entourages that is translation-invariant. The canonical uniformity on any TVS is translation-invariant.

The binary operator ${\displaystyle \circ }$ satisfies all of the following:

• ${\displaystyle (\mathrm{\Phi }\circ \mathrm{\Psi }{)}^{\mathrm{op}}={\mathrm{\Psi }}^{\mathrm{op}}\circ {\mathrm{\Phi }}^{\mathrm{op}}.}$
• If ${\displaystyle \mathrm{\Phi }\subseteq {\mathrm{\Phi }}_2}$ and ${\displaystyle \mathrm{\Psi }\subseteq {\mathrm{\Psi }}_2}$ then ${\displaystyle \mathrm{\Phi }\circ \mathrm{\Psi }\subseteq {\mathrm{\Phi }}_2\circ {\mathrm{\Psi }}_2.}$
• Associativity: ${\displaystyle \mathrm{\Phi }\circ (\mathrm{\Psi }\circ \mathrm{\Omega })=(\mathrm{\Phi }\circ \mathrm{\Psi })\circ \mathrm{\Omega }.}$
• Identity: ${\displaystyle \mathrm{\Phi }\circ {\mathrm{\Delta }}_X=\mathrm{\Phi }={\mathrm{\Delta }}_X\circ \mathrm{\Phi }.}$
• Zero: ${\displaystyle \mathrm{\Phi }\circ \varnothing =\varnothing =\varnothing \circ \mathrm{\Phi }}$

Symmetric entourages

Call a subset ${\displaystyle \mathrm{\Phi }\subseteq X\times X}$ symmetric if ${\displaystyle \mathrm{\Phi }={\mathrm{\Phi }}^{\mathrm{op}},}$ which is equivalent to ${\displaystyle {\mathrm{\Phi }}^{\mathrm{op}}\subseteq \mathrm{\Phi }.}$ This equivalence follows from the identity ${\displaystyle {\left({\mathrm{\Phi }}^{\mathrm{op}}\right)}^{\mathrm{op}}=\mathrm{\Phi }}$ and the fact that if ${\displaystyle \mathrm{\Psi }\subseteq X\times X,}$ then ${\displaystyle \mathrm{\Phi }\subseteq \mathrm{\Psi }}$ if and only if ${\displaystyle {\mathrm{\Phi }}^{\mathrm{op}}\subseteq {\mathrm{\Psi }}^{\mathrm{op}}.}$ For example, the set ${\displaystyle {\mathrm{\Phi }}^{\mathrm{op}}\cap \mathrm{\Phi }}$ is always symmetric for every ${\displaystyle \mathrm{\Phi }\subseteq X\times X.}$ And because ${\displaystyle (\mathrm{\Phi }\cap \mathrm{\Psi }{)}^{\mathrm{op}}={\mathrm{\Phi }}^{\mathrm{op}}\cap {\mathrm{\Psi }}^{\mathrm{op}},}$ if ${\displaystyle \mathrm{\Phi }}$ and ${\displaystyle \mathrm{\Psi }}$ are symmetric then so is ${\displaystyle \mathrm{\Phi }\cap \mathrm{\Psi }.}$

Topology generated by a uniformity

Relatives

Let ${\displaystyle \mathrm{\Phi }\subseteq X\times X}$ be arbitrary and let ${\displaystyle {\mathrm{Pr}}_1,{\mathrm{Pr}}_2:X\times X\to X}$ be the canonical projections onto the first and second coordinates, respectively.

For any ${\displaystyle S\subseteq X,}$ define $${\displaystyle S\cdot \mathrm{\Phi }\stackrel{\text{def}}{=}\{y\in X:\mathrm{\Phi }\cap (S\times \{x\})\ne \varnothing \}={\mathrm{Pr}}_2(\mathrm{\Phi }\cap (S\times X))}$$ $${\displaystyle \mathrm{\Phi }\cdot S\stackrel{\text{def}}{=}\{x\in X:\mathrm{\Phi }\cap (\{x\}\times S)\ne \varnothing \}={\mathrm{Pr}}_1(\mathrm{\Phi }\cap (X\times S))=S\cdot \left({\mathrm{\Phi }}^{\mathrm{op}}\right)}$$ where ${\displaystyle \mathrm{\Phi }\cdot S}$ (respectively, ${\displaystyle S\cdot \mathrm{\Phi }}$) is called the set of left (respectively, right) ${\displaystyle \mathrm{\Phi }}$-relatives of (points in) ${\displaystyle S.}$ Denote the special case where ${\displaystyle S=\{p\}}$ is a singleton set for some ${\displaystyle p\in X}$ by: $${\displaystyle p\cdot \mathrm{\Phi }\stackrel{\text{def}}{=}\{p\}\cdot \mathrm{\Phi }=\{y\in X:(p,y)\in \mathrm{\Phi }\}}$$ $${\displaystyle \mathrm{\Phi }\cdot p\stackrel{\text{def}}{=}\mathrm{\Phi }\cdot \{p\}=\{x\in X:(x,p)\in \mathrm{\Phi }\}=p\cdot \left({\mathrm{\Phi }}^{\mathrm{op}}\right)}$$ If ${\displaystyle \mathrm{\Phi },\mathrm{\Psi }\subseteq X\times X}$ then $(\mathrm{\Phi }\circ \mathrm{\Psi })\cdot S=\mathrm{\Phi }\cdot (\mathrm{\Psi }\cdot S).$ Moreover, ${\displaystyle \cdot }$ right distributes over both unions and intersections, meaning that if ${\displaystyle R,S\subseteq X}$ then ${\displaystyle (R\cup S)\cdot \mathrm{\Phi }=(R\cdot \mathrm{\Phi })\cup (S\cdot \mathrm{\Phi })}$ and ${\displaystyle (R\cap S)\cdot \mathrm{\Phi }\subseteq (R\cdot \mathrm{\Phi })\cap (S\cdot \mathrm{\Phi }).}$

