Metrizable topological vector space

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 $$X$$ is a map $$d : X \times X \rarr \R$$ satisfying the following properties:  $$d(x, x) = 0 \text{ for all } x \in X$$; Symmetry: $$d(x, y) = d(y, x) \text{ for all } x, y \in X$$; Subadditivity: $$d(x, z) \leq d(x, y) + d(y, z) \text{ for all } x, y, z \in X.$$ 

A pseudometric is called a metric if it satisfies: Identity of indiscernibles: for all $$x, y \in X,$$ if $$d(x, y) = 0$$ then $$x = y.$$ 

Ultrapseudometric

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 $$d.$$


 * 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 $$d.$$

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:  Translation invariance: $$d(x + z, y + z) = d(x, y) \text{ for all } x, y, z \in X$$;</li> <li>$$d(x, y) = d(x - y, 0) \text{ for all } x, y \in X.$$</li> </ol>

Value/G-seminorm
If $$X$$ is a topological group the a value or G-seminorm on $$X$$ (the G stands for Group) is a real-valued map $$p : X \rarr \R$$ with the following properties: <ol> <li>Non-negative: $$p \geq 0.$$</li> <li>Subadditive: $$p(x + y) \leq p(x) + p(y) \text{ for all } x, y \in X$$;</li> <li>$$p(0) = 0..$$</li> <li>Symmetric: $$p(-x) = p(x) \text{ for all } x \in X.$$</li> </ol>

where we call a G-seminorm a G-norm if it satisfies the additional condition: <li>Total/Positive definite: If $$p(x) = 0$$ then $$x = 0.$$</li> </ol>

Properties of values
If $$p$$ is a value on a vector space $$X$$ then: <ul> <li>$$|p(x) - p(y)| \leq p(x - y) \text{ for all } x, y \in X.$$</li> <li>$$p(n x) \leq n p(x)$$ and $$\frac{1}{n} p(x) \leq p(x / n)$$ for all $$x \in X$$ and positive integers $$n.$$</li> <li>The set $$\{ x \in X : p(x) = 0 \}$$ is an additive subgroup of $$X.$$</li> </ul>

Equivalence on topological groups
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Pseudometrizable topological groups
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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 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 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 if for every $$N \in \mathcal{N},$$ there exists some $$U \in \mathcal{N}$$ such that $$U + U \subseteq N.$$

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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.

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Assume that $$n_{\bull} = \left(n_1, \ldots, n_k\right)$$ always denotes a finite sequence of non-negative integers and use the notation: $$\sum 2^{- n_{\bull}} := 2^{- n_1} + \cdots + 2^{- n_k} \quad \text{ and } \quad \sum U_{n_{\bull}} := U_{n_1} + \cdots + U_{n_k}.$$

For any integers $$n \geq 0$$ and $$d > 2,$$ $$U_n \supseteq U_{n+1} + U_{n+1} \supseteq U_{n+1} + U_{n+2} + U_{n+2} \supseteq U_{n+1} + U_{n+2} + \cdots + U_{n+d} + U_{n+d+1} + U_{n+d+1}.$$

From this it follows that if $$n_{\bull} = \left(n_1, \ldots, n_k\right)$$ consists of distinct positive integers then $$\sum U_{n_{\bull}} \subseteq U_{-1 + \min \left(n_{\bull}\right)}.$$

