In functional analysis and related areas of mathematics, a metrizable (resp. pseudometrizable) topological vector space (TVS) is a TVS whose topology is induced by a metric (resp. pseudometric). An LM-space is an inductive limit of a sequence of locally convex metrizable TVS.

## Pseudometrics and metrics

A pseudometric on a set ${\displaystyle X}$ is a map ${\displaystyle d:X\times X\rightarrow \mathbb {R} }$ satisfying the following properties:

1. ${\displaystyle d(x,x)=0{\text{ for all ))x\in X}$;
2. Symmetry: ${\displaystyle d(x,y)=d(y,x){\text{ for all ))x,y\in X}$;
3. Subadditivity: ${\displaystyle d(x,z)\leq d(x,y)+d(y,z){\text{ for all ))x,y,z\in X.}$

A pseudometric is called a metric if it satisfies:

1. Identity of indiscernibles: for all ${\displaystyle x,y\in X,}$ if ${\displaystyle d(x,y)=0}$ then ${\displaystyle x=y.}$

Ultrapseudometric

A pseudometric ${\displaystyle d}$ on ${\displaystyle X}$ is called a ultrapseudometric or a strong pseudometric if it satisfies:

1. Strong/Ultrametric triangle inequality: ${\displaystyle d(x,z)\leq \max\{d(x,y),d(y,z)\}{\text{ for all ))x,y,z\in X.}$

Pseudometric space

A pseudometric space is a pair ${\displaystyle (X,d)}$ consisting of a set ${\displaystyle X}$ and a pseudometric ${\displaystyle d}$ on ${\displaystyle X}$ such that ${\displaystyle X}$'s topology is identical to the topology on ${\displaystyle X}$ induced by ${\displaystyle d.}$ We call a pseudometric space ${\displaystyle (X,d)}$ a metric space (resp. ultrapseudometric space) when ${\displaystyle d}$ is a metric (resp. ultrapseudometric).

### Topology induced by a pseudometric

If ${\displaystyle d}$ is a pseudometric on a set ${\displaystyle X}$ then collection of open balls:

${\displaystyle B_{r}(z):=\{x\in X:d(x,z)
as ${\displaystyle z}$ ranges over ${\displaystyle X}$ and ${\displaystyle r>0}$ ranges over the positive real numbers, forms a basis for a topology on ${\displaystyle X}$ that is called the ${\displaystyle d}$-topology or the pseudometric topology on ${\displaystyle X}$ induced by ${\displaystyle d.}$

Convention: If ${\displaystyle (X,d)}$ is a pseudometric space and ${\displaystyle X}$ is treated as a topological space, then unless indicated otherwise, it should be assumed that ${\displaystyle X}$ is endowed with the topology induced by ${\displaystyle d.}$

Pseudometrizable space

A topological space ${\displaystyle (X,\tau )}$ is called pseudometrizable (resp. metrizable, ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) ${\displaystyle d}$ on ${\displaystyle X}$ such that ${\displaystyle \tau }$ is equal to the topology induced by ${\displaystyle d.}$[1]

## Pseudometrics and values on topological groups

An additive topological group is an additive group endowed with a topology, called a group topology, under which addition and negation become continuous operators.

A topology ${\displaystyle \tau }$ on a real or complex vector space ${\displaystyle X}$ is called a vector topology or a TVS topology if it makes the operations of vector addition and scalar multiplication continuous (that is, if it makes ${\displaystyle X}$ into a topological vector space).

Every topological vector space (TVS) ${\displaystyle X}$ is an additive commutative topological group but not all group topologies on ${\displaystyle X}$ are vector topologies. This is because despite it making addition and negation continuous, a group topology on a vector space ${\displaystyle X}$ may fail to make scalar multiplication continuous. For instance, the discrete topology on any non-trivial vector space makes addition and negation continuous but do not make scalar multiplication continuous.

### Translation invariant pseudometrics

If ${\displaystyle X}$ is an additive group then we say that a pseudometric ${\displaystyle d}$ on ${\displaystyle X}$ is translation invariant or just invariant if it satisfies any of the following equivalent conditions:

1. Translation invariance: ${\displaystyle d(x+z,y+z)=d(x,y){\text{ for all ))x,y,z\in X}$;
2. ${\displaystyle d(x,y)=d(x-y,0){\text{ for all ))x,y\in X.}$

### Value/G-seminorm

If ${\displaystyle X}$ is a topological group the a value or G-seminorm on ${\displaystyle X}$ (the G stands for Group) is a real-valued map ${\displaystyle p:X\rightarrow \mathbb {R} }$ with the following properties:[2]

1. Non-negative: ${\displaystyle p\geq 0.}$
2. Subadditive: ${\displaystyle p(x+y)\leq p(x)+p(y){\text{ for all ))x,y\in X}$;
3. ${\displaystyle p(0)=0..}$
4. Symmetric: ${\displaystyle p(-x)=p(x){\text{ for all ))x\in X.}$

where we call a G-seminorm a G-norm if it satisfies the additional condition:

1. Total/Positive definite: If ${\displaystyle p(x)=0}$ then ${\displaystyle x=0.}$

#### Properties of values

If ${\displaystyle p}$ is a value on a vector space ${\displaystyle X}$ then:

• ${\displaystyle |p(x)-p(y)|\leq p(x-y){\text{ for all ))x,y\in X.}$[3]
• ${\displaystyle p(nx)\leq np(x)}$ and ${\displaystyle {\frac {1}{n))p(x)\leq p(x/n)}$ for all ${\displaystyle x\in X}$ and positive integers ${\displaystyle n.}$[4]
• The set ${\displaystyle \{x\in X:p(x)=0\))$ is an additive subgroup of ${\displaystyle X.}$[3]

### Equivalence on topological groups

Theorem[2] — Suppose that ${\displaystyle X}$ is an additive commutative group. If ${\displaystyle d}$ is a translation invariant pseudometric on ${\displaystyle X}$ then the map ${\displaystyle p(x):=d(x,0)}$ is a value on ${\displaystyle X}$ called the value associated with ${\displaystyle d}$, and moreover, ${\displaystyle d}$ generates a group topology on ${\displaystyle X}$ (i.e. the ${\displaystyle d}$-topology on ${\displaystyle X}$ makes ${\displaystyle X}$ into a topological group). Conversely, if ${\displaystyle p}$ is a value on ${\displaystyle X}$ then the map ${\displaystyle d(x,y):=p(x-y)}$ is a translation-invariant pseudometric on ${\displaystyle X}$ and the value associated with ${\displaystyle d}$ is just ${\displaystyle p.}$

### Pseudometrizable topological groups

Theorem[2] — If ${\displaystyle (X,\tau )}$ is an additive commutative topological group then the following are equivalent:

1. ${\displaystyle \tau }$ is induced by a pseudometric; (i.e. ${\displaystyle (X,\tau )}$ is pseudometrizable);
2. ${\displaystyle \tau }$ is induced by a translation-invariant pseudometric;
3. the identity element in ${\displaystyle (X,\tau )}$ has a countable neighborhood basis.

If ${\displaystyle (X,\tau )}$ is Hausdorff then the word "pseudometric" in the above statement may be replaced by the word "metric." A commutative topological group is metrizable if and only if it is Hausdorff and pseudometrizable.

### An invariant pseudometric that doesn't induce a vector topology

Let ${\displaystyle X}$ be a non-trivial (i.e. ${\displaystyle X\neq \{0\))$) real or complex vector space and let ${\displaystyle d}$ be the translation-invariant trivial metric on ${\displaystyle X}$ defined by ${\displaystyle d(x,x)=0}$ and ${\displaystyle d(x,y)=1{\text{ for all ))x,y\in X}$ such that ${\displaystyle x\neq y.}$ The topology ${\displaystyle \tau }$ that ${\displaystyle d}$ induces on ${\displaystyle X}$ is the discrete topology, which makes ${\displaystyle (X,\tau )}$ into a commutative topological group under addition but does not form a vector topology on ${\displaystyle X}$ because ${\displaystyle (X,\tau )}$ is disconnected but every vector topology is connected. What fails is that scalar multiplication isn't continuous on ${\displaystyle (X,\tau ).}$

This example shows that a translation-invariant (pseudo)metric is not enough to guarantee a vector topology, which leads us to define paranorms and F-seminorms.

