A pseudometric $d$ on $X$ is called a ultrapseudometric or a strong pseudometric if it satisfies:
Strong/Ultrametric triangle inequality: $d(x,z)\leq \max\{d(x,y),d(y,z)\}{\text{ for all ))x,y,z\in X.$
Pseudometric space
A pseudometric space is a pair $(X,d)$ consisting of a set $X$ and a pseudometric $d$ on $X$ such that $X$'s topology is identical to the topology on $X$ induced by $d.$ We call a pseudometric space $(X,d)$ a metric space (resp. ultrapseudometric space) when $d$ is a metric (resp. ultrapseudometric).
Topology induced by a pseudometric
If $d$ is a pseudometric on a set $X$ then collection of open balls:
$B_{r}(z):=\{x\in X:d(x,z)<r\))$
as $z$ ranges over $X$ and $r>0$ ranges over the positive real numbers,
forms a basis for a topology on $X$ that is called the $d$-topology or the pseudometric topology on $X$ induced by $d.$
Convention: If $(X,d)$ is a pseudometric space and $X$ is treated as a topological space, then unless indicated otherwise, it should be assumed that $X$ is endowed with the topology induced by $d.$
Pseudometrizable space
A topological space $(X,\tau )$ is called pseudometrizable (resp. metrizable, ultrapseudometrizable) if there exists a pseudometric (resp. metric, ultrapseudometric) $d$ on $X$ such that $\tau$ is equal to the topology induced by $d.$^{[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 $\tau$ on a real or complex vector space $X$ 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 $X$ into a topological vector space).
Every topological vector space (TVS) $X$ is an additive commutative topological group but not all group topologies on $X$ are vector topologies.
This is because despite it making addition and negation continuous, a group topology on a vector space $X$ 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 $X$ is an additive group then we say that a pseudometric $d$ on $X$ is translation invariant or just invariant if it satisfies any of the following equivalent conditions:
If $X$ is a topological group the a value or G-seminorm on $X$ (the G stands for Group) is a real-valued map $p:X\rightarrow \mathbb {R}$ with the following properties:^{[2]}
Non-negative: $p\geq 0.$
Subadditive: $p(x+y)\leq p(x)+p(y){\text{ for all ))x,y\in X$;
$p(0)=0..$
Symmetric: $p(-x)=p(x){\text{ for all ))x\in X.$
where we call a G-seminorm a G-norm if it satisfies the additional condition:
Total/Positive definite: If $p(x)=0$ then $x=0.$
Properties of values
If $p$ is a value on a vector space $X$ then:
$|p(x)-p(y)|\leq p(x-y){\text{ for all ))x,y\in X.$^{[3]}
$p(nx)\leq np(x)$ and ${\frac {1}{n))p(x)\leq p(x/n)$ for all $x\in X$ and positive integers $n.$^{[4]}
The set $\{x\in X:p(x)=0\))$ is an additive subgroup of $X.$^{[3]}
Equivalence on topological groups
Theorem^{[2]} — Suppose that $X$ is an additive commutative group.
If $d$ is a translation invariant pseudometric on $X$ then the map $p(x):=d(x,0)$ is a value on $X$ called the value associated with $d$, and moreover, $d$ generates a group topology on $X$ (i.e. the $d$-topology on $X$ makes $X$ into a topological group).
Conversely, if $p$ is a value on $X$ then the map $d(x,y):=p(x-y)$ is a translation-invariant pseudometric on $X$ and the value associated with $d$ is just $p.$
Pseudometrizable topological groups
Theorem^{[2]} — If $(X,\tau )$ is an additive commutative topological group then the following are equivalent:
$\tau$ is induced by a pseudometric; (i.e. $(X,\tau )$ is pseudometrizable);
$\tau$ is induced by a translation-invariant pseudometric;
the identity element in $(X,\tau )$ has a countable neighborhood basis.
If $(X,\tau )$ 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 $X$ be a non-trivial (i.e. $X\neq \{0\))$) real or complex vector space and let $d$ be the translation-invariant trivial metric on $X$ defined by $d(x,x)=0$ and $d(x,y)=1{\text{ for all ))x,y\in X$ such that $x\neq y.$
The topology $\tau$ that $d$ induces on $X$ is the discrete topology, which makes $(X,\tau )$ into a commutative topological group under addition but does not form a vector topology on $X$ because $(X,\tau )$ is disconnected but every vector topology is connected.