Neighborhoods and open sets

Two points ${\displaystyle x}$ and ${\displaystyle y}$ are ${\displaystyle \mathrm{\Phi }}$-close if ${\displaystyle (x,y)\in \mathrm{\Phi }}$ and a subset ${\displaystyle S\subseteq X}$ is called ${\displaystyle \mathrm{\Phi }}$-small if ${\displaystyle S\times S\subseteq \mathrm{\Phi }.}$

Let ${\displaystyle \mathcal{B}\subseteq \mathrm{\wp }(X\times X)}$ be a base of entourages on ${\displaystyle X.}$ The neighborhood prefilter at a point ${\displaystyle p\in X}$ and, respectively, on a subset ${\displaystyle S\subseteq X}$ are the families of sets: $${\displaystyle \mathcal{B}\cdot p\stackrel{\text{def}}{=}\mathcal{B}\cdot \{p\}=\{\mathrm{\Phi }\cdot p:\mathrm{\Phi }\in \mathcal{B}\}\text{ and }\mathcal{B}\cdot S\stackrel{\text{def}}{=}\{\mathrm{\Phi }\cdot S:\mathrm{\Phi }\in \mathcal{B}\}}$$ and the filters on ${\displaystyle X}$ that each generates is known as the neighborhood filter of ${\displaystyle p}$ (respectively, of ${\displaystyle S}$). Assign to every ${\displaystyle x\in X}$ the neighborhood prefilter $${\displaystyle \mathcal{B}\cdot x\stackrel{\text{def}}{=}\{\mathrm{\Phi }\cdot x:\mathrm{\Phi }\in \mathcal{B}\}}$$ and use the neighborhood definition of "open set" to obtain a topology on ${\displaystyle X}$ called the topology induced by ${\displaystyle \mathcal{B}}$ or the induced topology. Explicitly, a subset ${\displaystyle U\subseteq X}$ is open in this topology if and only if for every ${\displaystyle u\in U}$ there exists some ${\displaystyle N\in \mathcal{B}\cdot u}$ such that ${\displaystyle N\subseteq U;}$ that is, ${\displaystyle U}$ is open if and only if for every ${\displaystyle u\in U}$ there exists some ${\displaystyle \mathrm{\Phi }\in \mathcal{B}}$ such that ${\displaystyle \mathrm{\Phi }\cdot u\stackrel{\text{def}}{=}\{x\in X:(x,u)\in \mathrm{\Phi }\}\subseteq U.}$

The closure of a subset ${\displaystyle S\subseteq X}$ in this topology is: $${\displaystyle {\mathrm{cl}}_{X}S=\bigcap _{\mathrm{\Phi }\in \mathcal{B}}(\mathrm{\Phi }\cdot S)=\bigcap _{\mathrm{\Phi }\in \mathcal{B}}(S\cdot \mathrm{\Phi }).}$$

Cauchy prefilters and complete uniformities

A prefilter ${\displaystyle \mathcal{F}\subseteq \mathrm{\wp }(X)}$ on a uniform space ${\displaystyle X}$ with uniformity ${\displaystyle \mathcal{U}}$ is called a Cauchy prefilter if for every entourage ${\displaystyle N\in \mathcal{U},}$ there exists some ${\displaystyle F\in \mathcal{F}}$ such that ${\displaystyle F\times F\subseteq N.}$