## Additive sequences

A collection ${\displaystyle {\mathcal {N))}$ of subsets of a vector space is called additive[5] if for every ${\displaystyle N\in {\mathcal {N)),}$ there exists some ${\displaystyle U\in {\mathcal {N))}$ such that ${\displaystyle U+U\subseteq N.}$

Continuity of addition at 0 — If ${\displaystyle (X,+)}$ is a group (as all vector spaces are), ${\displaystyle \tau }$ is a topology on ${\displaystyle X,}$ and ${\displaystyle X\times X}$ is endowed with the product topology, then the addition map ${\displaystyle X\times X\to X}$ (i.e. the map ${\displaystyle (x,y)\mapsto x+y}$) is continuous at the origin of ${\displaystyle X\times X}$ if and only if the set of neighborhoods of the origin in ${\displaystyle (X,\tau )}$ is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."[5]

All of the above conditions are consequently a necessary for a topology to form a vector topology. Additive sequences of sets have the particularly nice property that they define non-negative continuous real-valued subadditive functions. These functions can then be used to prove many of the basic properties of topological vector spaces and also show that a Hausdorff TVS with a countable basis of neighborhoods is metrizable. The following theorem is true more generally for commutative additive topological groups.

Theorem — Let ${\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty ))$ be a collection of subsets of a vector space such that ${\displaystyle 0\in U_{i))$ and ${\displaystyle U_{i+1}+U_{i+1}\subseteq U_{i))$ for all ${\displaystyle i\geq 0.}$ For all ${\displaystyle u\in U_{0},}$ let

${\displaystyle \mathbb {S} (u):=\left\{n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)~:~k\geq 1,n_{i}\geq 0{\text{ for all ))i,{\text{ and ))u\in U_{n_{1))+\cdots +U_{n_{k))\right\}.}$

Define ${\displaystyle f:X\to [0,1]}$ by ${\displaystyle f(x)=1}$ if ${\displaystyle x\not \in U_{0))$ and otherwise let

${\displaystyle f(x):=\inf _{}\left\{2^{-n_{1))+\cdots 2^{-n_{k))~:~n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)\in \mathbb {S} (x)\right\}.}$

Then ${\displaystyle f}$ is subadditive (meaning ${\displaystyle f(x+y)\leq f(x)+f(y){\text{ for all ))x,y\in X}$) and ${\displaystyle f=0}$ on ${\displaystyle \bigcap _{i\geq 0}U_{i},}$ so in particular ${\displaystyle f(0)=0.}$ If all ${\displaystyle U_{i))$ are symmetric sets then ${\displaystyle f(-x)=f(x)}$ and if all ${\displaystyle U_{i))$ are balanced then ${\displaystyle f(sx)\leq f(x)}$ for all scalars ${\displaystyle s}$ such that ${\displaystyle |s|\leq 1}$ and all ${\displaystyle x\in X.}$ If ${\displaystyle X}$ is a topological vector space and if all ${\displaystyle U_{i))$ are neighborhoods of the origin then ${\displaystyle f}$ is continuous, where if in addition ${\displaystyle X}$ is Hausdorff and ${\displaystyle U_{\bullet ))$ forms a basis of balanced neighborhoods of the origin in ${\displaystyle X}$ then ${\displaystyle d(x,y):=f(x-y)}$ is a metric defining the vector topology on ${\displaystyle X.}$

Proof

Assume that ${\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)}$ always denotes a finite sequence of non-negative integers and use the notation:

${\displaystyle \sum 2^{-n_{\bullet )):=2^{-n_{1))+\cdots +2^{-n_{k))\quad {\text{ and ))\quad \sum U_{n_{\bullet )):=U_{n_{1))+\cdots +U_{n_{k)).}$

For any integers ${\displaystyle n\geq 0}$ and ${\displaystyle d>2,}$

${\displaystyle U_{n}\supseteq U_{n+1}+U_{n+1}\supseteq U_{n+1}+U_{n+2}+U_{n+2}\supseteq U_{n+1}+U_{n+2}+\cdots +U_{n+d}+U_{n+d+1}+U_{n+d+1}.}$

From this it follows that if ${\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)}$ consists of distinct positive integers then ${\displaystyle \sum U_{n_{\bullet ))\subseteq U_{-1+\min \left(n_{\bullet }\right)}.}$

It will now be shown by induction on ${\displaystyle k}$ that if ${\displaystyle n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)}$ consists of non-negative integers such that ${\displaystyle \sum 2^{-n_{\bullet ))\leq 2^{-M))$ for some integer ${\displaystyle M\geq 0}$ then ${\displaystyle \sum U_{n_{\bullet ))\subseteq U_{M}.}$ This is clearly true for ${\displaystyle k=1}$ and ${\displaystyle k=2}$ so assume that ${\displaystyle k>2,}$ which implies that all ${\displaystyle n_{i))$ are positive. If all ${\displaystyle n_{i))$ are distinct then this step is done, and otherwise pick distinct indices ${\displaystyle i such that ${\displaystyle n_{i}=n_{j))$ and construct ${\displaystyle m_{\bullet }=\left(m_{1},\ldots ,m_{k-1}\right)}$ from ${\displaystyle n_{\bullet ))$ by replacing each ${\displaystyle n_{i))$ with ${\displaystyle n_{i}-1}$ and deleting the ${\displaystyle j^{\text{th))}$ element of ${\displaystyle n_{\bullet ))$ (all other elements of ${\displaystyle n_{\bullet ))$ are transferred to ${\displaystyle m_{\bullet ))$ unchanged). Observe that ${\displaystyle \sum 2^{-n_{\bullet ))=\sum 2^{-m_{\bullet ))}$ and ${\displaystyle \sum U_{n_{\bullet ))\subseteq \sum U_{m_{\bullet ))}$ (because ${\displaystyle U_{n_{i))+U_{n_{j))\subseteq U_{n_{i}-1))$) so by appealing to the inductive hypothesis we conclude that ${\displaystyle \sum U_{n_{\bullet ))\subseteq \sum U_{m_{\bullet ))\subseteq U_{M},}$ as desired.

It is clear that ${\displaystyle f(0)=0}$ and that ${\displaystyle 0\leq f\leq 1}$ so to prove that ${\displaystyle f}$ is subadditive, it suffices to prove that ${\displaystyle f(x+y)\leq f(x)+f(y)}$ when ${\displaystyle x,y\in X}$ are such that ${\displaystyle f(x)+f(y)<1,}$ which implies that ${\displaystyle x,y\in U_{0}.}$ This is an exercise. If all ${\displaystyle U_{i))$ are symmetric then ${\displaystyle x\in \sum U_{n_{\bullet ))}$ if and only if ${\displaystyle -x\in \sum U_{n_{\bullet ))}$ from which it follows that ${\displaystyle f(-x)\leq f(x)}$ and ${\displaystyle f(-x)\geq f(x).}$ If all ${\displaystyle U_{i))$ are balanced then the inequality ${\displaystyle f(sx)\leq f(x)}$ for all unit scalars ${\displaystyle s}$ such that ${\displaystyle |s|\leq 1}$ is proved similarly. Because ${\displaystyle f}$ is a nonnegative subadditive function satisfying ${\displaystyle f(0)=0,}$ as described in the article on sublinear functionals, ${\displaystyle f}$ is uniformly continuous on ${\displaystyle X}$ if and only if ${\displaystyle f}$ is continuous at the origin. If all ${\displaystyle U_{i))$ are neighborhoods of the origin then for any real ${\displaystyle r>0,}$ pick an integer ${\displaystyle M>1}$ such that ${\displaystyle 2^{-M} so that ${\displaystyle x\in U_{M))$ implies ${\displaystyle f(x)\leq 2^{-M} If the set of all ${\displaystyle U_{i))$ form basis of balanced neighborhoods of the origin then it may be shown that for any ${\displaystyle n>1,}$ there exists some ${\displaystyle 0 such that ${\displaystyle f(x) implies ${\displaystyle x\in U_{n}.}$ ${\displaystyle \blacksquare }$