What fails is that scalar multiplication isn't continuous on $(X,\tau ).$
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 ${\mathcal {N))$ of subsets of a vector space is called additive^{[5]} if for every $N\in {\mathcal {N)),$ there exists some $U\in {\mathcal {N))$ such that $U+U\subseteq N.$
Continuity of addition at 0 — If $(X,+)$ is a group (as all vector spaces are), $\tau$ is a topology on $X,$ and $X\times X$ is endowed with the product topology, then the addition map $X\times X\to X$ (i.e. the map $(x,y)\mapsto x+y$) is continuous at the origin of $X\times X$ if and only if the set of neighborhoods of the origin in $(X,\tau )$ 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 $U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty ))$ be a collection of subsets of a vector space such that $0\in U_{i))$ and $U_{i+1}+U_{i+1}\subseteq U_{i))$ for all $i\geq 0.$
For all $u\in U_{0},$ let
$\mathbb {S} (u):=\left\{n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)~:~k\geq 1,n_{i}\geq 0{\text{ for all ))i,{\text{ and ))u\in U_{n_{1))+\cdots +U_{n_{k))\right\}.$
Define $f:X\to [0,1]$ by $f(x)=1$ if $x\not \in U_{0))$ and otherwise let
Then $f$ is subadditive (meaning $f(x+y)\leq f(x)+f(y){\text{ for all ))x,y\in X$) and $f=0$ on $\bigcap _{i\geq 0}U_{i},$ so in particular $f(0)=0.$
If all $U_{i))$ are symmetric sets then $f(-x)=f(x)$ and if all $U_{i))$ are balanced then $f(sx)\leq f(x)$ for all scalars $s$ such that $|s|\leq 1$ and all $x\in X.$
If $X$ is a topological vector space and if all $U_{i))$ are neighborhoods of the origin then $f$ is continuous, where if in addition $X$ is Hausdorff and $U_{\bullet ))$ forms a basis of balanced neighborhoods of the origin in $X$ then $d(x,y):=f(x-y)$ is a metric defining the vector topology on $X.$
Proof
Assume that $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ always denotes a finite sequence of non-negative integers and use the notation:
From this it follows that if $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ consists of distinct positive integers then $\sum U_{n_{\bullet ))\subseteq U_{-1+\min \left(n_{\bullet }\right)}.$
It will now be shown by induction on $k$ that if $n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)$ consists of non-negative integers such that $\sum 2^{-n_{\bullet ))\leq 2^{-M))$ for some integer $M\geq 0$ then $\sum U_{n_{\bullet ))\subseteq U_{M}.$
This is clearly true for $k=1$ and $k=2$ so assume that $k>2,$ which implies that all $n_{i))$ are positive.
If all $n_{i))$ are distinct then this step is done, and otherwise pick distinct indices $i<j$ such that $n_{i}=n_{j))$ and construct $m_{\bullet }=\left(m_{1},\ldots ,m_{k-1}\right)$ from $n_{\bullet ))$ by replacing each $n_{i))$ with $n_{i}-1$ and deleting the $j^{\text{th))$ element of $n_{\bullet ))$ (all other elements of $n_{\bullet ))$ are transferred to $m_{\bullet ))$ unchanged).
Observe that $\sum 2^{-n_{\bullet ))=\sum 2^{-m_{\bullet ))$ and $\sum U_{n_{\bullet ))\subseteq \sum U_{m_{\bullet ))$ (because $U_{n_{i))+U_{n_{j))\subseteq U_{n_{i}-1))$) so by appealing to the inductive hypothesis we conclude that $\sum U_{n_{\bullet ))\subseteq \sum U_{m_{\bullet ))\subseteq U_{M},$ as desired.
It is clear that $f(0)=0$ and that $0\leq f\leq 1$ so to prove that $f$ is subadditive, it suffices to prove that $f(x+y)\leq f(x)+f(y)$ when $x,y\in X$ are such that $f(x)+f(y)<1,$ which implies that $x,y\in U_{0}.$
This is an exercise.
If all $U_{i))$ are symmetric then $x\in \sum U_{n_{\bullet ))$ if and only if $-x\in \sum U_{n_{\bullet ))$ from which it follows that $f(-x)\leq f(x)$ and $f(-x)\geq f(x).$
If all $U_{i))$ are balanced then the inequality $f(sx)\leq f(x)$ for all unit scalars $s$ such that $|s|\leq 1$ is proved similarly.
Because $f$ is a nonnegative subadditive function satisfying $f(0)=0,$ as described in the article on sublinear functionals, $f$ is uniformly continuous on $X$ if and only if $f$ is continuous at the origin.