It will now be shown by induction on $$k$$ that if $$n_{\bull} = \left(n_1, \ldots, n_k\right)$$ consists of non-negative integers such that $$\sum 2^{- n_{\bull}} \leq 2^{- M}$$ for some integer $$M \geq 0$$ then $$\sum U_{n_{\bull}} \subseteq U_M.$$ 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 $$m_{\bull} = \left(m_1, \ldots, m_{k-1}\right)$$ from $$n_{\bull}$$ by replacing each $$n_i$$ with $$n_i - 1$$ and deleting the $$j^{\text{th}}$$ element of $$n_{\bull}$$ (all other elements of $$n_{\bull}$$ are transferred to $$m_{\bull}$$ unchanged). Observe that $$\sum 2^{- n_{\bull}} = \sum 2^{- m_{\bull}}$$ and $$\sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}}$$ (because $$U_{n_i} + U_{n_j} \subseteq U_{n_i - 1}$$) so by appealing to the inductive hypothesis we conclude that $$\sum U_{n_{\bull}} \subseteq \sum U_{m_{\bull}} \subseteq U_M,$$ 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 $$x \in \sum U_{n_{\bull}}$$ if and only if $$- x \in \sum U_{n_{\bull}}$$ 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 $$f(s x) \leq f(x)$$ 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) $$p : X \rarr \R$$ on $$X$$ that satisfies any of the following additional conditions, each of which begins with "for all sequences $$x_{\bull} = \left(x_i\right)_{i=1}^{\infty}$$ in $$X$$ and all convergent sequences of scalars $$s_{\bull} = \left(s_i\right)_{i=1}^{\infty}$$": <ol> <li>Continuity of multiplication: if $$s$$ is a scalar and $$x \in X$$ are such that $$p\left(x_i - x\right) \to 0$$ and $$s_{\bull} \to s,$$ then $$p\left(s_i x_i - s x\right) \to 0.$$</li> <li>Both of the conditions: <li>Both of the conditions: <li>Separate continuity: </ol>
 * if $$s_{\bull} \to 0$$ 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 $$p\left(x_{\bull}\right) \to 0$$ then $$p\left(s x_i\right) \to 0$$ for every scalar $$s.$$</li>
 * if $$p\left(x_{\bull}\right) \to 0$$ and $$s_{\bull} \to s$$ for some scalar $$s$$ then $$p\left(s_i x_i\right) \to 0$$;
 * if $$s_{\bull} \to 0$$ then $$p\left(s_i x\right) \to 0 \text{ for all } x \in X.$$</li>
 * if $$s_{\bull} \to s$$ for some scalar $$s$$ then $$p\left(s x_i - s x\right) \to 0$$ for every $$x \in X$$;
 * if $$s$$ is a scalar, $$x \in X,$$ and $$p\left(x_i - x\right) \to 0$$ then $$p\left(s x_i - s x\right) \to 0$$ .</li>

A paranorm is called total if in addition it satisfies: <ul> <li>Total/Positive definite: $$p(x) = 0$$ implies $$x = 0.$$</li> </ul>

Properties of paranorms
If $$p$$ is a paranorm on a vector space $$X$$ then the map $$d : X \times X \rarr \R$$ defined by $$d(x, y) := p(x - y)$$ is a translation-invariant pseudometric on $$X$$ that defines a on $$X.$$

If $$p$$ is a paranorm on a vector space $$X$$ then: <ul> <li>the set $$\{ x \in X : p(x) = 0 \}$$ is a vector subspace of $$X.$$</li> <li>$$p(x + n) = p(x) \text{ for all } x, n \in X$$ with $$p(n) = 0.$$</li> <li>If a paranorm $$p$$ satisfies $$p(s x) \leq |s| p(x) \text{ for all } x \in X$$ and scalars $$s,$$ then $$p$$ is absolutely homogeneity (i.e. equality holds) and thus $$p$$ is a seminorm.</li> </ul>