## Paranorms

If ${\displaystyle X}$ is a vector space over the real or complex numbers then a paranorm on ${\displaystyle X}$ is a G-seminorm (defined above) ${\displaystyle p:X\rightarrow \mathbb {R} }$ on ${\displaystyle X}$ that satisfies any of the following additional conditions, each of which begins with "for all sequences ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle X}$ and all convergent sequences of scalars ${\displaystyle s_{\bullet }=\left(s_{i}\right)_{i=1}^{\infty ))$":[6]

1. Continuity of multiplication: if ${\displaystyle s}$ is a scalar and ${\displaystyle x\in X}$ are such that ${\displaystyle p\left(x_{i}-x\right)\to 0}$ and ${\displaystyle s_{\bullet }\to s,}$ then ${\displaystyle p\left(s_{i}x_{i}-sx\right)\to 0.}$
2. Both of the conditions:
• if ${\displaystyle s_{\bullet }\to 0}$ and if ${\displaystyle x\in X}$ is such that ${\displaystyle p\left(x_{i}-x\right)\to 0}$ then ${\displaystyle p\left(s_{i}x_{i}\right)\to 0}$;
• if ${\displaystyle p\left(x_{\bullet }\right)\to 0}$ then ${\displaystyle p\left(sx_{i}\right)\to 0}$ for every scalar ${\displaystyle s.}$
3. Both of the conditions:
• if ${\displaystyle p\left(x_{\bullet }\right)\to 0}$ and ${\displaystyle s_{\bullet }\to s}$ for some scalar ${\displaystyle s}$ then ${\displaystyle p\left(s_{i}x_{i}\right)\to 0}$;
• if ${\displaystyle s_{\bullet }\to 0}$ then ${\displaystyle p\left(s_{i}x\right)\to 0{\text{ for all ))x\in X.}$
4. Separate continuity:[7]
• if ${\displaystyle s_{\bullet }\to s}$ for some scalar ${\displaystyle s}$ then ${\displaystyle p\left(sx_{i}-sx\right)\to 0}$ for every ${\displaystyle x\in X}$;
• if ${\displaystyle s}$ is a scalar, ${\displaystyle x\in X,}$ and ${\displaystyle p\left(x_{i}-x\right)\to 0}$ then ${\displaystyle p\left(sx_{i}-sx\right)\to 0}$ .

A paranorm is called total if in addition it satisfies:

• Total/Positive definite: ${\displaystyle p(x)=0}$ implies ${\displaystyle x=0.}$

### Properties of paranorms

If ${\displaystyle p}$ is a paranorm on a vector space ${\displaystyle X}$ then the map ${\displaystyle d:X\times X\rightarrow \mathbb {R} }$ defined by ${\displaystyle d(x,y):=p(x-y)}$ is a translation-invariant pseudometric on ${\displaystyle X}$ that defines a vector topology on ${\displaystyle X.}$[8]

If ${\displaystyle p}$ is a paranorm on a vector space ${\displaystyle X}$ then:

• the set ${\displaystyle \{x\in X:p(x)=0\))$ is a vector subspace of ${\displaystyle X.}$[8]
• ${\displaystyle p(x+n)=p(x){\text{ for all ))x,n\in X}$ with ${\displaystyle p(n)=0.}$[8]
• If a paranorm ${\displaystyle p}$ satisfies ${\displaystyle p(sx)\leq |s|p(x){\text{ for all ))x\in X}$ and scalars ${\displaystyle s,}$ then ${\displaystyle p}$ is absolutely homogeneity (i.e. equality holds)[8] and thus ${\displaystyle p}$ is a seminorm.

### Examples of paranorms

• If ${\displaystyle d}$ is a translation-invariant pseudometric on a vector space ${\displaystyle X}$ that induces a vector topology ${\displaystyle \tau }$ on ${\displaystyle X}$ (i.e. ${\displaystyle (X,\tau )}$ is a TVS) then the map ${\displaystyle p(x):=d(x-y,0)}$ defines a continuous paranorm on ${\displaystyle (X,\tau )}$; moreover, the topology that this paranorm ${\displaystyle p}$ defines in ${\displaystyle X}$ is ${\displaystyle \tau .}$[8]
• If ${\displaystyle p}$ is a paranorm on ${\displaystyle X}$ then so is the map ${\displaystyle q(x):=p(x)/[1+p(x)].}$[8]
• Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).
• Every seminorm is a paranorm.[8]
• The restriction of an paranorm (resp. total paranorm) to a vector subspace is an paranorm (resp. total paranorm).[9]
• The sum of two paranorms is a paranorm.[8]
• If ${\displaystyle p}$ and ${\displaystyle q}$ are paranorms on ${\displaystyle X}$ then so is ${\displaystyle (p\wedge q)(x):=\inf _{}\{p(y)+q(z):x=y+z{\text{ with ))y,z\in X\}.}$ Moreover, ${\displaystyle (p\wedge q)\leq p}$ and ${\displaystyle (p\wedge q)\leq q.}$ This makes the set of paranorms on ${\displaystyle X}$ into a conditionally complete lattice.[8]
• Each of the following real-valued maps are paranorms on ${\displaystyle X:=\mathbb {R} ^{2))$:
• ${\displaystyle (x,y)\mapsto |x|}$
• ${\displaystyle (x,y)\mapsto |x|+|y|}$
• The real-valued maps ${\displaystyle (x,y)\mapsto {\sqrt {\left|x^{2}-y^{2}\right|))}$ and ${\displaystyle (x,y)\mapsto \left|x^{2}-y^{2}\right|^{3/2))$ are not a paranorms on ${\displaystyle X:=\mathbb {R} ^{2}.}$[8]
• If ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I))$ is a Hamel basis on a vector space ${\displaystyle X}$ then the real-valued map that sends ${\displaystyle x=\sum _{i\in I}s_{i}x_{i}\in X}$ (where all but finitely many of the scalars ${\displaystyle s_{i))$ are 0) to ${\displaystyle \sum _{i\in I}{\sqrt {\left|s_{i}\right|))}$ is a paranorm on ${\displaystyle X,}$ which satisfies ${\displaystyle p(sx)={\sqrt {|s|))p(x)}$ for all ${\displaystyle x\in X}$ and scalars ${\displaystyle s.}$[8]
• The function ${\displaystyle p(x):=|\sin(\pi x)|+\min\{2,|x|\))$is a paranorm on ${\displaystyle \mathbb {R} }$ that is not balanced but nevertheless equivalent to the usual norm on ${\displaystyle R.}$ Note that the function ${\displaystyle x\mapsto |\sin(\pi x)|}$ is subadditive.[10]
• Let ${\displaystyle X_{\mathbb {C} ))$ be a complex vector space and let ${\displaystyle X_{\mathbb {R} ))$ denote ${\displaystyle X_{\mathbb {C} ))$ considered as a vector space over ${\displaystyle \mathbb {R} .}$ Any paranorm on ${\displaystyle X_{\mathbb {C} ))$ is also a paranorm on ${\displaystyle X_{\mathbb {R} }.}$[9]

## F-seminorms

If ${\displaystyle X}$ is a vector space over the real or complex numbers then an F-seminorm on ${\displaystyle X}$ (the ${\displaystyle F}$ stands for Fréchet) is a real-valued map ${\displaystyle p:X\to \mathbb {R} }$ with the following four properties: [11]

1. Non-negative: ${\displaystyle p\geq 0.}$
2. Subadditive: ${\displaystyle p(x+y)\leq p(x)+p(y)}$ for all ${\displaystyle x,y\in X}$
3. Balanced: ${\displaystyle p(ax)\leq p(x)}$ for ${\displaystyle x\in X}$ all scalars ${\displaystyle a}$ satisfying ${\displaystyle |a|\leq 1;}$
• This condition guarantees that each set of the form ${\displaystyle \{z\in X:p(z)\leq r\))$ or ${\displaystyle \{z\in X:p(z) for some ${\displaystyle r\geq 0}$ is a balanced set.
4. For every ${\displaystyle x\in X,}$ ${\displaystyle p\left({\tfrac {1}{n))x\right)\to 0}$ as ${\displaystyle n\to \infty }$
• The sequence ${\displaystyle \left({\tfrac {1}{n))\right)_{n=1}^{\infty ))$ can be replaced by any positive sequence converging to the zero.[12]