If all $U_{i))$ are neighborhoods of the origin then for any real $r>0,$ pick an integer $M>1$ such that $2^{-M}<r$ so that $x\in U_{M))$ implies $f(x)\leq 2^{-M}<r.$
If the set of all $U_{i))$ form basis of balanced neighborhoods of the origin then it may be shown that for any $n>1,$ there exists some $0<r\leq 2^{-n))$ such that $f(x)<r$ implies $x\in U_{n}.$$\blacksquare$
Paranorms
If $X$ is a vector space over the real or complex numbers then a paranorm on $X$ is a G-seminorm (defined above) $p:X\rightarrow \mathbb {R}$ on $X$ that satisfies any of the following additional conditions, each of which begins with "for all sequences $x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty ))$ in $X$ and all convergent sequences of scalars $s_{\bullet }=\left(s_{i}\right)_{i=1}^{\infty ))$":^{[6]}
Continuity of multiplication: if $s$ is a scalar and $x\in X$ are such that $p\left(x_{i}-x\right)\to 0$ and $s_{\bullet }\to s,$ then $p\left(s_{i}x_{i}-sx\right)\to 0.$
Both of the conditions:
if $s_{\bullet }\to 0$ and if $x\in X$ is such that $p\left(x_{i}-x\right)\to 0$ then $p\left(s_{i}x_{i}\right)\to 0$;
if $p\left(x_{\bullet }\right)\to 0$ then $p\left(sx_{i}\right)\to 0$ for every scalar $s.$
Both of the conditions:
if $p\left(x_{\bullet }\right)\to 0$ and $s_{\bullet }\to s$ for some scalar $s$ then $p\left(s_{i}x_{i}\right)\to 0$;
if $s_{\bullet }\to 0$ then $p\left(s_{i}x\right)\to 0{\text{ for all ))x\in X.$
Separate continuity:^{[7]}
if $s_{\bullet }\to s$ for some scalar $s$ then $p\left(sx_{i}-sx\right)\to 0$ for every $x\in X$;
if $s$ is a scalar, $x\in X,$ and $p\left(x_{i}-x\right)\to 0$ then $p\left(sx_{i}-sx\right)\to 0$ .
A paranorm is called total if in addition it satisfies:
Total/Positive definite: $p(x)=0$ implies $x=0.$
Properties of paranorms
If $p$ is a paranorm on a vector space $X$ then the map $d:X\times X\rightarrow \mathbb {R}$ defined by $d(x,y):=p(x-y)$ is a translation-invariant pseudometric on $X$ that defines a vector topology on $X.$^{[8]}
If $p$ is a paranorm on a vector space $X$ then:
the set $\{x\in X:p(x)=0\))$ is a vector subspace of $X.$^{[8]}
$p(x+n)=p(x){\text{ for all ))x,n\in X$ with $p(n)=0.$^{[8]}
If a paranorm $p$ satisfies $p(sx)\leq |s|p(x){\text{ for all ))x\in X$ and scalars $s,$ then $p$ is absolutely homogeneity (i.e. equality holds)^{[8]} and thus $p$ is a seminorm.
Examples of paranorms
If $d$ is a translation-invariant pseudometric on a vector space $X$ that induces a vector topology $\tau$ on $X$ (i.e. $(X,\tau )$ is a TVS) then the map $p(x):=d(x-y,0)$ defines a continuous paranorm on $(X,\tau )$; moreover, the topology that this paranorm $p$ defines in $X$ is $\tau .$^{[8]}
If $p$ is a paranorm on $X$ then so is the map $q(x):=p(x)/[1+p(x)].$^{[8]}
Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).
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 $p$ and $q$ are paranorms on $X$ then so is $(p\wedge q)(x):=\inf _{}\{p(y)+q(z):x=y+z{\text{ with ))y,z\in X\}.$ Moreover, $(p\wedge q)\leq p$ and $(p\wedge q)\leq q.$ This makes the set of paranorms on $X$ into a conditionally complete lattice.^{[8]}
Each of the following real-valued maps are paranorms on $X:=\mathbb {R} ^{2))$:
$(x,y)\mapsto |x|$
$(x,y)\mapsto |x|+|y|$
The real-valued maps $(x,y)\mapsto {\sqrt {\left|x^{2}-y^{2}\right|))$ and $(x,y)\mapsto \left|x^{2}-y^{2}\right|^{3/2))$ are not a paranorms on $X:=\mathbb {R} ^{2}.$^{[8]}
If $x_{\bullet }=\left(x_{i}\right)_{i\in I))$ is a Hamel basis on a vector space $X$ then the real-valued map that sends $x=\sum _{i\in I}s_{i}x_{i}\in X$ (where all but finitely many of the scalars $s_{i))$ are 0) to $\sum _{i\in I}{\sqrt {\left|s_{i}\right|))$ is a paranorm on $X,$ which satisfies $p(sx)={\sqrt {|s|))p(x)$ for all $x\in X$ and scalars $s.$^{[8]}
The function $p(x):=|\sin(\pi x)|+\min\{2,|x|\))$is a paranorm on $\mathbb {R}$ that is not balanced but nevertheless equivalent to the usual norm on $R.$ Note that the function $x\mapsto |\sin(\pi x)|$ is subadditive.^{[10]}
Let $X_{\mathbb {C} ))$ be a complex vector space and let $X_{\mathbb {R} ))$ denote $X_{\mathbb {C} ))$ considered as a vector space over $\mathbb {R} .$ Any paranorm on $X_{\mathbb {C} ))$ is also a paranorm on $X_{\mathbb {R} }.$^{[9]}
F-seminorms
If $X$ is a vector space over the real or complex numbers then an F-seminorm on $X$ (the $F$ stands for Fréchet) is a real-valued map $p:X\to \mathbb {R}$ with the following four properties: ^{[11]}
Non-negative: $p\geq 0.$
Subadditive: $p(x+y)\leq p(x)+p(y)$ for all $x,y\in X$
Balanced: $p(ax)\leq p(x)$ for $x\in X$ all scalars $a$ satisfying $|a|\leq 1;$
This condition guarantees that each set of the form $\{z\in X:p(z)\leq r\))$ or $\{z\in X:p(z)<r\))$ for some $r\geq 0$ is a balanced set.