A uniform space ${\displaystyle (X,\mathcal{U})}$ is called a complete uniform space (respectively, a sequentially complete uniform space) if every Cauchy prefilter (respectively, every elementary Cauchy prefilter) on ${\displaystyle X}$ converges to at least one point of ${\displaystyle X}$ when ${\displaystyle X}$ is endowed with the topology induced by ${\displaystyle \mathcal{U}.}$

Case of a topological vector space

If ${\displaystyle (X,\tau )}$ is a topological vector space then for any ${\displaystyle S\subseteq X}$ and ${\displaystyle x\in X,}$ $${\displaystyle {\mathrm{\Delta }}_X(N)\cdot S=S+N\text{ and }{\mathrm{\Delta }}_X(N)\cdot x=x+N,}$$ and the topology induced on ${\displaystyle X}$ by the canonical uniformity is the same as the topology that ${\displaystyle X}$ started with (that is, it is ${\displaystyle \tau }$).

Uniform continuity

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be TVSs, ${\displaystyle D\subseteq X,}$ and ${\displaystyle f:D\to Y}$ be a map. Then ${\displaystyle f:D\to Y}$ is uniformly continuous if for every neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X,}$ there exists a neighborhood ${\displaystyle V}$ of the origin in ${\displaystyle Y}$ such that for all ${\displaystyle x,y\in D,}$ if ${\displaystyle y-x\in U}$ then ${\displaystyle f(y)-f(x)\in V.}$

Suppose that ${\displaystyle f:D\to Y}$ is uniformly continuous. If ${\displaystyle x_{\bullet }={\left(x_i\right)}_{i\in I}}$ is a Cauchy net in ${\displaystyle D}$ then ${\displaystyle f\circ x_{\bullet }={\left(f\left(x_i\right)\right)}_{i\in I}}$ is a Cauchy net in ${\displaystyle Y.}$ If ${\displaystyle \mathcal{B}}$ is a Cauchy prefilter in ${\displaystyle D}$ (meaning that ${\displaystyle \mathcal{B}}$ is a family of subsets of ${\displaystyle D}$ that is Cauchy in ${\displaystyle X}$) then ${\displaystyle f\left(\mathcal{B}\right)}$ is a Cauchy prefilter in ${\displaystyle Y.}$ However, if ${\displaystyle \mathcal{B}}$ is a Cauchy filter on ${\displaystyle D}$ then although ${\displaystyle f\left(\mathcal{B}\right)}$ will be a Cauchy prefilter, it will be a Cauchy filter in ${\displaystyle Y}$ if and only if ${\displaystyle f:D\to Y}$ 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 ${\displaystyle p}$ is a metric if and only if ${\displaystyle p(x,y)=0}$ implies ${\displaystyle x=y.}$ Thus every metric space is a pseudometric space and a pseudometric space ${\displaystyle (X,p)}$ is a metric space if and only if ${\displaystyle p}$ is a metric.

If ${\displaystyle S}$ is a subset of a pseudometric space ${\displaystyle (X,d)}$ then the diameter of ${\displaystyle S}$ is defined to be $${\displaystyle \mathrm{diam}(S)\stackrel{\text{def}}{=}\underset{}{sup}\{d(s,t):s,t\in S\}.}$$

A prefilter ${\displaystyle \mathcal{B}}$ on a pseudometric space ${\displaystyle (X,d)}$ is called a ${\displaystyle d}$-Cauchy prefilter or simply a Cauchy prefilter if for each real ${\displaystyle r>0,}$ there is some ${\displaystyle B\in \mathcal{B}}$ such that the diameter of ${\displaystyle B}$ is less than ${\displaystyle r.}$

Suppose ${\displaystyle (X,d)}$ is a pseudometric space. A net ${\displaystyle x_{\bullet }={\left(x_i\right)}_{i\in I}}$ in ${\displaystyle X}$ is called a ${\displaystyle d}$-Cauchy net or simply a Cauchy net if ${\displaystyle \mathrm{Tails}\left(x_{\bullet }\right)}$ is a Cauchy prefilter, which happens if and only if

for every ${\displaystyle r>0}$ there is some ${\displaystyle i\in I}$ such that if ${\displaystyle j,k\in I}$ with ${\displaystyle j\ge i}$ and ${\displaystyle k\ge i}$ then ${\displaystyle d\left(x_j,x_k\right)<r}$

or equivalently, if and only if ${\displaystyle {\left(d\left(x_j,x_k\right)\right)}_{(i,j)\in I\times I}\to 0}$ in ${\displaystyle \mathbb{R}.}$ This is analogous to the following characterization of the converge of ${\displaystyle x_{\bullet }}$ to a point: if ${\displaystyle x\in X,}$ then ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle (X,d)}$ if and only if ${\displaystyle {\left(x_i,x\right)}_{i\in I}\to 0}$ in ${\displaystyle \mathbb{R}.}$

A Cauchy sequence is a sequence that is also a Cauchy net.