Examples of paranorms
<ul> <li>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 $$\tau.$$</li> <li>If $$p$$ is a paranorm on $$X$$ then so is the map $$q(x) := p(x) / [1 + p(x)].$$</li> <li>Every positive scalar multiple of a paranorm (resp. total paranorm) is again such a paranorm (resp. total paranorm).</li> <li>Every seminorm is a paranorm.</li> <li>The restriction of an paranorm (resp. total paranorm) to a vector subspace is an paranorm (resp. total paranorm).</li> <li>The sum of two paranorms is a paranorm.</li> <li>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.</li> <li>Each of the following real-valued maps are paranorms on $$X := \R^2$$: <li>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 a paranorms on $$X := \R^2.$$</li> <li>If $$x_{\bull} = \left(x_i\right)_{i \in I}$$ 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 $$s.$$</li> <li>The function $$p(x) := |\sin (\pi x)| + \min \{ 2, |x| \}$$is a paranorm on $$\R$$ that is balanced but nevertheless equivalent to the usual norm on $$R.$$ Note that the function $$x \mapsto |\sin (\pi x)|$$ is subadditive.</li> <li>Let $$X_{\Complex}$$ be a complex vector space and let $$X_{\R}$$ denote $$X_{\Complex}$$ considered as a vector space over $$\R.$$ Any paranorm on $$X_{\Complex}$$ is also a paranorm on $$X_{\R}.$$</li> <li> </ul>
 * $$(x, y) \mapsto |x|$$
 * $$(x, y) \mapsto |x| + |y|$$</li>

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 $$p : X \to \Reals$$ with the following four properties: <ol> <li>Non-negative: $$p \geq 0.$$</li> <li>Subadditive: $$p(x + y) \leq p(x) + p(y)$$ for all $$x, y \in X$$</li> <li>Balanced: $$p(a x) \leq p(x)$$ for $$x \in X$$ all scalars $$a$$ satisfying $$|a| \leq 1;$$ <li>For every $$x \in X,$$ $$p\left(\tfrac{1}{n} x\right) \to 0$$ as $$n \to \infty$$ </ol>
 * 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.</li>
 * The sequence $$\left(\tfrac{1}{n}\right)_{n=1}^\infty$$ can be replaced by any positive sequence converging to the zero.</li>

An F-seminorm is called an F-norm if in addition it satisfies: <li>Total/Positive definite: $$p(x) = 0$$ implies $$x = 0.$$</li> </ol>

An F-seminorm is called monotone if it satisfies: <li>Monotone: $$p(r x) < p(s x)$$ for all non-zero $$x \in X$$ and all real $$s$$ and $$t$$ such that $$s < t.$$</li> </ol>

F-seminormed spaces
An F-seminormed space (resp. F-normed space) 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 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.

Examples of F-seminorms
<ul> <li>Every positive scalar multiple of an F-seminorm (resp. F-norm, seminorm) is again an F-seminorm (resp. F-norm, seminorm).</li> <li>The sum of finitely many F-seminorms (resp. F-norms) is an F-seminorm (resp. F-norm).</li> <li>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 $$X.$$</li> <li>The restriction of an F-seminorm (resp. F-norm) to a vector subspace is an F-seminorm (resp. F-norm).</li> <li>A non-negative real-valued function on $$X$$ 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. In particular, every seminorm is an F-seminorm.</li> <li>For any $$0 < p < 1,$$ the map $$f$$ on $$\Reals^n$$ defined by $$[f\left(x_1, \ldots, x_n\right)]^p = \left|x_1\right|^p + \cdots \left|x_n\right|^p$$ is an F-norm that is not a norm.</li> <li>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 $$X.$$</li> <li>Let $$X_\Complex$$ be a complex vector space and let $$X_\Reals$$ denote $$X_\Complex$$ considered as a vector space over $$\Reals.$$ Any F-seminorm on $$X_\Complex$$ is also an F-seminorm on $$X_\Reals.$$</li> </ul>

Properties of F-seminorms
Every F-seminorm is a paranorm and every paranorm is equivalent to some F-seminorm. 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
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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 $$U_{\mathcal{F}, r} := \bigcap_{p \in \mathcal{F}} \{x \in X : p(x) < r\}.$$

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 $$\tau_{\mathcal{L}}.$$ Each $$U_{\mathcal{F}, r}$$ is a balanced and absorbing subset of $$X.$$  These sets satisfy $$U_{\mathcal{F}, r/2} + U_{\mathcal{F}, r/2} \subseteq U_{\mathcal{F}, r}.$$