For every $x\in X,$$p\left({\tfrac {1}{n))x\right)\to 0$ as $n\to \infty$
The sequence $\left({\tfrac {1}{n))\right)_{n=1}^{\infty ))$ 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:
Total/Positive definite: $p(x)=0$ implies $x=0.$
An F-seminorm is called monotone if it satisfies:
Monotone: $p(rx)<p(sx)$ for all non-zero $x\in X$ and all real $s$ and $t$ such that $s<t.$^{[12]}
F-seminormed spaces
An F-seminormed space (resp. F-normed space)^{[12]} is a pair $(X,p)$ consisting of a vector space $X$ and an F-seminorm (resp. F-norm) $p$ on $X.$
If $(X,p)$ and $(Z,q)$ are F-seminormed spaces then a map $f:X\to Z$ is called an isometric embedding^{[12]} if $q(f(x)-f(y))=p(x,y){\text{ for all ))x,y\in X.$
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 $p$ and $q$ are F-seminorms on $X$ then so is their pointwise supremum $x\mapsto \sup\{p(x),q(x)\}.$ The same is true of the supremum of any non-empty finite family of F-seminorms on $X.$^{[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 $X$ is a seminorm if and only if it is a convexF-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 $0<p<1,$ the map $f$ on $\mathbb {R} ^{n))$ defined by
If $L:X\to Y$ is a linear map and if $q$ is an F-seminorm on $Y,$ then $q\circ L$ is an F-seminorm on $X.$^{[12]}
Let $X_{\mathbb {C} ))$ be a complex vector space and let $X_{\mathbb {R} ))$ denote $X_{\mathbb {C} ))$ considered as a vector space over $\mathbb {R} .$ Any F-seminorm on $X_{\mathbb {C} ))$ is also an F-seminorm on $X_{\mathbb {R} }.$^{[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 $X$ is a value on $X.$ In particular, $p(x)=0,$ and $p(x)=p(-x)$ for all $x\in X.$
Topology induced by a single F-seminorm
Theorem^{[11]} — Let $p$ be an F-seminorm on a vector space $X.$
Then the map $d:X\times X\to \mathbb {R}$ defined by
$d(x,y):=p(x-y)$
is a translation invariant pseudometric on $X$ that defines a vector topology $\tau$ on $X.$
If $p$ is an F-norm then $d$ is a metric.
When $X$ is endowed with this topology then $p$ is a continuous map on $X.$
The balanced sets $\{x\in X~:~p(x)\leq r\},$ as $r$ ranges over the positive reals, form a neighborhood basis at the origin for this topology consisting of closed set.
Similarly, the balanced sets $\{x\in X~:~p(x)<r\},$ as $r$ 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 ${\mathcal {L))$ is a non-empty collection of F-seminorms on a vector space $X$ and for any finite subset ${\mathcal {F))\subseteq {\mathcal {L))$ and any $r>0,$ let
The set $\left\{U_((\mathcal {F)),r}~:~r>0,{\mathcal {F))\subseteq {\mathcal {L)),{\mathcal {F)){\text{ finite ))\right\))$ forms a filter base on $X$ that also forms a neighborhood basis at the origin for a vector topology on $X$ denoted by $\tau _{\mathcal {L)).$^{[12]} Each $U_((\mathcal {F)),r))$ is a balanced and absorbing subset of $X.$^{[12]} These sets satisfy^{[12]}
$\tau _{\mathcal {L))$ is the coarsest vector topology on $X$ making each $p\in {\mathcal {L))$ continuous.^{[12]}
$\tau _{\mathcal {L))$ is Hausdorff if and only if for every non-zero $x\in X,$ there exists some $p\in {\mathcal {L))$ such that $p(x)>0.$^{[12]}
If ${\mathcal {F))$ is the set of all continuous F-seminorms on $\left(X,\tau _{\mathcal {L))\right)$ then $\tau _{\mathcal {L))=\tau _{\mathcal {F)).$^{[12]}
If ${\mathcal {F))$ is the set of all pointwise suprema of non-empty finite subsets of ${\mathcal {F))$ of ${\mathcal {L))$ then ${\mathcal {F))$ is a directed family of F-seminorms and $\tau _{\mathcal {L))=\tau _{\mathcal {F)).$^{[12]}
Fréchet combination
Suppose that $p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is a family of non-negative subadditive functions on a vector space $X.$
The Fréchet combination^{[8]} of $p_{\bullet ))$ is defined to be the real-valued map
Assume that $p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is an increasing sequence of seminorms on $X$ and let $p$ be the Fréchet combination of $p_{\bullet }.$
Then $p$ is an F-seminorm on $X$ that induces the same locally convex topology as the family $p_{\bullet ))$ of seminorms.^{[13]}
Since $p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is increasing, a basis of open neighborhoods of the origin consists of all sets of the form $\left\{x\in X~:~p_{i}(x)<r\right\))$ as $i$ ranges over all positive integers and $r>0$ ranges over all positive real numbers.