Every pseudometric ${\displaystyle p}$ on a set ${\displaystyle X}$ induces the usual canonical topology on ${\displaystyle X,}$ which we'll denote by ${\displaystyle {\tau }_p}$; it also induces a canonical uniformity on ${\displaystyle X,}$ which we'll denote by ${\displaystyle {\mathcal{U}}_p.}$ The topology on ${\displaystyle X}$ induced by the uniformity ${\displaystyle {\mathcal{U}}_p}$ is equal to ${\displaystyle {\tau }_p.}$ A net ${\displaystyle x_{\bullet }={\left(x_i\right)}_{i\in I}}$ in ${\displaystyle X}$ is Cauchy with respect to ${\displaystyle p}$ if and only if it is Cauchy with respect to the uniformity ${\displaystyle {\mathcal{U}}_p.}$ The pseudometric space ${\displaystyle (X,p)}$ is a complete (resp. a sequentially complete) pseudometric space if and only if ${\displaystyle \left(X,{\mathcal{U}}_p\right)}$ is a complete (resp. a sequentially complete) uniform space. Moreover, the pseudometric space ${\displaystyle (X,p)}$ (resp. the uniform space ${\displaystyle \left(X,{\mathcal{U}}_p\right)}$) is complete if and only if it is sequentially complete.

A pseudometric space ${\displaystyle (X,d)}$ (for example, a metric space) is called complete and ${\displaystyle d}$ is called a complete pseudometric if any of the following equivalent conditions hold:

1. Every Cauchy prefilter on ${\displaystyle X}$ converges to at least one point of ${\displaystyle X.}$
2. The above statement but with the word "prefilter" replaced by "filter."
3. Every Cauchy net in ${\displaystyle X}$ converges to at least one point of ${\displaystyle X.}$
   • If ${\displaystyle d}$ is a metric on ${\displaystyle X}$ then any limit point is necessarily unique and the same is true for limits of Cauchy prefilters on ${\displaystyle X.}$
4. Every Cauchy sequence in ${\displaystyle X}$ converges to at least one point of ${\displaystyle X.}$
   • Thus to prove that ${\displaystyle (X,d)}$ is complete, it suffices to only consider Cauchy sequences in ${\displaystyle X}$ (and it is not necessary to consider the more general Cauchy nets).
5. The canonical uniformity on ${\displaystyle X}$ induced by the pseudometric ${\displaystyle d}$ is a complete uniformity.

And if addition ${\displaystyle d}$ is a metric then we may add to this list:

1. Every decreasing sequence of closed balls whose diameters shrink to ${\displaystyle 0}$ 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 ${\displaystyle d}$ on a vector space ${\displaystyle X}$ is said to be a translation invariant pseudometric if ${\displaystyle d(x,y)=d(x+z,y+z)}$ for all vectors ${\displaystyle x,y,z\in X.}$

Suppose ${\displaystyle (X,\tau )}$ is pseudometrizable TVS (for example, a metrizable TVS) and that ${\displaystyle p}$ is any pseudometric on ${\displaystyle X}$ such that the topology on ${\displaystyle X}$ induced by ${\displaystyle p}$ is equal to ${\displaystyle \tau .}$ If ${\displaystyle p}$ is translation-invariant, then ${\displaystyle (X,\tau )}$ is a complete TVS if and only if ${\displaystyle (X,p)}$ is a complete pseudometric space. If ${\displaystyle p}$ is not translation-invariant, then may be possible for ${\displaystyle (X,\tau )}$ to be a complete TVS but ${\displaystyle (X,p)}$ to not be a complete pseudometric space (see this footnote for an example).

Theorem (Klee)—Let ${\displaystyle d}$ be any metric on a vector space ${\displaystyle X}$ such that the topology ${\displaystyle \tau }$ induced by ${\displaystyle d}$ on ${\displaystyle X}$ makes ${\displaystyle (X,\tau )}$ into a topological vector space. If ${\displaystyle (X,d)}$ is a complete metric space then ${\displaystyle (X,\tau )}$ 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 ${\displaystyle p}$ and ${\displaystyle q}$ are two equivalent norms on a vector space ${\displaystyle X}$ then the normed space ${\displaystyle (X,p)}$ is a Banach space if and only if ${\displaystyle (X,q)}$ 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. 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.

Completions

A completion of a TVS ${\displaystyle X}$ is a complete TVS that contains a dense vector subspace that is TVS-isomorphic to ${\displaystyle X.}$ In other words, it is a complete TVS ${\displaystyle C}$ into which ${\displaystyle X}$ 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.

