# Topological vector space

(Redirected from Topological vector spaces)

In mathematics, a topological vector space (also called a linear topological space and commonly abbreviated TVS or t.v.s.) is one of the basic structures investigated in functional analysis.

A topological vector space is a vector space (an algebraic structure) which is also a topological space, the latter thereby admitting a notion of continuity. More specifically, its topological space has a uniform topological structure, allowing a notion of uniform convergence.

The elements of topological vector spaces are typically functions or linear operators acting on topological vector spaces, and the topology is often defined so as to capture a particular notion of convergence of sequences of functions.

Banach spaces and Hilbert spaces, are well-known examples.

Unless stated otherwise, the underlying field of a topological vector space is assumed to be either the complex numbers ℂ or the real numbers ℝ.

## Motivation

Normed spaces

Every normed vector space has a natural topological structure: the norm induces a metric and the metric induces a topology. This is a topological vector space because:

1. The vector addition + : X × XX is jointly continuous with respect to this topology. This follows directly from the triangle inequality obeyed by the norm.
2. The scalar multiplication · : 𝕂 × XX, where 𝕂 is the underlying scalar field of X, is jointly continuous. This follows from the triangle inequality and homogeneity of the norm.

Thus all Banach spaces and Hilbert spaces are examples of topological vector spaces.

Non-normed spaces

There are topological vector spaces whose topology is not induced by a norm, but are still of interest in analysis. Examples of such spaces are spaces of holomorphic functions on an open domain, spaces of infinitely differentiable functions, the Schwartz spaces, and spaces of test functions and the spaces of distributions on them. These are all examples of Montel spaces. An infinite-dimensional Montel space is never normable. The existence of a norm for a given topological vector space is characterized by Kolmogorov's normability criterion.

A topological field is a topological vector space over each of its subfields.

## Definition

A family of neighborhoods of the origin with the above two properties determines uniquely a topological vector space. The system of neighborhoods of any other point in the vector space is obtained by translation.
Definition: A topological vector space (TVS) X is a vector space over a topological field 𝕂 (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that vector addition + : X × XX and scalar multiplication · : 𝕂 × XX are continuous functions (where the domains of these functions are endowed with product topologies). Such a topology is called a vector topology or a TVS topology on X.

Every topological vector space is also a commutative topological group under addition.

Hausdorff assumption

Some authors (e.g., Walter Rudin) require the topology on X to be T1; it then follows that the space is Hausdorff, and even Tychonoff. A topological vector space is said to be separated if it is Hausdorff (note that "separated" does not mean separable). The topological and linear algebraic structures can be tied together even more closely with additional assumptions, the most common of which are listed below.

Category and morphisms

The category of topological vector spaces over a given topological field 𝕂 is commonly denoted TVS𝕂 or TVect𝕂. The objects are the topological vector spaces over 𝕂 and the morphisms are the continuous 𝕂-linear maps from one object to another.

Definition:[1][2] A TVS homomorphism or topological homomorphism is a continuous linear map u : XY between topological vector spaces (TVSs) such that the induced map u : X → Im u is an open mapping when Im u, which is the range or image of u, is given the subspace topology induced by Y.
Definition:[1] A TVS embedding or a topological monomorphism is an injective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also a topological embedding.
Definition:[1] A TVS isomorphism or an isomorphism in the category of TVSs is a bijective linear homeomorphism. Equivalently, it is a surjective TVS embedding.

Many properties of TVSs that are studied, such as local convexity, metrizability, completeness, and normability, are invariant under TVS isomorphisms.

A necessary condition for a vector topology
Definition:[3] A collection 𝒩 of subsets of a vector space is called additive if for every N ∈ 𝒩, there exists some U ∈ 𝒩 such that U + UN.