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 $p_{i))$ is a paranorm then so is $p$ and moreover, $p$ induces the same topology on $X$ as the family $p_{\bullet ))$ of paranorms.^{[8]}
This is also true of the following paranorms on $X$:
$q(x):=\inf _{}\left\{\sum _{i=1}^{n}p_{i}(x)+{\frac {1}{n))~:~n>0{\text{ is an integer ))\right\}.$^{[8]}
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 $R:[0,\infty )\to [0,\infty )$ that has a bounded range, is subadditive (meaning that $R(s+t)\leq R(s)+R(t)$ for all $s,t\geq 0$), and satisfies $R(s)=0$ if and only if $s=0.$
Examples of bounded remetrization functions include $\arctan t,$$\tanh t,$$t\mapsto \min\{t,1\},$ and $t\mapsto {\frac {t}{1+t)).$^{[15]}
If $d$ is a pseudometric (respectively, metric) on $X$ and $R$ is a bounded remetrization function then $R\circ d$ is a bounded pseudometric (respectively, bounded metric) on $X$ that is uniformly equivalent to $d.$^{[15]}
Suppose that $p_{\bullet }=\left(p_{i}\right)_{i=1}^{\infty ))$ is a family of non-negative F-seminorm on a vector space $X,$$R$ is a bounded remetrization function, and $r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty ))$ is a sequence of positive real numbers whose sum is finite.
Then
defines a bounded F-seminorm that is uniformly equivalent to the $p_{\bullet }.$^{[16]}
It has the property that for any net $x_{\bullet }=\left(x_{a}\right)_{a\in A))$ in $X,$$p\left(x_{\bullet }\right)\to 0$ if and only if $p_{i}\left(x_{\bullet }\right)\to 0$ for all $i.$^{[16]}$p$ is an F-norm if and only if the $p_{\bullet ))$ separate points on $X.$^{[16]}
Characterizations
Of (pseudo)metrics induced by (semi)norms
A pseudometric (resp. metric) $d$ is induced by a seminorm (resp. norm) on a vector space $X$ if and only if $d$ is translation invariant and absolutely homogeneous, which means that for all scalars $s$ and all $x,y\in X,$ in which case the function defined by $p(x):=d(x,0)$ is a seminorm (resp. norm) and the pseudometric (resp. metric) induced by $p$ is equal to $d.$
Of pseudometrizable TVS
If $(X,\tau )$ is a topological vector space (TVS) (where note in particular that $\tau$ is assumed to be a vector topology) then the following are equivalent:^{[11]}
$X$ is pseudometrizable (i.e. the vector topology $\tau$ is induced by a pseudometric on $X$).
$X$ has a countable neighborhood base at the origin.
The topology on $X$ is induced by a translation-invariant pseudometric on $X.$
The topology on $X$ is induced by an F-seminorm.
The topology on $X$ is induced by a paranorm.
Of metrizable TVS
If $(X,\tau )$ is a TVS then the following are equivalent:
$X$ has a countable neighborhood base at the origin consisting of convex sets.
The topology of $X$ is induced by a countable family of (continuous) seminorms.