An F-seminorm is called an F-norm if in addition it satisfies:

1. Total/Positive definite: ${\displaystyle p(x)=0}$ implies ${\displaystyle x=0.}$

An F-seminorm is called monotone if it satisfies:

1. Monotone: ${\displaystyle p(rx) for all non-zero ${\displaystyle x\in X}$ and all real ${\displaystyle s}$ and ${\displaystyle t}$ such that ${\displaystyle s[12]

### F-seminormed spaces

An F-seminormed space (resp. F-normed space)[12] is a pair ${\displaystyle (X,p)}$ consisting of a vector space ${\displaystyle X}$ and an F-seminorm (resp. F-norm) ${\displaystyle p}$ on ${\displaystyle X.}$

If ${\displaystyle (X,p)}$ and ${\displaystyle (Z,q)}$ are F-seminormed spaces then a map ${\displaystyle f:X\to Z}$ is called an isometric embedding[12] if ${\displaystyle q(f(x)-f(y))=p(x,y){\text{ for all ))x,y\in X.}$

Every isometric embedding of one F-seminormed space into another is a topological embedding, but the converse is not true in general.[12]

### Examples of F-seminorms

• Every positive scalar multiple of an F-seminorm (resp. F-norm, seminorm) is again an F-seminorm (resp. F-norm, seminorm).
• The sum of finitely many F-seminorms (resp. F-norms) is an F-seminorm (resp. F-norm).
• If ${\displaystyle p}$ and ${\displaystyle q}$ are F-seminorms on ${\displaystyle X}$ then so is their pointwise supremum ${\displaystyle x\mapsto \sup\{p(x),q(x)\}.}$ The same is true of the supremum of any non-empty finite family of F-seminorms on ${\displaystyle X.}$[12]
• The restriction of an F-seminorm (resp. F-norm) to a vector subspace is an F-seminorm (resp. F-norm).[9]
• A non-negative real-valued function on ${\displaystyle X}$ is a seminorm if and only if it is a convex F-seminorm, or equivalently, if and only if it is a convex balanced G-seminorm.[10] In particular, every seminorm is an F-seminorm.
• For any ${\displaystyle 0 the map ${\displaystyle f}$ on ${\displaystyle \mathbb {R} ^{n))$ defined by
${\displaystyle [f\left(x_{1},\ldots ,x_{n}\right)]^{p}=\left|x_{1}\right|^{p}+\cdots \left|x_{n}\right|^{p))$
is an F-norm that is not a norm.
• If ${\displaystyle L:X\to Y}$ is a linear map and if ${\displaystyle q}$ is an F-seminorm on ${\displaystyle Y,}$ then ${\displaystyle q\circ L}$ is an F-seminorm on ${\displaystyle X.}$[12]
• Let ${\displaystyle X_{\mathbb {C} ))$ be a complex vector space and let ${\displaystyle X_{\mathbb {R} ))$ denote ${\displaystyle X_{\mathbb {C} ))$ considered as a vector space over ${\displaystyle \mathbb {R} .}$ Any F-seminorm on ${\displaystyle X_{\mathbb {C} ))$ is also an F-seminorm on ${\displaystyle X_{\mathbb {R} }.}$[9]

### Properties of F-seminorms

Every F-seminorm is a paranorm and every paranorm is equivalent to some F-seminorm.[7] Every F-seminorm on a vector space ${\displaystyle X}$ is a value on ${\displaystyle X.}$ In particular, ${\displaystyle p(x)=0,}$ and ${\displaystyle p(x)=p(-x)}$ for all ${\displaystyle x\in X.}$

### Topology induced by a single F-seminorm

Theorem[11] — Let ${\displaystyle p}$ be an F-seminorm on a vector space ${\displaystyle X.}$ Then the map ${\displaystyle d:X\times X\to \mathbb {R} }$ defined by ${\displaystyle d(x,y):=p(x-y)}$ is a translation invariant pseudometric on ${\displaystyle X}$ that defines a vector topology ${\displaystyle \tau }$ on ${\displaystyle X.}$ If ${\displaystyle p}$ is an F-norm then ${\displaystyle d}$ is a metric. When ${\displaystyle X}$ is endowed with this topology then ${\displaystyle p}$ is a continuous map on ${\displaystyle X.}$

The balanced sets ${\displaystyle \{x\in X~:~p(x)\leq r\},}$ as ${\displaystyle r}$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set. Similarly, the balanced sets ${\displaystyle \{x\in X~:~p(x) as ${\displaystyle r}$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of open sets.

### Topology induced by a family of F-seminorms

Suppose that ${\displaystyle {\mathcal {L))}$ is a non-empty collection of F-seminorms on a vector space ${\displaystyle X}$ and for any finite subset ${\displaystyle {\mathcal {F))\subseteq {\mathcal {L))}$ and any ${\displaystyle r>0,}$ let

${\displaystyle U_((\mathcal {F)),r}:=\bigcap _{p\in {\mathcal {F))}\{x\in X:p(x)

The set ${\displaystyle \left\{U_((\mathcal {F)),r}~:~r>0,{\mathcal {F))\subseteq {\mathcal {L)),{\mathcal {F)){\text{ finite ))\right\))$ forms a filter base on ${\displaystyle X}$ that also forms a neighborhood basis at the origin for a vector topology on ${\displaystyle X}$ denoted by ${\displaystyle \tau _{\mathcal {L)).}$[12] Each ${\displaystyle U_((\mathcal {F)),r))$ is a balanced and absorbing subset of ${\displaystyle X.}$[12] These sets satisfy[12]

${\displaystyle U_((\mathcal {F)),r/2}+U_((\mathcal {F)),r/2}\subseteq U_((\mathcal {F)),r}.}$

• ${\displaystyle \tau _{\mathcal {L))}$ is the coarsest vector topology on ${\displaystyle X}$ making each ${\displaystyle p\in {\mathcal {L))}$ continuous.[12]
• ${\displaystyle \tau _{\mathcal {L))}$ is Hausdorff if and only if for every non-zero ${\displaystyle x\in X,}$ there exists some ${\displaystyle p\in {\mathcal {L))}$ such that ${\displaystyle p(x)>0.}$[12]
• If ${\displaystyle {\mathcal {F))}$ is the set of all continuous F-seminorms on ${\displaystyle \left(X,\tau _{\mathcal {L))\right)}$ then ${\displaystyle \tau _{\mathcal {L))=\tau _{\mathcal {F)).}$[12]
• If ${\displaystyle {\mathcal {F))}$ is the set of all pointwise suprema of non-empty finite subsets of ${\displaystyle {\mathcal {F))}$ of ${\displaystyle {\mathcal {L))}$ then ${\displaystyle {\mathcal {F))}$ is a directed family of F-seminorms and ${\displaystyle \tau _{\mathcal {L))=\tau _{\mathcal {F)).}$[12]

## Fréchet combination

Suppose that ${\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is a family of non-negative subadditive functions on a vector space ${\displaystyle X.}$

The Fréchet combination[8] of ${\displaystyle p_{\bullet ))$ is defined to be the real-valued map

${\displaystyle p(x):=\sum _{i=1}^{\infty }{\frac {p_{i}(x)}{2^{i}\left[1+p_{i}(x)\right])).}$

### As an F-seminorm

Assume that ${\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is an increasing sequence of seminorms on ${\displaystyle X}$ and let ${\displaystyle p}$ be the Fréchet combination of ${\displaystyle p_{\bullet }.}$ Then ${\displaystyle p}$ is an F-seminorm on ${\displaystyle X}$ that induces the same locally convex topology as the family ${\displaystyle p_{\bullet ))$ of seminorms.[13]

Since ${\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is increasing, a basis of open neighborhoods of the origin consists of all sets of the form ${\displaystyle \left\{x\in X~:~p_{i}(x) as ${\displaystyle i}$ ranges over all positive integers and ${\displaystyle r>0}$ ranges over all positive real numbers.