<ul> <li>$$\tau_{\mathcal{L}}$$ is the coarsest vector topology on $$X$$ making each $$p \in \mathcal{L}$$ continuous.</li> <li>$$\tau_{\mathcal{L}}$$ is Hausdorff if and only if for every non-zero $$x \in X,$$ there exists some $$p \in \mathcal{L}$$ such that $$p(x) > 0.$$</li> <li>If $$\mathcal{F}$$ is the set of all continuous F-seminorms on $$\left(X, \tau_{\mathcal{L}}\right)$$ then $$\tau_{\mathcal{L}} = \tau_{\mathcal{F}}.$$</li> <li>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 $$\tau_{\mathcal{L}} = \tau_{\mathcal{F}}.$$</li> </ul>

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

The Fréchet combination of $$p_{\bull}$$ is defined to be the real-valued map $$p(x) := \sum_{i=1}^{\infty} \frac{p_i(x)}{2^{i} \left[ 1 + p_i(x)\right]}.$$

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

Since $$p_{\bull} = \left(p_i\right)_{i=1}^{\infty}$$ 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.

The translation invariant pseudometric on $$X$$ induced by this F-seminorm $$p$$ is $$d(x, y) = \sum^{\infty}_{i=1} \frac{1}{2^i} \frac{p_i( x - y )}{1 + p_i( x - y )}.$$

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

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 $$p_{\bull}$$ of paranorms. This is also true of the following paranorms on $$X$$: <ul> <li>$$q(x) := \inf_{} \left\{ \sum_{i=1}^n p_i(x) + \frac{1}{n} ~:~ n > 0 \text{ is an integer }\right\}.$$</li> <li>$$r(x) := \sum_{n=1}^{\infty} \min \left\{ \frac{1}{2^n}, p_n(x)\right\}.$$</li> </ul>

Generalization
The Fréchet combination can be generalized by use of a bounded remetrization function.

A 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 $$t \mapsto \frac{t}{1 + t}.$$ 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 $$d.$$

Suppose that $$p_\bull = \left(p_i\right)_{i=1}^\infty$$ is a family of non-negative F-seminorm on a vector space $$X,$$ $$R$$ is a bounded remetrization function, and $$r_\bull = \left(r_i\right)_{i=1}^\infty$$ is a sequence of positive real numbers whose sum is finite. Then $$p(x) := \sum_{i=1}^\infty r_i R\left(p_i(x)\right)$$ defines a bounded F-seminorm that is uniformly equivalent to the $$p_\bull.$$ It has the property that for any net $$x_\bull = \left(x_a\right)_{a \in A}$$ in $$X,$$ $$p\left(x_\bull\right) \to 0$$ if and only if $$p_i\left(x_\bull\right) \to 0$$ for all $$i.$$ $$p$$ is an F-norm if and only if the $$p_\bull$$ separate points on $$X.$$

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: <ol> <li>$$X$$ is pseudometrizable (i.e. the vector topology $$\tau$$ is induced by a pseudometric on $$X$$).</li> <li>$$X$$ has a countable neighborhood base at the origin.</li> <li>The topology on $$X$$ is induced by a translation-invariant pseudometric on $$X.$$</li> <li>The topology on $$X$$ is induced by an F-seminorm.</li> <li>The topology on $$X$$ is induced by a paranorm.</li> </ol>

Of metrizable TVS
If $$(X, \tau)$$ is a TVS then the following are equivalent: <ol> <li>$$X$$ is metrizable.</li> <li>$$X$$ is Hausdorff and pseudometrizable.</li> <li>$$X$$ is Hausdorff and has a countable neighborhood base at the origin.</li> <li>The topology on $$X$$ is induced by a translation-invariant metric on $$X.$$</li> <li>The topology on $$X$$ is induced by an F-norm.</li> <li>The topology on $$X$$ is induced by a monotone F-norm.</li> <li>The topology on $$X$$ is induced by a total paranorm.</li> </ol>