Characterization of continuity of addition at 0[3] — If (X, +) is a group (as all vector spaces are), τ is a topology on X, and X × X is endowed with the product topology, then the addition map X × XX (i.e. the map (x, y) ↦ x + y) is continuous at the origin of X × X if and only if the set of neighborhoods of the origin in (X, τ) is additive. This statement remains true if the word "neighborhood" is replaced by "open neighborhood."

All of the above conditions are consequently a necessity for a topology to form a vector topology.

### Defining topologies using neighborhoods of the origin

Since every vector topology is translation invariant (i.e. for all x0X, the map XX defined by xx0 + x is a homeomorphism), to define a vector topology it suffices to define a neighborhood basis (or subbasis) for it at the origin.

Theorem[4] (Neighborhood filter of the origin) — Suppose that X is a real or complex vector space. If be a non-empty additive collection of balanced and absorbing subsets of X then is a neighborhood base at 0 for a vector topology on X. That is, the assumptions are that is a filter base that satisfies the following conditions:

1. Every B ∈ ℬ is balanced and absorbing,
2. is additive: For every B ∈ ℬ there exists a U ∈ ℬ such that U + UB,

If satisfies the above two conditions but is not a filter base then it will form a neighborhood subbasis at 0 (rather than a neighborhood basis) for a vector topology on X.

Note that in general, the set of all balanced and absorbing subsets of a vector space does not satisfy the conditions of this theorem and does not form a neighborhood basis at the origin for any vector topology.[3]

### Defining topologies using strings

Definitions:[5][6][7] Let X be a vector space and let U = (Ui)
i=1
be a sequence of subsets of X. Each set in the sequence U is called a knot of U and for every index i, Ui is called the ith knot of U. We call U1 the beginning of U.

We say that the sequence U is/is a:

• Summative if Ui+1 + Ui+1  ⊆  Ui for every index i.
• Balanced (resp. absorbing, closed,[note 1] convex, open, symmetric, barrelled, absolutely convex/disked, etc.) if this is true of every Ui.
• String if U is summative, absorbing, and balanced.
• Topological string or a neighborhood string in a TVS X if U is a string and each of its knows is a neighborhood of the origin in X.

If U is an absorbing disk in a vector space X then the sequence defined by Ui := 21 - i U forms a string beginning with U1 = U. This is called the natural string of U[5] Moreover, if a vector space X has countable dimension then every string contains an absolutely convex string.

Summative 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.

Theorem (ℝ-valued function induced by a string) — Let U = (Ui)
i=0
be a collection of subsets of a vector space such that 0 ∈ Ui and Ui+1 + Ui+1Ui for all i ≥ 0. For all uU0, let

𝕊(u) := { n = (n1, ⋅⋅⋅, nk) : k ≥ 1, ni ≥ 0 for all i, and uUn1 + ⋅⋅⋅ + Unk}.

Define f : X → [0, 1] by f (x) = 1 if xU0 and otherwise let

f (x) := inf { 2- n1 + ⋅⋅⋅ + 2- nk : n = (n1, ⋅⋅⋅, nk) ∈ 𝕊(x)  }.

Then f is subadditive (i.e. f (x + y) ≤ f (x) + f (y) for all x, yX) and f = 0 on Ui, so in particular f (0) = 0. If all Ui are symmetric sets then f (- x) = f (x) and if all Ui are balanced then f (s x) ≤ f (x) for all scalars s such that |s| ≤ 1 and all xX. If X is a topological vector space and if all Ui are neighborhoods of the origin then f is continuous, where if in addition X is Hausdorff and U 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

We also assume that n = (n1, ⋅⋅⋅, nk) always denotes a finite sequence of non-negative integers and we will use the notation:

2- n  :=  2- n1 + ⋅⋅⋅ + 2- nk    and    Un  :=  Un1 + ⋅⋅⋅ + Unk.

Observe that for any integers n ≥ 0 and d > 2,

Un    Un+1 + Un+1    Un+1 + Un+2 + Un+2    Un+1 + Un+2 +  ⋅⋅⋅  + Un+d + Un+d+1 + Un+d+1.