The topology of $X$ is induced by a countable increasing sequence of (continuous) seminorms $\left(p_{i}\right)_{i=1}^{\infty ))$ (increasing means that for all $i,$$p_{i}\geq p_{i+1}.$
The topology of $X$ is induced by an F-seminorm of the form:
where $\left(p_{i}\right)_{i=1}^{\infty ))$ are (continuous) seminorms on $X.$^{[18]}
Quotients
Let $M$ be a vector subspace of a topological vector space $(X,\tau ).$
If $X$ is a pseudometrizable TVS then so is $X/M.$^{[11]}
If $X$ is a complete pseudometrizable TVS and $M$ is a closed vector subspace of $X$ then $X/M$ is complete.^{[11]}
If $X$ is metrizable TVS and $M$ is a closed vector subspace of $X$ then $X/M$ is metrizable.^{[11]}
If $p$ is an F-seminorm on $X,$ then the map $P:X/M\to \mathbb {R}$ defined by
$P(x+M):=\inf _{}\{p(x+m):m\in M\))$
is an F-seminorm on $X/M$ that induces the usual quotient topology on $X/M.$^{[11]} If in addition $p$ is an F-norm on $X$ and if $M$ is a closed vector subspace of $X$ then $P$ is an F-norm on $X.$^{[11]}
Examples and sufficient conditions
Every seminormed space$(X,p)$ is pseudometrizable with a canonical pseudometric given by $d(x,y):=p(x-y)$ for all $x,y\in X.$^{[19]}.
If $(X,d)$ is pseudometric TVS with a translation invariant pseudometric $d,$ then $p(x):=d(x,0)$ defines a paranorm.^{[20]} However, if $d$ is a translation invariant pseudometric on the vector space $X$ (without the addition condition that $(X,d)$ is pseudometric TVS), then $d$ 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]}
If $X$ is Hausdorff locally convex TVS then $X$ with the strong topology, $\left(X,b\left(X,X^{\prime }\right)\right),$ is metrizable if and only if there exists a countable set ${\mathcal {B))$ of bounded subsets of $X$ such that every bounded subset of $X$ is contained in some element of ${\mathcal {B)).$^{[22]}
The strong dual space$X_{b}^{\prime ))$ of a metrizable locally convex space (such as a Fréchet space^{[23]}) $X$ 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 $X$ is a metrizable locally convex space then its strong dual $X_{b}^{\prime ))$ 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 convexcomplete TVS, being TVS-isomorphic to Euclidean space. Consequently, any metrizable TVS that is not normable must be infinite dimensional.
If $M$ is a metrizable locally convex TVS that possess a countable fundamental system of bounded sets, then $M$ is normable.^{[27]}
and if this locally convex space $X$ is also metrizable, then the following may be appended to this list:
the strong dual space of $X$ is metrizable.^{[28]}
the strong dual space of $X$ is a Fréchet–Urysohn locally convex space.^{[23]}
In particular, if a metrizable locally convex space $X$ (such as a Fréchet space) is not normable then its strong dual space$X_{b}^{\prime ))$ is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space $X_{b}^{\prime ))$ is also neither metrizable nor normable.
Another consequence of this is that if $X$ is a reflexivelocally convex TVS whose strong dual $X_{b}^{\prime ))$ is metrizable then $X_{b}^{\prime ))$ is necessarily a reflexive Fréchet space, $X$ is a DF-space, both $X$ and $X_{b}^{\prime ))$ are necessarily complete Hausdorff ultrabornologicaldistinguishedwebbed spaces, and moreover, $X_{b}^{\prime ))$ is normable if and only if $X$ is normable if and only if $X$ is Fréchet–Urysohn if and only if $X$ is metrizable. In particular, such a space $X$ is either a Banach space or else it is not even a Fréchet–Urysohn space.
Metrically bounded sets and bounded sets
Suppose that $(X,d)$ is a pseudometric space and $B\subseteq X.$
The set $B$ is metrically bounded or $d$-bounded if there exists a real number $R>0$ such that $d(x,y)\leq R$ for all $x,y\in B$;
the smallest such $R$ is then called the diameter or $d$-diameter of $B.$^{[14]}
If $B$ is bounded in a pseudometrizable TVS $X$ 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 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 $X$ is a metrizable locally convex space, then the strong dual of $X$ is bornological if and only if it is barreled, if and only if it is infrabarreled.^{[26]}
If $X$ is a complete pseudometrizable TVS and $M$ is a closed vector subspace of $X,$ then $X/M$ is complete.^{[11]}
If $(X,\tau )$ and $(X,\nu )$ are complete metrizable TVSs (i.e. F-spaces) and if $\nu$ is coarser than $\tau$ then $\tau =\nu$;^{[33]} this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete.^{[34]} Said differently, if $(X,\tau )$ and $(X,\nu )$ are both F-spaces but with different topologies, then neither one of $\tau$ and $\nu$ contains the other as a subset. One particular consequence of this is, for example, that if $(X,p)$ is a Banach space and $(X,q)$ is some other normed space whose norm-induced topology is finer than (or alternatively, is coarser than) that of $(X,p)$ (i.e. if $p\leq Cq$ or if $q\leq Cp$ for some constant $C>0$), then the only way that $(X,q)$ can be a Banach space (i.e. also be complete) is if these two norms $p$ and $q$ are equivalent; if they are not equivalent, then $(X,q)$ can not be a Banach space.