Examples of completions

For example, the vector space consisting of scalar-valued simple functions ${\displaystyle f}$ for which ${\displaystyle |f{|}_p<\mathrm{\infty }}$ (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 ${\displaystyle L^p}$-space (endowed with the usual complete ${\displaystyle \Vert \cdot {\Vert }_p}$ 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 ${\displaystyle {\ell }^1(S)}$ with a complete Hausdorff locally convex TVS ${\displaystyle Y}$ results in a complete TVS that is TVS-isomorphic to a "generalized" ${\displaystyle {\ell }^1(S;Y)}$-space consisting ${\displaystyle Y}$-valued functions on ${\displaystyle S}$ (where this "generalized" TVS is defined analogously to original space ${\displaystyle {\ell }^1(S)}$ of scalar-valued functions on ${\displaystyle S}$). Similarly, the completion of the injective tensor product of the space of scalar-valued ${\displaystyle C^k}$-test functions with such a TVS ${\displaystyle Y}$ is TVS-isomorphic to the analogously defined TVS of ${\displaystyle Y}$-valued ${\displaystyle C^k}$ 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 (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 ${\displaystyle C}$ denote any complete TVS and let ${\displaystyle I}$ denote any TVS endowed with the indiscrete topology, which recall makes ${\displaystyle I}$ into a complete TVS. Since both ${\displaystyle I}$ and ${\displaystyle C}$ are complete TVSs, so is their product ${\displaystyle I\times C.}$ If ${\displaystyle U}$ and ${\displaystyle V}$ are non-empty open subsets of ${\displaystyle I}$ and ${\displaystyle C,}$ respectively, then ${\displaystyle U=I}$ and ${\displaystyle (U\times V)\cap (\{0\}\times C)=\{0\}\times V\ne \varnothing ,}$ which shows that ${\displaystyle \{0\}\times C}$ is a dense subspace of ${\displaystyle I\times C.}$ Thus by definition of "completion," ${\displaystyle I\times C}$ is a completion of ${\displaystyle \{0\}\times C}$ (it doesn't matter that ${\displaystyle \{0\}\times C}$ is already complete). So by identifying ${\displaystyle \{0\}\times C}$ with ${\displaystyle C,}$ if ${\displaystyle X\subseteq C}$ is a dense vector subspace of ${\displaystyle C,}$ then ${\displaystyle X}$ has both ${\displaystyle C}$ and ${\displaystyle I\times C}$ as completions.

Hausdorff 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 ${\displaystyle X}$ and ${\displaystyle C}$ are Hausdorff TVSs with ${\displaystyle C}$ complete. Suppose that ${\displaystyle E:X\to C}$ is a TVS-embedding onto a dense vector subspace of ${\displaystyle C.}$ Then

Universal property: for every continuous linear map ${\displaystyle f:X\to Z}$ into a complete Hausdorff TVS ${\displaystyle Z,}$ there exists a unique continuous linear map ${\displaystyle F:C\to Z}$ such that ${\displaystyle f=F\circ E.}$

If ${\displaystyle E_2:X\to C_2}$ is a TVS embedding onto a dense vector subspace of a complete Hausdorff TVS ${\displaystyle C_2}$ having the above universal property, then there exists a unique (bijective) TVS-isomorphism ${\displaystyle I:C\to C_2}$ such that ${\displaystyle E_2=I\circ E.}$

Corollary—Suppose ${\displaystyle C}$ is a complete Hausdorff TVS and ${\displaystyle X}$ is a dense vector subspace of ${\displaystyle C.}$ Then every continuous linear map ${\displaystyle f:X\to Z}$ into a complete Hausdorff TVS ${\displaystyle Z}$ has a unique continuous linear extension to a map ${\displaystyle C\to Z.}$

Existence of Hausdorff completions

A Cauchy filter ${\displaystyle \mathcal{B}}$ on a TVS ${\displaystyle X}$ is called a minimal Cauchy filter if there does not exist a Cauchy filter on ${\displaystyle X}$ that is strictly coarser than ${\displaystyle \mathcal{B}}$ (that is, "strictly coarser than ${\displaystyle \mathcal{B}}$" means contained as a proper subset of ${\displaystyle \mathcal{B}}$).

If ${\displaystyle \mathcal{B}}$ is a Cauchy filter on ${\displaystyle X}$ then the filter generated by the following prefilter: $${\displaystyle \left\{B+N:B\in \mathcal{B}\text{ and }N\text{ is a neighborhood of }0\text{ in }X\right\}}$$ is the unique minimal Cauchy filter on ${\displaystyle X}$ that is contained as a subset of ${\displaystyle \mathcal{B}.}$ In particular, for any ${\displaystyle x\in X,}$ the neighborhood filter at ${\displaystyle x}$ is a minimal Cauchy filter.