From this it follows that if n = (n1, ⋅⋅⋅, nk) consists of distinct positive integers then UnU-1 + min (n).

We show by induction on k that if n = (n1, ⋅⋅⋅, nk) consists of non-negative integers such that 2- n ≤ 2- M for some integer M ≥ 0 then UnUM. This is clearly true for k = 1 and k = 2 so assume that k > 2, which implies that all ni are positive. If all ni are distinct then we're done, otherwise pick distinct indices i < j such that ni = nj and construct m = (m1, ⋅⋅⋅, mk-1) from n by replacing ni with ni - 1 and deleting the jth element of n (all other elements of n are transferred to m unchanged). Observe that 2- n = 2- m and Un Um (since Uni + UnjUni - 1) so by appealing to the inductive hypothesis we conclude that Un UmUM, as desired.

It is clear that f (0) = 0 and that 0 ≤ f ≤ 1 so to prove that f is subadditive, it suffices to prove that f (x + y) ≤ f (x) + f (y) when x, yX are such that f (x) + f (y) < 1, which implies that x, yU0. This is an exercise. If all Ui are symmetric then x Un if and only if - x Un from which it follows that f (-x) ≤ f (x) and f (-x) ≥ f (x). If all Ui are balanced then the inequality f (s x) ≤ f (x) for all unit scalars s is proved similarly. Since f is a nonnegative subadditive function satisfying f (0) = 0, f is uniformly continuous on X if and only if f is continuous at 0. If all Ui are neighborhoods of the origin then for any real r > 0, pick an integer M > 1 such that 2- M < r so that xUM implies f (x) ≤ 2- M < r. If all Ui form basis of balanced neighborhoods of the origin then one may show that for any n > 0, there exists some 0 < r ≤ 2- n such that f (x) < r implies xUn. ∎

Definitions:[5] If U = (Ui)i ∈ ℕ and V = (Vi)i ∈ ℕ are two collections of subsets of a vector space X and if s is a scalar, then define:
• V contains U:  U  ⊆  V if and only if Ui  ⊆  Vi for every index i.
• Set of knots:  Knots (U)  :=  { Ui  :  i ∈ ℕ }.
• Kernel:  ker U  :=  Ui.
• Scalar multiple:  s U  :=  (s Ui)i ∈ ℕ.
• Sum:  U + V  :=  (Ui + Vi)i ∈ ℕ.
• Intersection:  UV  :=  (UiVi)i ∈ ℕ.

Definition (Directed): If 𝕊 is a collection sequences of subsets of X, then we say that 𝕊 directed (downwards) under inclusion or simply directed if 𝕊 is not empty and for all U, V ∈ 𝕊 there exists some W ∈ 𝕊 such that WU and WV (said differently, if and only if 𝕊 is a prefilter with respect to the containment defined above).

Notation: Let Knots (𝕊)  :=  Knots (U) be the set of all knots of all strings in 𝕊.

Defining vector topologies using collections of strings is particularly useful for defining classes of TVSs that are not necessarily locally convex.

Theorem[5] (Topology induced by strings) — If (X, 𝜏) is a topological vector space then there exists a set 𝕊[proof 1] of neighborhood strings in X that is directed downward and such that the set of all knots of all strings in 𝕊 is a neighborhood basis at the origin for (X, 𝜏). We say that such a collection of strings is 𝜏 fundamental.

Conversely, if X is a vector space and if 𝕊 is a collection of strings in X that is directed downward, then the set Knots (𝕊) of all knots of all strings in 𝕊 forms a neighborhood basis at the origin for a vector topology on X. In this case, we will denote this topology 𝜏𝕊 and say it is the topology generated by 𝕊.