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Of locally convex pseudometrizable TVS
If $$(X, \tau)$$ is TVS then the following are equivalent: <ol> <li>$$X$$ is locally convex and pseudometrizable.</li> <li>$$X$$ has a countable neighborhood base at the origin consisting of convex sets.</li> <li>The topology of $$X$$ is induced by a countable family of (continuous) seminorms.</li> <li>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 $$i,$$ $$p_i \geq p_{i+1}.$$</li> <li>The topology of $$X$$ is induced by an F-seminorm of the form: $$p(x) = \sum_{n=1}^{\infty} 2^{-n} \operatorname{arctan} p_n(x)$$ where $$\left(p_i\right)_{i=1}^{\infty}$$ are (continuous) seminorms on $$X.$$</li> </ol>

Quotients
Let $$M$$ be a vector subspace of a topological vector space $$(X, \tau).$$ <ul> <li>If $$X$$ is a pseudometrizable TVS then so is $$X / M.$$</li> <li>If $$X$$ is a complete pseudometrizable TVS and $$M$$ is a closed vector subspace of $$X$$ then $$X / M$$ is complete.</li> <li>If $$X$$ is metrizable TVS and $$M$$ is a closed vector subspace of $$X$$ then $$X / M$$ is metrizable.</li> <li>If $$p$$ is an F-seminorm on $$X,$$ then the map $$P : X / M \to \R$$ 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 $$X / M.$$ 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 $$X.$$</li> </ul>

Examples and sufficient conditions
<ul> <li>Every seminormed space $$(X, p)$$ is pseudometrizable with a canonical pseudometric given by $$d(x, y) := p(x - y)$$ for all $$x, y \in X.$$.</li> <li>If $$(X, d)$$ is pseudometric TVS with a translation invariant pseudometric $$d,$$ then $$p(x) := d(x, 0)$$ defines a paranorm. However, if $$d$$ is a translation invariant pseudometric on the vector space $$X$$ (without the addition condition that $$(X, d)$$ is ), then $$d$$ need not be either an F-seminorm nor a paranorm.</li> <li>If a TVS has a bounded neighborhood of the origin then it is pseudometrizable; the converse is in general false.</li> <li>If a Hausdorff TVS has a bounded neighborhood of the origin then it is metrizable.</li> <li>Suppose $$X$$ is either a DF-space or an LM-space. If $$X$$ is a sequential space then it is either metrizable or else a Montel DF-space.</li> </ul>

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 $$\mathcal{B}.$$

The strong dual space $$X_b^{\prime}$$ of a metrizable locally convex space (such as a Fréchet space ) $$X$$ is a DF-space. The strong dual of a DF-space is a Fréchet space. The strong dual of a reflexive Fréchet space is a bornological space. The strong bidual (that is, the strong dual space of the strong dual space) of a metrizable locally convex space is a Fréchet space. 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.

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. 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 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.

If $$X$$ is a Hausdorff locally convex space then the following are equivalent:

<ol> <li>$$X$$ is normable.</li> <li>$$X$$ has a (von Neumann) bounded neighborhood of the origin.</li> <li>the strong dual space $$X^{\prime}_b$$ of $$X$$ is normable.</li> </ol>

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

<li>the strong dual space of $$X$$ is metrizable.</li> <li>the strong dual space of $$X$$ is a Fréchet–Urysohn locally convex space. </li> </ol>

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

Another consequence of this is that if $$X$$ is a reflexive locally convex TVS whose strong dual $$X^{\prime}_b$$ is metrizable then $$X^{\prime}_b$$ is necessarily a reflexive Fréchet space, $$X$$ is a DF-space, both $$X$$ and $$X^{\prime}_b$$ are necessarily complete Hausdorff ultrabornological distinguished webbed spaces, and moreover, $$X^{\prime}_b$$ 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 $$B.$$ 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.