Let ${\displaystyle \mathbb{M}}$ be the set of all minimal Cauchy filters on ${\displaystyle X}$ and let ${\displaystyle E:X\to \mathbb{M}}$ be the map defined by sending ${\displaystyle x\in X}$ to the neighborhood filter of ${\displaystyle x}$ in ${\displaystyle X.}$ Endow ${\displaystyle \mathbb{M}}$ with the following vector space structure: Given ${\displaystyle \mathcal{B},\mathcal{C}\in \mathbb{M}}$ and a scalar ${\displaystyle s,}$ let ${\displaystyle \mathcal{B}+\mathcal{C}}$ (resp. ${\displaystyle s\mathcal{B}}$) denote the unique minimal Cauchy filter contained in the filter generated by ${\displaystyle \left\{B+C:B\in \mathcal{B},C\in \mathcal{C}\right\}}$ (resp. ${\displaystyle \{sB:B\in \mathcal{B}\}}$).

For every balanced neighborhood ${\displaystyle N}$ of the origin in ${\displaystyle X,}$ let $${\displaystyle \mathbb{U}(N)\stackrel{\text{def}}{=}\left\{\mathcal{B}\in \mathbb{M}:\text{ there exist }B\in \mathcal{B}\text{ and a neighborhood }V\text{ of the origin in }X\text{ such that }B+V\subseteq N\right\}}$$

If ${\displaystyle X}$ is Hausdorff then the collection of all sets ${\displaystyle \mathbb{U}(N),}$ as ${\displaystyle N}$ ranges over all balanced neighborhoods of the origin in ${\displaystyle X,}$ forms a vector topology on ${\displaystyle \mathbb{M}}$ making ${\displaystyle \mathbb{M}}$ into a complete Hausdorff TVS. Moreover, the map ${\displaystyle E:X\to \mathbb{M}}$ is a TVS-embedding onto a dense vector subspace of ${\displaystyle \mathbb{M}.}$

If ${\displaystyle X}$ is a metrizable TVS then a Hausdorff completion of ${\displaystyle X}$ 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 ${\displaystyle X}$ 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 ${\displaystyle I=\mathrm{cl}\{0\}}$ denote the closure of the origin in ${\displaystyle X,}$ where ${\displaystyle I}$ is endowed with its subspace topology induced by ${\displaystyle X}$ (so that ${\displaystyle I}$ has the indiscrete topology). Since ${\displaystyle I}$ has the trivial topology, it is easily shown that every vector subspace of ${\displaystyle X}$ that is an algebraic complement of ${\displaystyle I}$ in ${\displaystyle X}$ is necessarily a topological complement of ${\displaystyle I}$ in ${\displaystyle X.}$ Let ${\displaystyle H}$ denote any topological complement of ${\displaystyle I}$ in ${\displaystyle X,}$ which is necessarily a Hausdorff TVS (since it is TVS-isomorphic to the quotient TVS ${\displaystyle X/I}$). Since ${\displaystyle X}$ is the topological direct sum of ${\displaystyle I}$ and ${\displaystyle H}$ (which means that ${\displaystyle X=I\oplus H}$ in the category of TVSs), the canonical map $${\displaystyle I\times H\to I\oplus H=X\text{ given by }(x,y)\mapsto x+y}$$ is a TVS-isomorphism. Let ${\displaystyle A:X=I\oplus H\to I\times H}$ denote the inverse of this canonical map. (As a side note, it follows that every open and every closed subset ${\displaystyle U}$ of ${\displaystyle X}$ satisfies ${\displaystyle U=I+U.}$)

The Hausdorff TVS ${\displaystyle H}$ can be TVS-embedded, say via the map ${\displaystyle {\mathrm{In}}_H:H\to C,}$ onto a dense vector subspace of its completion ${\displaystyle C.}$ Since ${\displaystyle I}$ and ${\displaystyle C}$ are complete, so is their product ${\displaystyle I\times C.}$ Let ${\displaystyle {\mathrm{Id}}_I:I\to I}$ denote the identity map and observe that the product map ${\displaystyle {\mathrm{Id}}_I\times {\mathrm{In}}_H:I\times H\to I\times C}$ is a TVS-embedding whose image is dense in ${\displaystyle I\times C.}$ Define the map $${\displaystyle B:X=I\oplus H\to I\times C\text{ by }B\stackrel{\text{def}}{=}\left({\mathrm{Id}}_I\times {\mathrm{In}}_H\right)\circ A}$$ which is a TVS-embedding of ${\displaystyle X=I\oplus H}$ onto a dense vector subspace of the complete TVS ${\displaystyle I\times C.}$ Moreover, observe that the closure of the origin in ${\displaystyle I\times C}$ is equal to ${\displaystyle I\times \{0\},}$ and that ${\displaystyle I\times \{0\}}$ and ${\displaystyle \{0\}\times C}$ are topological complements in ${\displaystyle I\times C.}$

