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Metrizable topological vector space

## Summary

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.

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

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 ${\displaystyle T:X\to Y}$