The translation invariant pseudometric on ${\displaystyle X}$ induced by this F-seminorm ${\displaystyle p}$ is

${\displaystyle d(x,y)=\sum _{i=1}^{\infty }{\frac {1}{2^{i))}{\frac {p_{i}(x-y)}{1+p_{i}(x-y))).}$

This metric was discovered by Fréchet in his 1906 thesis for the spaces of real and complex sequences with pointwise operations.[14]

### As a paranorm

If each ${\displaystyle p_{i))$ is a paranorm then so is ${\displaystyle p}$ and moreover, ${\displaystyle p}$ induces the same topology on ${\displaystyle X}$ as the family ${\displaystyle p_{\bullet ))$ of paranorms.[8] This is also true of the following paranorms on ${\displaystyle X}$:

• ${\displaystyle q(x):=\inf _{}\left\{\sum _{i=1}^{n}p_{i}(x)+{\frac {1}{n))~:~n>0{\text{ is an integer ))\right\}.}$[8]
• ${\displaystyle r(x):=\sum _{n=1}^{\infty }\min \left$$(\frac {1}{2^{n))},p_{n}(x)\right\}.}$[8] ### Generalization The Fréchet combination can be generalized by use of a bounded remetrization function. A bounded remetrization function[15] is a continuous non-negative non-decreasing map ${\displaystyle R:[0,\infty )\to [0,\infty )}$ that has a bounded range, is subadditive (meaning that ${\displaystyle R(s+t)\leq R(s)+R(t)}$ for all ${\displaystyle s,t\geq 0}$), and satisfies ${\displaystyle R(s)=0}$ if and only if ${\displaystyle s=0.}$ Examples of bounded remetrization functions include ${\displaystyle \arctan t,}$ ${\displaystyle \tanh t,}$ ${\displaystyle t\mapsto \min\{t,1\},}$ and ${\displaystyle t\mapsto {\frac {t}{1+t)).}$[15] If ${\displaystyle d}$ is a pseudometric (respectively, metric) on ${\displaystyle X}$ and ${\displaystyle R}$ is a bounded remetrization function then ${\displaystyle R\circ d}$ is a bounded pseudometric (respectively, bounded metric) on ${\displaystyle X}$ that is uniformly equivalent to ${\displaystyle d.}$[15] Suppose that ${\displaystyle p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is a family of non-negative F-seminorm on a vector space ${\displaystyle X,}$ ${\displaystyle R}$ is a bounded remetrization function, and ${\displaystyle r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty ))$ is a sequence of positive real numbers whose sum is finite. Then ${\displaystyle p(x):=\sum _{i=1}^{\infty }r_{i}R\left(p_{i}(x)\right)}$ defines a bounded F-seminorm that is uniformly equivalent to the ${\displaystyle p_{\bullet }.}$[16] It has the property that for any net ${\displaystyle x_{\bullet }=\left(x_{a}\right)_{a\in A))$ in ${\displaystyle X,}$ ${\displaystyle p\left(x_{\bullet }\right)\to 0}$ if and only if ${\displaystyle p_{i}\left(x_{\bullet }\right)\to 0}$ for all ${\displaystyle i.}$[16] ${\displaystyle p}$ is an F-norm if and only if the ${\displaystyle p_{\bullet ))$ separate points on ${\displaystyle X.}$[16] ## Characterizations ### Of (pseudo)metrics induced by (semi)norms A pseudometric (resp. metric) ${\displaystyle d}$ is induced by a seminorm (resp. norm) on a vector space ${\displaystyle X}$ if and only if ${\displaystyle d}$ is translation invariant and absolutely homogeneous, which means that for all scalars ${\displaystyle s}$ and all ${\displaystyle x,y\in X,}$ in which case the function defined by ${\displaystyle p(x):=d(x,0)}$ is a seminorm (resp. norm) and the pseudometric (resp. metric) induced by ${\displaystyle p}$ is equal to ${\displaystyle d.}$ ### Of pseudometrizable TVS If ${\displaystyle (X,\tau )}$ is a topological vector space (TVS) (where note in particular that ${\displaystyle \tau }$ is assumed to be a vector topology) then the following are equivalent:[11] 1. ${\displaystyle X}$ is pseudometrizable (i.e. the vector topology ${\displaystyle \tau }$ is induced by a pseudometric on ${\displaystyle X}$). 2. ${\displaystyle X}$ has a countable neighborhood base at the origin. 3. The topology on ${\displaystyle X}$ is induced by a translation-invariant pseudometric on ${\displaystyle X.}$ 4. The topology on ${\displaystyle X}$ is induced by an F-seminorm. 5. The topology on ${\displaystyle X}$ is induced by a paranorm. ### Of metrizable TVS If ${\displaystyle (X,\tau )}$ is a TVS then the following are equivalent: 1. ${\displaystyle X}$ is metrizable. 2. ${\displaystyle X}$ is Hausdorff and pseudometrizable. 3. ${\displaystyle X}$ is Hausdorff and has a countable neighborhood base at the origin.[11][12] 4. The topology on ${\displaystyle X}$ is induced by a translation-invariant metric on ${\displaystyle X.}$[11] 5. The topology on ${\displaystyle X}$ is induced by an F-norm.[11][12] 6. The topology on ${\displaystyle X}$ is induced by a monotone F-norm.[12] 7. The topology on ${\displaystyle X}$ is induced by a total paranorm. Birkhoff–Kakutani theorem — If ${\displaystyle (X,\tau )}$ is a topological vector space then the following three conditions are equivalent:[17][note 1] 1. The origin ${\displaystyle \{0$$)$ is closed in ${\displaystyle X,}$ and there is a countable basis of neighborhoods for ${\displaystyle 0}$ in ${\displaystyle X.}$
2. ${\displaystyle (X,\tau )}$ is metrizable (as a topological space).
3. There is a translation-invariant metric on ${\displaystyle X}$ that induces on ${\displaystyle X}$ the topology ${\displaystyle \tau ,}$ which is the given topology on ${\displaystyle X.}$

By the Birkhoff–Kakutani theorem, it follows that there is an equivalent metric that is translation-invariant.

### Of locally convex pseudometrizable TVS

If ${\displaystyle (X,\tau )}$ is TVS then the following are equivalent:[13]

1. ${\displaystyle X}$ is locally convex and pseudometrizable.
2. ${\displaystyle X}$ has a countable neighborhood base at the origin consisting of convex sets.
3. The topology of ${\displaystyle X}$ is induced by a countable family of (continuous) seminorms.
4. The topology of ${\displaystyle X}$ is induced by a countable increasing sequence of (continuous) seminorms ${\displaystyle \left(p_{i}\right)_{i=1}^{\infty ))$ (increasing means that for all ${\displaystyle i,}$ ${\displaystyle p_{i}\geq p_{i+1}.}$
5. The topology of ${\displaystyle X}$ is induced by an F-seminorm of the form:
${\displaystyle p(x)=\sum _{n=1}^{\infty }2^{-n}\operatorname {arctan} p_{n}(x)}$
where ${\displaystyle \left(p_{i}\right)_{i=1}^{\infty ))$ are (continuous) seminorms on ${\displaystyle X.}$[18]

## Quotients

Let ${\displaystyle M}$ be a vector subspace of a topological vector space ${\displaystyle (X,\tau ).}$

• If ${\displaystyle X}$ is a pseudometrizable TVS then so is ${\displaystyle X/M.}$[11]
• If ${\displaystyle X}$ is a complete pseudometrizable TVS and ${\displaystyle M}$ is a closed vector subspace of ${\displaystyle X}$ then ${\displaystyle X/M}$ is complete.[11]
• If ${\displaystyle X}$ is metrizable TVS and ${\displaystyle M}$ is a closed vector subspace of ${\displaystyle X}$ then ${\displaystyle X/M}$ is metrizable.[11]
• If ${\displaystyle p}$ is an F-seminorm on ${\displaystyle X,}$ then the map ${\displaystyle P:X/M\to \mathbb {R} }$ defined by
${\displaystyle P(x+M):=\inf _{}\{p(x+m):m\in M\))$
is an F-seminorm on ${\displaystyle X/M}$ that induces the usual quotient topology on ${\displaystyle X/M.}$[11] If in addition ${\displaystyle p}$ is an F-norm on ${\displaystyle X}$ and if ${\displaystyle M}$ is a closed vector subspace of ${\displaystyle X}$ then ${\displaystyle P}$ is an F-norm on ${\displaystyle X.}$[11]