To summarize, given any algebraic (and thus topological) complement ${\displaystyle H}$ of ${\displaystyle I\stackrel{\text{def}}{=}\mathrm{cl}\{0\}}$ in ${\displaystyle X}$ and given any completion ${\displaystyle C}$ of the Hausdorff TVS ${\displaystyle H}$ such that ${\displaystyle H\subseteq C,}$ then the natural inclusion $${\displaystyle {\mathrm{In}}_H:X=I\oplus H\to I\oplus C}$$ is a well-defined TVS-embedding of ${\displaystyle X}$ onto a dense vector subspace of the complete TVS ${\displaystyle I\oplus C}$ where moreover, $${\displaystyle X=I\oplus H\subseteq I\oplus C\cong I\times C.}$$

Topology of a completion

Theorem (Topology of a completion)—Let ${\displaystyle C}$ be a complete TVS and let ${\displaystyle X}$ be a dense vector subspace of ${\displaystyle X.}$ If ${\displaystyle {\mathcal{N}}_X(0)}$ is any neighborhood base of the origin in ${\displaystyle X}$ then the set $${\displaystyle {\mathcal{N}}_X(0)\stackrel{\text{def}}{=}\left\{{\mathrm{cl}}_{C}N:N\in {\mathcal{N}}_X(0)\right\}}$$ is a neighborhood of the origin in the completion ${\displaystyle C}$ of ${\displaystyle X.}$

If ${\displaystyle X}$ is locally convex and ${\displaystyle \mathcal{P}}$ is a family of continuous seminorms on ${\displaystyle X}$ that generate the topology of ${\displaystyle X,}$ then the family of all continuous extensions to ${\displaystyle C}$ of all members of ${\displaystyle \mathcal{P}}$ is a generating family of seminorms for ${\displaystyle C.}$

Said differently, if ${\displaystyle C}$ is a completion of a TVS ${\displaystyle X}$ with ${\displaystyle X\subseteq C}$ and if ${\displaystyle \mathcal{N}}$ is a neighborhood base of the origin in ${\displaystyle X,}$ then the family of sets $${\displaystyle \left\{{\mathrm{cl}}_{C}N:N\in \mathcal{N}\right\}}$$ is a neighborhood basis at the origin in ${\displaystyle C.}$

Theorem (Completions of quotients)—Let ${\displaystyle M}$ be a metrizable topological vector space and let ${\displaystyle N}$ be a closed vector subspace of ${\displaystyle M.}$ Suppose that ${\displaystyle C}$ is a completion of ${\displaystyle M.}$ Then the completion of ${\displaystyle M/N}$ is TVS-isomorphic to ${\displaystyle C/{\mathrm{cl}}_{C}N.}$ If in addition ${\displaystyle M}$ is a normed space, then this TVS-isomorphism is also an isometry.