Properties of pseudometrizable TVS
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<ul> <li>Every metrizable locally convex TVS is a quasibarrelled space, bornological space, and a Mackey space.</li> <li>Every complete metrizable TVS is a barrelled space and a Baire space (and hence non-meager). However, there exist metrizable Baire spaces that are not complete.</li> <li>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.</li> <li>If $$X$$ is a complete pseudometrizable TVS and $$M$$ is a closed vector subspace of $$X,$$ then $$X / M$$ is complete.</li> <li>The strong dual of a locally convex metrizable TVS is a webbed space.</li> <li>If $$(X, \tau)$$ and $$(X, \nu)$$ are complete metrizable TVSs (i.e. F-spaces) and if $$\nu$$ is coarser than $$\tau$$ then $$\tau = \nu$$; this is no longer guaranteed to be true if any one of these metrizable TVSs is not complete. 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 $$p \leq C q$$ or if $$q \leq C p$$ 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 $$p : (X, \nu) \to \R$$ is continuous if and only if the Fréchet space $$(X, \nu)$$ 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). </li> <li>A metrizable locally convex space is normable if and only if its strong dual space is a Fréchet–Urysohn locally convex space. </li> <li>Any product of complete metrizable TVSs is a Baire space.</li> <li>A product of metrizable TVSs is metrizable if and only if it all but at most countably many of these TVSs have dimension $$0.$$</li> <li>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.</li> <li>Every complete metrizable TVS is a barrelled space and a Baire space (and thus non-meager).</li> <li>The dimension of a complete metrizable TVS is either finite or uncountable.</li> </ul>

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 $$X$$ 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 a complete metric (such metrics exist even for $$X = \R$$). 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:

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

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

Subsets and subsequences
<ul> <li>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 $$S \subseteq \operatorname{cl}_C R.$$</li> <li>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 $$0.$$</li> <li>In a pseudometrizable TVS, every bornivore is a neighborhood of the origin.</li> <li>If $$d$$ is a translation invariant metric on a vector space $$X,$$ then $$d(n x, 0) \leq n d(x, 0)$$ for all $$x \in X$$ and every positive integer $$n.$$</li> <li>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 $$\left(r_i x_i\right)_{i=1}^{\infty} \to 0.$$</li> <li>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.</li> <li>If $$X$$ is a metrizable locally convex TVS then for every bounded subset $$B$$ of $$X,$$ there exists a bounded disk $$D$$ in $$X$$ such that $$B \subseteq X_D,$$ and both $$X$$ and the auxiliary normed space $$X_D$$ induce the same subspace topology on $$B.$$</li> </ul>

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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 $$I = \N$$ and $$X = \Reals,$$ for instance, then the generalized series $$\textstyle\sum\limits_{i \in \N} r_i$$ 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); 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. Discontinuous linear functionals exist on any infinite-dimensional pseudometrizable TVS. Thus, a pseudometrizable TVS is finite-dimensional if and only if its continuous dual space is equal to its algebraic dual space.

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

<ol> <li>$$F$$ is continuous;</li> <li>$$F$$ is a (locally) bounded map (that is, $$F$$ maps (von Neumann) bounded subsets of $$X$$ to bounded subsets of $$Y$$);</li> <li>$$F$$ is sequentially continuous;</li> <li>the image under $$F$$ of every null sequence in $$X$$ is a bounded set where by definition, a  is a sequence that converges to the origin.</li> <li>$$F$$ maps null sequences to null sequences;</li> </ol>

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.


 * 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.


 * 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.


 * 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.

Hahn-Banach extension property
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 $$X.$$ Say that a TVS $$X$$ has the Hahn-Banach extension property (HBEP) if every vector subspace of $$X$$ has the extension property.

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

$$

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.