## Examples and sufficient conditions

• Every seminormed space ${\displaystyle (X,p)}$ is pseudometrizable with a canonical pseudometric given by ${\displaystyle d(x,y):=p(x-y)}$ for all ${\displaystyle x,y\in X.}$[19].
• If ${\displaystyle (X,d)}$ is pseudometric TVS with a translation invariant pseudometric ${\displaystyle d,}$ then ${\displaystyle p(x):=d(x,0)}$ defines a paranorm.[20] However, if ${\displaystyle d}$ is a translation invariant pseudometric on the vector space ${\displaystyle X}$ (without the addition condition that ${\displaystyle (X,d)}$ is pseudometric TVS), then ${\displaystyle d}$ need not be either an F-seminorm[21] nor a paranorm.
• If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.[14]
• If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.[14]
• Suppose ${\displaystyle X}$ is either a DF-space or an LM-space. If ${\displaystyle X}$ is a sequential space then it is either metrizable or else a Montel DF-space.

If ${\displaystyle X}$ is Hausdorff locally convex TVS then ${\displaystyle X}$ with the strong topology, ${\displaystyle \left(X,b\left(X,X^{\prime }\right)\right),}$ is metrizable if and only if there exists a countable set ${\displaystyle {\mathcal {B))}$ of bounded subsets of ${\displaystyle X}$ such that every bounded subset of ${\displaystyle X}$ is contained in some element of ${\displaystyle {\mathcal {B)).}$[22]

The strong dual space ${\displaystyle X_{b}^{\prime ))$ of a metrizable locally convex space (such as a Fréchet space[23]) ${\displaystyle X}$ is a DF-space.[24] The strong dual of a DF-space is a Fréchet space.[25] The strong dual of a reflexive Fréchet space is a bornological space.[24] The strong bidual (that is, the strong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space.[26] If ${\displaystyle X}$ is a metrizable locally convex space then its strong dual ${\displaystyle X_{b}^{\prime ))$ has one of the following properties, if and only if it has all of these properties: (1) bornological, (2) infrabarreled, (3) barreled.[26]

### Normability

A topological vector space is seminormable if and only if it has a convex bounded neighborhood of the origin. Moreover, a TVS is normable if and only if it is Hausdorff and seminormable.[14] Every metrizable TVS on a finite-dimensional vector space is a normable locally convex complete TVS, being TVS-isomorphic to Euclidean space. Consequently, any metrizable TVS that is not normable must be infinite dimensional.

If ${\displaystyle M}$ is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then ${\displaystyle M}$ is normable.[27]

If ${\displaystyle X}$ is a Hausdorff locally convex space then the following are equivalent:

1. ${\displaystyle X}$ is normable.
2. ${\displaystyle X}$ has a (von Neumann) bounded neighborhood of the origin.
3. the strong dual space ${\displaystyle X_{b}^{\prime ))$ of ${\displaystyle X}$ is normable.[28]

and if this locally convex space ${\displaystyle X}$ is also metrizable, then the following may be appended to this list:

1. the strong dual space of ${\displaystyle X}$ is metrizable.[28]
2. the strong dual space of ${\displaystyle X}$ is a Fréchet–Urysohn locally convex space.[23]

In particular, if a metrizable locally convex space ${\displaystyle X}$ (such as a Fréchet space) is not normable then its strong dual space ${\displaystyle X_{b}^{\prime ))$ is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space ${\displaystyle X_{b}^{\prime ))$ is also neither metrizable nor normable.

Another consequence of this is that if ${\displaystyle X}$ is a reflexive locally convex TVS whose strong dual ${\displaystyle X_{b}^{\prime ))$ is metrizable then ${\displaystyle X_{b}^{\prime ))$ is necessarily a reflexive Fréchet space, ${\displaystyle X}$ is a DF-space, both ${\displaystyle X}$ and ${\displaystyle X_{b}^{\prime ))$ are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, ${\displaystyle X_{b}^{\prime ))$ is normable if and only if ${\displaystyle X}$ is normable if and only if ${\displaystyle X}$ is Fréchet–Urysohn if and only if ${\displaystyle X}$ is metrizable. In particular, such a space ${\displaystyle X}$ is either a Banach space or else it is not even a Fréchet–Urysohn space.

## Metrically bounded sets and bounded sets

Suppose that ${\displaystyle (X,d)}$ is a pseudometric space and ${\displaystyle B\subseteq X.}$ The set ${\displaystyle B}$ is metrically bounded or ${\displaystyle d}$-bounded if there exists a real number ${\displaystyle R>0}$ such that ${\displaystyle d(x,y)\leq R}$ for all ${\displaystyle x,y\in B}$; the smallest such ${\displaystyle R}$ is then called the diameter or ${\displaystyle d}$-diameter of ${\displaystyle B.}$[14] If ${\displaystyle B}$ is bounded in a pseudometrizable TVS ${\displaystyle X}$ then it is metrically bounded; the converse is in general false but it is true for locally convex metrizable TVSs.[14]

## Properties of pseudometrizable TVS

Theorem[29] — All infinite-dimensional separable complete metrizable TVS are homeomorphic.

• Every metrizable locally convex TVS is a quasibarrelled space,[30] bornological space, and a Mackey space.
• Every complete pseudometrizable TVS is a barrelled space and a Baire space (and hence non-meager).[31] However, there exist metrizable Baire spaces that are not complete.[31]
• If ${\displaystyle X}$ is a metrizable locally convex space, then the strong dual of ${\displaystyle X}$ is bornological if and only if it is barreled, if and only if it is infrabarreled.[26]
• If ${\displaystyle X}$ is a complete pseudometrizable TVS and ${\displaystyle M}$ is a closed vector subspace of ${\displaystyle X,}$ then ${\displaystyle X/M}$ is complete.[11]
• The strong dual of a locally convex metrizable TVS is a webbed space.[32]
• If ${\displaystyle (X,\tau )}$ and ${\displaystyle (X,\nu )}$ are complete metrizable TVSs (i.e. F-spaces) and if ${\displaystyle \nu }$ is coarser than ${\displaystyle \tau }$ then ${\displaystyle \tau =\nu }$;[33] this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete.[34] Said differently, if ${\displaystyle (X,\tau )}$ and ${\displaystyle (X,\nu )}$ are both F-spaces but with different topologies, then neither one of ${\displaystyle \tau }$ and ${\displaystyle \nu }$ contains the other as a subset. One particular consequence of this is, for example, that if ${\displaystyle (X,p)}$ is a Banach space and ${\displaystyle (X,q)}$ is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of ${\displaystyle (X,p)}$ (i.e. if ${\displaystyle p\leq Cq}$ or if ${\displaystyle q\leq Cp}$ for some constant ${\displaystyle C>0}$), then the only way that ${\displaystyle (X,q)}$ can be a Banach space (i.e. also be complete) is if these two norms ${\displaystyle p}$ and ${\displaystyle q}$ are equivalent; if they are not equivalent, then ${\displaystyle (X,q)}$ can not be a Banach space. As another consequence, if ${\displaystyle (X,p)}$ is a Banach space and ${\displaystyle (X,\nu )}$ is a Fréchet space, then the map ${\displaystyle p:(X,\nu )\to \mathbb {R} }$ is continuous if and only if the Fréchet space ${\displaystyle (X,\nu )}$ is the TVS ${\displaystyle (X,p)}$ (here, the Banach space ${\displaystyle (X,p)}$ is being considered as a TVS, which means that its norm is "forgetten" but its topology is remembered).
• A metrizable locally convex space is normable if and only if its strong dual space is a Fréchet–Urysohn locally convex space.[23]
• Any product of complete metrizable TVSs is a Baire space.[31]
• A product of metrizable TVSs is metrizable if and only if it all but at most countably many of these TVSs have dimension ${\displaystyle 0.}$[35]
• A product of pseudometrizable TVSs is pseudometrizable if and only if it all but at most countably many of these TVSs have the trivial topology.
• Every complete pseudometrizable TVS is a barrelled space and a Baire space (and thus non-meager).[31]
• The dimension of a complete metrizable TVS is either finite or uncountable.[35]