Grothendieck's Completeness Theorem

Let ${\displaystyle \mathcal{E}}$ denote the equicontinuous compactology on the continuous dual space ${\displaystyle X^{\mathrm{\prime }},}$ which by definition consists of all equicontinuous weak-* closed and weak-* bounded absolutely convex subsets of ${\displaystyle X^{\mathrm{\prime }}}$ (which are necessarily weak-* compact subsets of ${\displaystyle X^{\mathrm{\prime }}}$). Assume that every ${\displaystyle E^{\mathrm{\prime }}\in \mathcal{E}}$ is endowed with the weak-* topology. A filter ${\displaystyle \mathcal{B}}$ on ${\displaystyle X^{\mathrm{\prime }}}$ is said to converge continuously to ${\displaystyle x^{\mathrm{\prime }}\in X^{\mathrm{\prime }}}$ if there exists some ${\displaystyle E^{\mathrm{\prime }}\in \mathcal{E}\cap \mathcal{B}}$ containing ${\displaystyle x^{\mathrm{\prime }}}$ (that is, ${\displaystyle x^{\mathrm{\prime }}\in E^{\mathrm{\prime }}}$) such that the trace of ${\displaystyle \mathcal{B}}$ on ${\displaystyle E^{\mathrm{\prime }},}$ which is the family ${\displaystyle \mathcal{B}{|}_{E^{\mathrm{\prime }}}\stackrel{\text{def}}{=}\left\{B\cap E^{\mathrm{\prime }}:B\in \mathcal{B}\right\},}$ converges to ${\displaystyle x^{\mathrm{\prime }}}$ in ${\displaystyle E^{\mathrm{\prime }}}$ (that is, if ${\displaystyle \mathcal{B}{|}_{E^{\mathrm{\prime }}}\to x^{\mathrm{\prime }}}$ in the given weak-* topology). The filter ${\displaystyle \mathcal{B}}$ converges continuously to ${\displaystyle x^{\mathrm{\prime }}}$ if and only if ${\displaystyle \mathcal{B}-x^{\mathrm{\prime }}}$ converges continuously to the origin, which happens if and only if for every ${\displaystyle x\in X,}$ the filter ${\displaystyle ⟨\mathcal{B},x+\mathcal{N}⟩\to ⟨x^{\mathrm{\prime }},x⟩}$ in the scalar field (which is ${\displaystyle \mathbb{R}}$ or ${\displaystyle \mathbb{C}}$) where ${\displaystyle \mathcal{N}}$ denotes any neighborhood basis at the origin in ${\displaystyle X,}$ ${\displaystyle ⟨\cdot ,\cdot ⟩}$ denotes the duality pairing, and ${\displaystyle ⟨\mathcal{B},x+\mathcal{N}⟩}$ denotes the filter generated by ${\displaystyle \{⟨B,x+N⟩:B\in \mathcal{B},N\in \mathcal{N}\}.}$ A map ${\displaystyle f:X^{\mathrm{\prime }}\to T}$ into a topological space (such as ${\displaystyle \mathbb{R}}$ or ${\displaystyle \mathbb{C}}$) is said to be ${\displaystyle \gamma }$-continuous if whenever a filter ${\displaystyle \mathcal{B}}$ on ${\displaystyle X^{\mathrm{\prime }}}$ converges continuously to ${\displaystyle x^{\mathrm{\prime }}\in X^{\mathrm{\prime }},}$ then ${\displaystyle f(\mathcal{B})\to f\left(x^{\mathrm{\prime }}\right).}$

Grothendieck's Completeness Theorem—If ${\displaystyle X}$ is a Hausdorff topological vector space then its completion is linearly isomorphic to the set of all ${\displaystyle \gamma }$-continuous linear functions on ${\displaystyle X^{\mathrm{\prime }}.}$

Properties preserved by completions

If a TVS ${\displaystyle X}$ has any of the following properties then so does its completion:

• Hausdorff
• Locally convex
• Pseudometrizable
• Metrizable
• Seminormable
• Normable
  • Moreover, if ${\displaystyle (X,p)}$ is a normed space, then the completion can be chosen to be a Banach space ${\displaystyle (B,q)}$ such that the TVS-embedding of ${\displaystyle (X,p)}$ into ${\displaystyle (B,q)}$ is an isometry.
• Hausdorff pre-Hilbert. That is, a TVS induced by an inner product.
• Nuclear
• Barrelled
• Mackey
• DF-space

Completions of Hilbert spaces

Every inner product space ${\displaystyle \left(H,⟨\cdot ,\cdot ⟩\right)}$ has a completion ${\displaystyle \left(\overline{H},⟨\cdot ,\cdot {⟩}_{\overline{H}}\right)}$ that is a Hilbert space, where the inner product ${\displaystyle ⟨\cdot ,\cdot {⟩}_{\overline{H}}}$ is the unique continuous extension to ${\displaystyle \overline{H}}$ of the original inner product ${\displaystyle ⟨\cdot ,\cdot ⟩.}$ The norm induced by ${\displaystyle \left(\overline{H},⟨\cdot ,\cdot {⟩}_{\overline{H}}\right)}$ is also the unique continuous extension to ${\displaystyle \overline{H}}$ of the norm induced by ${\displaystyle ⟨\cdot ,\cdot ⟩.}$

Other preserved properties

If ${\displaystyle X}$ is a Hausdorff TVS, then the continuous dual space of ${\displaystyle X}$ is identical to the continuous dual space of the completion of ${\displaystyle X.}$ The completion of a locally convex bornological space is a barrelled space. If ${\displaystyle X}$ and ${\displaystyle Y}$ 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 ${\displaystyle X=Y\oplus Z}$ (meaning that the addition map ${\displaystyle Y\times Z\to X}$ is a TVS-isomorphism) has a Hausdorff completion ${\displaystyle C}$ then ${\displaystyle \left({\mathrm{cl}}_{C}Y\right)+\left({\mathrm{cl}}_{C}Z\right)=C.}$ If in addition ${\displaystyle X}$ is an inner product space and ${\displaystyle Y}$ and ${\displaystyle Z}$ are orthogonal complements of each other in ${\displaystyle X}$ (that is, ${\displaystyle ⟨Y,Z⟩=\{0\}}$), then ${\displaystyle {\mathrm{cl}}_{C}Y}$ and ${\displaystyle {\mathrm{cl}}_{C}Z}$ are orthogonal complements in the Hilbert space ${\displaystyle C.}$