As another consequence, if $(X,p)$ is a Banach space and $(X,\nu )$ is a Fréchet space, then the map $p:(X,\nu )\to \mathbb {R}$ is continuous if and only if the Fréchet space $(X,\nu )$is the TVS $(X,p)$ (here, the Banach space $(X,p)$ is being considered as a TVS, which means that its norm is "forgetten" but its topology is remembered).
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 $X$ is a metrizable TVS and $d$ is a metric that defines $X$'s topology, then its possible that $X$ is complete as a TVS (i.e. relative to its uniformity) but the metric $d$ is not a complete metric (such metrics exist even for $X=\mathbb {R}$).
Thus, if $X$ is a TVS whose topology is induced by a pseudometric $d,$ then the notion of completeness of $X$ (as a TVS) and the notion of completeness of the pseudometric space $(X,d)$ are not always equivalent.
The next theorem gives a condition for when they are equivalent:
Theorem — If $X$ is a pseudometrizable TVS whose topology is induced by a translation invariant pseudometric $d,$ then $d$ is a complete pseudometric on $X$ if and only if $X$ is complete as a TVS.^{[36]}
Theorem^{[37]}^{[38]}(Klee) — Let $d$ be any^{[note 2]} metric on a vector space $X$ such that the topology $\tau$ induced by $d$ on $X$ makes $(X,\tau )$ into a topological vector space. If $(X,d)$ is a complete metric space then $(X,\tau )$ is a complete-TVS.
Theorem — If $X$ is a TVS whose topology is induced by a paranorm $p,$ then $X$ is complete if and only if for every sequence $\left(x_{i}\right)_{i=1}^{\infty ))$ in $X,$ if $\sum _{i=1}^{\infty }p\left(x_{i}\right)<\infty$ then $\sum _{i=1}^{\infty }x_{i))$ converges in $X.$^{[39]}
If $M$ is a closed vector subspace of a complete pseudometrizable TVS $X,$ then the quotient space $X/M$ is complete.^{[40]}
If $M$ is a complete vector subspace of a metrizable TVS $X$ and if the quotient space $X/M$ is complete then so is $X.$^{[40]} If $X$ is not complete then $M:=X,$ but not complete, vector subspace of $X.$
Let $M$ be a separable locally convex metrizable topological vector space and let $C$ be its completion. If $S$ is a bounded subset of $C$ then there exists a bounded subset $R$ of $X$ such that $S\subseteq \operatorname {cl} _{C}R.$^{[41]}
Every totally bounded subset of a locally convex metrizable TVS $X$ is contained in the closed convex balanced hull of some sequence in $X$ that converges to $0.$
In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.^{[42]}
If $d$ is a translation invariant metric on a vector space $X,$ then $d(nx,0)\leq nd(x,0)$ for all $x\in X$ and every positive integer $n.$^{[43]}
If $\left(x_{i}\right)_{i=1}^{\infty ))$ is a null sequence (that is, it converges to the origin) in a metrizable TVS then there exists a sequence $\left(r_{i}\right)_{i=1}^{\infty ))$ of positive real numbers diverging to $\infty$ such that $\left(r_{i}x_{i}\right)_{i=1}^{\infty }\to 0.$^{[43]}
A subset of a complete metric space is closed if and only if it is complete. If a space $X$ is not complete, then $X$ is a closed subset of $X$ that is not complete.
Banach-Saks theorem^{[45]} — If $\left(x_{n}\right)_{n=1}^{\infty ))$ is a sequence in a locally convex metrizable TVS $(X,\tau )$ that converges weakly to some $x\in X,$ then there exists a sequence $y_{\bullet }=\left(y_{i}\right)_{i=1}^{\infty ))$ in $X$ such that $y_{\bullet }\to x$ in $(X,\tau )$ and each $y_{i))$ is a convex combination of finitely many $x_{n}.$
Mackey's countability condition^{[14]} — Suppose that $X$ is a locally convex metrizable TVS and that $\left(B_{i}\right)_{i=1}^{\infty ))$ is a countable sequence of bounded subsets of $X.$
Then there exists a bounded subset $B$ of $X$ and a sequence $\left(r_{i}\right)_{i=1}^{\infty ))$ of positive real numbers such that $B_{i}\subseteq r_{i}B$ for all $i.$
Generalized series
As described in this article's section on generalized series, for any $I$-indexed family family $\left(r_{i}\right)_{i\in I))$ of vectors from a TVS $X,$ it is possible to define their sum $\textstyle \sum \limits _{i\in I}r_{i))$ as the limit of the net of finite partial sums $F\in \operatorname {FiniteSubsets} (I)\mapsto \textstyle \sum \limits _{i\in F}r_{i))$ where the domain $\operatorname {FiniteSubsets} (I)$ is directed by $\,\subseteq .\,$
If $I=\mathbb {N}$ and $X=\mathbb {R} ,$ for instance, then the generalized series $\textstyle \sum \limits _{i\in \mathbb {N} }r_{i))$ converges if and only if $\textstyle \sum \limits _{i=1}^{\infty }r_{i))$converges unconditionally in the usual sense (which for real numbers, is equivalent to absolute convergence).