### Completeness

 Main article: Complete topological vector space

Every topological vector space (and more generally, a topological group) has a canonical uniform structure, induced by its topology, which allows the notions of completeness and uniform continuity to be applied to it. If ${\displaystyle X}$ is a metrizable TVS and ${\displaystyle d}$ is a metric that defines ${\displaystyle X}$'s topology, then its possible that ${\displaystyle X}$ is complete as a TVS (i.e. relative to its uniformity) but the metric ${\displaystyle d}$ is not a complete metric (such metrics exist even for ${\displaystyle X=\mathbb {R} }$). Thus, if ${\displaystyle X}$ is a TVS whose topology is induced by a pseudometric ${\displaystyle d,}$ then the notion of completeness of ${\displaystyle X}$ (as a TVS) and the notion of completeness of the pseudometric space ${\displaystyle (X,d)}$ are not always equivalent. The next theorem gives a condition for when they are equivalent:

Theorem — If ${\displaystyle X}$ is a pseudometrizable TVS whose topology is induced by a translation invariant pseudometric ${\displaystyle d,}$ then ${\displaystyle d}$ is a complete pseudometric on ${\displaystyle X}$ if and only if ${\displaystyle X}$ is complete as a TVS.[36]

Theorem[37][38] (Klee) — Let ${\displaystyle d}$ be any[note 2] 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.

Theorem — If ${\displaystyle X}$ is a TVS whose topology is induced by a paranorm ${\displaystyle p,}$ then ${\displaystyle X}$ is complete if and only if for every sequence ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle X,}$ if ${\displaystyle \sum _{i=1}^{\infty }p\left(x_{i}\right)<\infty }$ then ${\displaystyle \sum _{i=1}^{\infty }x_{i))$ converges in ${\displaystyle X.}$[39]

If ${\displaystyle M}$ is a closed vector subspace of a complete pseudometrizable TVS ${\displaystyle X,}$ then the quotient space ${\displaystyle X/M}$ is complete.[40] If ${\displaystyle M}$ is a complete vector subspace of a metrizable TVS ${\displaystyle X}$ and if the quotient space ${\displaystyle X/M}$ is complete then so is ${\displaystyle X.}$[40] If ${\displaystyle X}$ is not complete then ${\displaystyle M:=X,}$ but not complete, vector subspace of ${\displaystyle X.}$

A Baire separable topological group is metrizable if and only if it is cosmic.[23]

### Subsets and subsequences

• Let ${\displaystyle M}$ be a separable locally convex metrizable topological vector space and let ${\displaystyle C}$ be its completion. If ${\displaystyle S}$ is a bounded subset of ${\displaystyle C}$ then there exists a bounded subset ${\displaystyle R}$ of ${\displaystyle X}$ such that ${\displaystyle S\subseteq \operatorname {cl} _{C}R.}$[41]
• Every totally bounded subset of a locally convex metrizable TVS ${\displaystyle X}$ is contained in the closed convex balanced hull of some sequence in ${\displaystyle X}$ that converges to ${\displaystyle 0.}$
• In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.[42]
• If ${\displaystyle d}$ is a translation invariant metric on a vector space ${\displaystyle X,}$ then ${\displaystyle d(nx,0)\leq nd(x,0)}$ for all ${\displaystyle x\in X}$ and every positive integer ${\displaystyle n.}$[43]
• If ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty ))$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence ${\displaystyle \left(r_{i}\right)_{i=1}^{\infty ))$ of positive real numbers diverging to ${\displaystyle \infty }$ such that ${\displaystyle \left(r_{i}x_{i}\right)_{i=1}^{\infty }\to 0.}$[43]
• A subset of a complete metric space is closed if and only if it is complete. If a space ${\displaystyle X}$ is not complete, then ${\displaystyle X}$ is a closed subset of ${\displaystyle X}$ that is not complete.
• If ${\displaystyle X}$ is a metrizable locally convex TVS then for every bounded subset ${\displaystyle B}$ of ${\displaystyle X,}$ there exists a bounded disk ${\displaystyle D}$ in ${\displaystyle X}$ such that ${\displaystyle B\subseteq X_{D},}$ and both ${\displaystyle X}$ and the auxiliary normed space ${\displaystyle X_{D))$ induce the same subspace topology on ${\displaystyle B.}$[44]

Banach-Saks theorem[45] — If ${\displaystyle \left(x_{n}\right)_{n=1}^{\infty ))$ is a sequence in a locally convex metrizable TVS ${\displaystyle (X,\tau )}$ that converges weakly to some ${\displaystyle x\in X,}$ then there exists a sequence ${\displaystyle y_{\bullet }=\left(y_{i}\right)_{i=1}^{\infty ))$ in ${\displaystyle X}$ such that ${\displaystyle y_{\bullet }\to x}$ in ${\displaystyle (X,\tau )}$ and each ${\displaystyle y_{i))$ is a convex combination of finitely many ${\displaystyle x_{n}.}$

Mackey's countability condition[14] — Suppose that ${\displaystyle X}$ is a locally convex metrizable TVS and that ${\displaystyle \left(B_{i}\right)_{i=1}^{\infty ))$ is a countable sequence of bounded subsets of ${\displaystyle X.}$ Then there exists a bounded subset ${\displaystyle B}$ of ${\displaystyle X}$ and a sequence ${\displaystyle \left(r_{i}\right)_{i=1}^{\infty ))$ of positive real numbers such that ${\displaystyle B_{i}\subseteq r_{i}B}$ for all ${\displaystyle i.}$

Generalized series

As described in this article's section on generalized series, for any ${\displaystyle I}$-indexed family family ${\displaystyle \left(r_{i}\right)_{i\in I))$ of vectors from a TVS ${\displaystyle X,}$ it is possible to define their sum ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i))$ as the limit of the net of finite partial sums ${\displaystyle F\in \operatorname {FiniteSubsets} (I)\mapsto \textstyle \sum \limits _{i\in F}r_{i))$ where the domain ${\displaystyle \operatorname {FiniteSubsets} (I)}$ is directed by ${\displaystyle \,\subseteq .\,}$ If ${\displaystyle I=\mathbb {N} }$ and ${\displaystyle X=\mathbb {R} ,}$ for instance, then the generalized series ${\displaystyle \textstyle \sum \limits _{i\in \mathbb {N} }r_{i))$ converges if and only if ${\displaystyle \textstyle \sum \limits _{i=1}^{\infty }r_{i))$ converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence). If a generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i))$ converges in a metrizable TVS, then the set ${\displaystyle \left\{i\in I:r_{i}\neq 0\right\))$ is necessarily countable (that is, either finite or countably infinite);[proof 1] in other words, all but at most countably many ${\displaystyle r_{i))$ will be zero and so this generalized series ${\displaystyle \textstyle \sum \limits _{i\in I}r_{i}~=~\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}\neq 0))r_{i))$ is actually a sum of at most countably many non-zero terms.

### Linear maps

If ${\displaystyle X}$ is a pseudometrizable TVS and ${\displaystyle A}$ maps bounded subsets of ${\displaystyle X}$ to bounded subsets of ${\displaystyle Y,}$ then ${\displaystyle A}$ is continuous.[14] Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS.[46] Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to its algebraic dual space.[46]

If ${\displaystyle F:X\to Y}$ is a linear map between TVSs and ${\displaystyle X}$ is metrizable then the following are equivalent:

1. ${\displaystyle F}$ is continuous;
2. ${\displaystyle F}$ is a (locally) bounded map (that is, ${\displaystyle F}$ maps (von Neumann) bounded subsets of ${\displaystyle X}$ to bounded subsets of ${\displaystyle Y}$);[12]
3. ${\displaystyle F}$ is sequentially continuous;[12]
4. the image under ${\displaystyle F}$ of every null sequence in ${\displaystyle X}$ is a bounded set[12] where by definition, a null sequence is a sequence that converges to the origin.
5. ${\displaystyle F}$ maps null sequences to null sequences;

Open and almost open maps

Theorem: If ${\displaystyle X}$ is a complete pseudometrizable TVS, ${\displaystyle Y}$ is a Hausdorff TVS, and ${\displaystyle T:X\to Y}$ is a closed and almost open linear surjection, then ${\displaystyle T}$ is an open map.[47]
Theorem: If ${\displaystyle T:X\to Y}$ is a surjective linear operator from a locally convex space ${\displaystyle X}$ onto a barrelled space ${\displaystyle Y}$ (e.g. every complete pseudometrizable space is barrelled) then ${\displaystyle T}$ is almost open.[47]
Theorem: If ${\displaystyle T:X\to Y}$ is a surjective linear operator from a TVS ${\displaystyle X}$ onto a Baire space ${\displaystyle Y}$ then ${\displaystyle T}$ is almost open.[47]
Theorem: Suppose ${\displaystyle T:X\to Y}$ is a continuous linear operator from a complete pseudometrizable TVS ${\displaystyle X}$ into a Hausdorff TVS ${\displaystyle Y.}$ If the image of ${\displaystyle T}$ is non-meager in ${\displaystyle Y}$ then ${\displaystyle T:X\to Y}$ is a surjective open map and ${\displaystyle Y}$ is a complete metrizable space.[47]

### Hahn-Banach extension property

 Main article: Hahn-Banach theorem

A vector subspace ${\displaystyle M}$ of a TVS ${\displaystyle X}$ has the extension property if any continuous linear functional on ${\displaystyle M}$ can be extended to a continuous linear functional on ${\displaystyle X.}$[22] Say that a TVS ${\displaystyle X}$ has the Hahn-Banach extension property (HBEP) if every vector subspace of ${\displaystyle X}$ has the extension property.[22]

The Hahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable TVSs there is a converse:

Theorem (Kalton) — Every complete metrizable TVS with the Hahn-Banach extension property is locally convex.[22]

If a vector space ${\displaystyle X}$ has uncountable dimension and if we endow it with the finest vector topology then this is a TVS with the HBEP that is neither locally convex or metrizable.[22]

## Notes

1. ^ In fact, this is true for topological group, for the proof doesn't use the scalar multiplications.
2. ^ Not assumed to be translation-invariant.

Proofs

1. ^ Suppose the net ${\textstyle \textstyle \sum \limits _{i\in I}r_{i}~{\stackrel {\scriptscriptstyle {\text{def))}{=))~{\textstyle \lim \limits _{A\in \operatorname {FiniteSubsets} (I)))\ \textstyle \sum \limits _{i\in A}r_{i}=\lim \left\{\textstyle \sum \limits _{i\in A}r_{i}\,:A\subseteq I,A{\text{ finite ))\right\))$ converges to some point in a metrizable TVS ${\displaystyle X,}$ where recall that this net's domain is the directed set ${\displaystyle (\operatorname {FiniteSubsets} (I),\subseteq ).}$ Like every convergent net, this convergent net of partial sums ${\displaystyle A\mapsto \textstyle \sum \limits _{i\in A}r_{i))$ is a Cauchy net, which for this particular net means (by definition) that for every neighborhood ${\displaystyle W}$ of the origin in ${\displaystyle X,}$ there exists a finite subset ${\displaystyle A_{0))$ of ${\displaystyle I}$ such that ${\textstyle \textstyle \sum \limits _{i\in B}r_{i}-\textstyle \sum \limits _{i\in C}r_{i}\in W}$ for all finite supersets ${\displaystyle B,C\supseteq A_{0};}$ this implies that ${\displaystyle r_{i}\in W}$ for every ${\displaystyle i\in I\setminus A_{0))$ (by taking ${\displaystyle B:=A_{0}\cup \{i\))$ and ${\displaystyle C:=A_{0))$). Since ${\displaystyle X}$ is metrizable, it has a countable neighborhood basis ${\displaystyle U_{1},U_{2},\ldots }$ at the origin, whose intersection is necessarily ${\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\))$ (since ${\displaystyle X}$ is a Hausdorff TVS). For every positive integer ${\displaystyle n\in \mathbb {N} ,}$ pick a finite subset ${\displaystyle A_{n}\subseteq I}$ such that ${\displaystyle r_{i}\in U_{n))$ for every ${\displaystyle i\in I\setminus A_{n}.}$ If ${\displaystyle i}$ belongs to ${\displaystyle (I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\cup A_{2}\cup \cdots \right)}$ then ${\displaystyle r_{i))$ belongs to ${\displaystyle U_{1}\cap U_{2}\cap \cdots =\{0\}.}$ Thus ${\displaystyle r_{i}=0}$ for every index ${\displaystyle i\in I}$ that does not belong to the countable set ${\displaystyle A_{1}\cup A_{2}\cup \cdots .}$ ${\displaystyle \blacksquare }$

## References

1. ^ Narici & Beckenstein 2011, pp. 1–18.
2. ^ a b c Narici & Beckenstein 2011, pp. 37–40.
3. ^ a b Swartz 1992, p. 15.
4. ^ Wilansky 2013, p. 17.
5. ^ a b Wilansky 2013, pp. 40–47.
6. ^ Wilansky 2013, p. 15.
7. ^ a b Schechter 1996, pp. 689–691.
8. Wilansky 2013, pp. 15–18.
9. ^ a b c d Schechter 1996, p. 692.
10. ^ a b Schechter 1996, p. 691.
11. Narici & Beckenstein 2011, pp. 91–95.
12. Jarchow 1981, pp. 38–42.
13. ^ a b Narici & Beckenstein 2011, p. 123.
14. Narici & Beckenstein 2011, pp. 156–175.
15. ^ a b c Schechter 1996, p. 487.
16. ^ a b c Schechter 1996, pp. 692–693.
17. ^ Köthe 1983, section 15.11
18. ^ Schechter 1996, p. 706.
19. ^ Narici & Beckenstein 2011, pp. 115–154.
20. ^ Wilansky 2013, pp. 15–16.
21. ^ Schaefer & Wolff 1999, pp. 91–92.
22. Narici & Beckenstein 2011, pp. 225–273.
23. ^ a b c d Gabriyelyan, S.S. "On topological spaces and topological groups with certain local countable networks (2014)
24. ^ a b Schaefer & Wolff 1999, p. 154.
25. ^ Schaefer & Wolff 1999, p. 196.
26. ^ a b c Schaefer & Wolff 1999, p. 153.
27. ^ Schaefer & Wolff 1999, pp. 68–72.
28. ^ a b Trèves 2006, p. 201.
29. ^ Wilansky 2013, p. 57.
30. ^ Jarchow 1981, p. 222.
31. ^ a b c d Narici & Beckenstein 2011, pp. 371–423.
32. ^ Narici & Beckenstein 2011, pp. 459–483.
33. ^ Köthe 1969, p. 168.
34. ^ Wilansky 2013, p. 59.
35. ^ a b Schaefer & Wolff 1999, pp. 12–35.
36. ^ Narici & Beckenstein 2011, pp. 47–50.
37. ^ Schaefer & Wolff 1999, p. 35.
38. ^ Klee, V. L. (1952). "Invariant metrics in groups (solution of a problem of Banach)" (PDF). Proc. Amer. Math. Soc. 3 (3): 484–487. doi:10.1090/s0002-9939-1952-0047250-4.
39. ^ Wilansky 2013, pp. 56–57.
40. ^ a b Narici & Beckenstein 2011, pp. 47–66.
41. ^ Schaefer & Wolff 1999, pp. 190–202.
42. ^ Narici & Beckenstein 2011, pp. 172–173.
43. ^ a b Rudin 1991, p. 22.
44. ^ Narici & Beckenstein 2011, pp. 441–457.
45. ^ Rudin 1991, p. 67.
46. ^ a b Narici & Beckenstein 2011, p. 125.
47. ^ a b c d Narici & Beckenstein 2011, pp. 466–468.