If a generalized series $\textstyle \sum \limits _{i\in I}r_{i))$ converges in a metrizable TVS, then the set $\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 $r_{i))$ will be zero and so this generalized series $\textstyle \sum \limits _{i\in I}r_{i}~=~\textstyle \sum \limits _{\stackrel {i\in I}{r_{i}\neq 0))r_{i))$ is actually a sum of at most countably many non-zero terms.
Linear maps
If $X$ is a pseudometrizable TVS and $A$ maps bounded subsets of $X$ to bounded subsets of $Y,$ then $A$ 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 $F:X\to Y$ is a linear map between TVSs and $X$ is metrizable then the following are equivalent:
$F$ is continuous;
$F$ is a (locally) bounded map (that is, $F$ maps (von Neumann) bounded subsets of $X$ to bounded subsets of $Y$);^{[12]}
the image under $F$ of every null sequence in $X$ is a bounded set^{[12]} where by definition, a null sequence is a sequence that converges to the origin.
$F$ maps null sequences to null sequences;
Open and almost open maps
Theorem: If $X$ is a complete pseudometrizable TVS, $Y$ is a Hausdorff TVS, and $T:X\to Y$ is a closed and almost open linear surjection, then $T$ is an open map.^{[47]}
Theorem: If $T:X\to Y$ is a surjective linear operator from a locally convex space $X$ onto a barrelled space$Y$ (e.g. every complete pseudometrizable space is barrelled) then $T$ is almost open.^{[47]}
Theorem: If $T:X\to Y$ is a surjective linear operator from a TVS $X$ onto a Baire space$Y$ then $T$ is almost open.^{[47]}
Theorem: Suppose $T:X\to Y$ is a continuous linear operator from a complete pseudometrizable TVS $X$ into a Hausdorff TVS $Y.$ If the image of $T$ is non-meager in $Y$ then $T:X\to Y$ is a surjective open map and $Y$ is a complete metrizable space.^{[47]}
A vector subspace $M$ of a TVS $X$ has the extension property if any continuous linear functional on $M$ can be extended to a continuous linear functional on $X.$^{[22]}
Say that a TVS $X$ has the Hahn-Banach extension property (HBEP) if every vector subspace of $X$ 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 $X$ 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]}
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^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 $X,$ where recall that this net's domain is the directed set$(\operatorname {FiniteSubsets} (I),\subseteq ).$
Like every convergent net, this convergent net of partial sums $A\mapsto \textstyle \sum \limits _{i\in A}r_{i))$ is a Cauchy net, which for this particular net means (by definition) that for every neighborhood $W$ of the origin in $X,$ there exists a finite subset $A_{0))$ of $I$ 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 $B,C\supseteq A_{0};$
this implies that $r_{i}\in W$ for every $i\in I\setminus A_{0))$ (by taking $B:=A_{0}\cup \{i\))$ and $C:=A_{0))$).
Since $X$ is metrizable, it has a countable neighborhood basis $U_{1},U_{2},\ldots$ at the origin, whose intersection is necessarily $U_{1}\cap U_{2}\cap \cdots =\{0\))$ (since $X$ is a Hausdorff TVS).
For every positive integer $n\in \mathbb {N} ,$ pick a finite subset $A_{n}\subseteq I$ such that $r_{i}\in U_{n))$ for every $i\in I\setminus A_{n}.$
If $i$ belongs to $(I\setminus A_{1})\cap (I\setminus A_{2})\cap \cdots =I\setminus \left(A_{1}\cup A_{2}\cup \cdots \right)$ then $r_{i))$ belongs to $U_{1}\cap U_{2}\cap \cdots =\{0\}.$
Thus $r_{i}=0$ for every index $i\in I$ that does not belong to the countable set $A_{1}\cup A_{2}\cup \cdots .$$\blacksquare$
Berberian, Sterling K. (1974). Lectures in Functional Analysis and Operator Theory. Graduate Texts in Mathematics. Vol. 15. New York: Springer. ISBN978-0-387-90081-0. OCLC878109401.
Köthe, Gottfried (1983) [1969]. Topological Vector Spaces I. Grundlehren der mathematischen Wissenschaften. Vol. 159. Translated by Garling, D.J.H. New York: Springer Science & Business Media. ISBN978-3-642-64988-2. MR0248498. OCLC840293704.