If 𝕊 is the set of all topological strings in a TVS (X, 𝜏) then 𝜏𝕊 = 𝜏.[5]

A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.[8]

## Topological structure

A vector space is an abelian group with respect to the operation of addition, and in a topological vector space the inverse operation is always continuous (since it is the same as multiplication by −1). Hence, every topological vector space is an abelian topological group. Every TVS is completely regular but a TVS need not be normal.[9]

Let X be a topological vector space. Given a subspace MX, the quotient space X/M with the usual quotient topology is a Hausdorff topological vector space if and only if M is closed.[note 2] This permits the following construction: given a topological vector space X (that is probably not Hausdorff), form the quotient space X / M where M is the closure of {0}. X / M is then a Hausdorff topological vector space that can be studied instead of X.

### Invariance of vector topologies

One of the most used properties of vector topologies is that every vector topology is translation invariant:

for all x0X, the map XX defined by xx0 + x is a homeomorphism, but if x0 ≠ 0 then it is not linear and so not a TVS-isomorphism.

Scalar multiplication by a non-zero scalar is a TVS-isomorphism. This means that if s ≠ 0 then the linear map XX defined by xs x is a homeomorphism. Using s = -1 produces the negation map XX defined by x ↦ -x, which is consequently a linear homeomorphism and thus a TVS-isomorphism.

If xX and any subset SX, then  cl(x + S) = x + cl(S)[4] and moreover, if 0 ∈ S then x + S is a neighborhood (resp. open neighborhood, closed neighborhood) of x in X if and only if the same is true of S at the origin.

### Local notions

A subset E of a vector space X is said to be

• absorbing (in X): if for every xX, there exists a real r > 0 such that c xE for any scalar c satisfying |c| ≤ r.
• balanced or circled: if tEE for every scalar |t| ≤ 1.
• convex: if tE + (1-t)EE for every real 0 ≤ t ≤ 1.
• a disk or absolutely convex: if E is convex and balanced.
• symmetric: if -EE, or equivalently, if -E = E.

Every neighborhood of 0 is an absorbing set and contains an open balanced neighborhood of 0[4] so every topological vector space has a local base of absorbing and balanced sets. The origin even has a neighborhood basis consisting of closed balanced neighborhoods of 0; if the space is locally convex then it also has a neighborhood basis consisting of closed convex balanced neighborhoods of 0.

Bounded subsets
Definition:[10] A subset E of a topological vector space X is bounded if for every neighborhood V of 0, then EtV when t is sufficiently large.

The definition of boundedness can be weakened a bit; E is bounded if and only if every countable subset of it is bounded. A set is bounded if and only if each of its subsequences is a bounded set.[11] Also, E is bounded if and only if for every balanced neighborhood V of 0, there exists t such that EtV. Moreover, when X is locally convex, the boundedness can be characterized by seminorms: the subset E is bounded iff every continuous semi-norm p is bounded on E.

Every totally bounded set is bounded.[11] If M is a vector subspace of a TVS X, then a subset of M is bounded in M if and only if it is bounded in X.[11]

### Metrizability

Birkhoff–Kakutani theorem — If (X, τ) is a topological vector space then the following three conditions are equivalent:[12][note 3]

1. The origin { 0 } is closed in X, and there is a countable basis of neighborhoods for 0 in X.
2. (X, τ) is metrizable (as a topological space).
3. There is a translation-invariant metric on X that induces on X the topology τ, which is the given topology on X.
4. (X, τ) is a metrizable topological vector space.[note 4]

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

A TVS is pseudometrizable if and only if it has a countable neighborhood basis at the origin, or equivalent, if and only if its topology is generated by an F-seminorm. A TVS is metrizable if and only if it is Hausdorff and pseudometrizable.

More strongly: a topological vector space is said to be normable if its topology can be induced by a norm. A topological vector space is normable if and only if it is Hausdorff and has a convex bounded neighborhood of 0.[13]

Let 𝕂 be a non-discrete locally compact topological field, for example the real or complex numbers. A Hausdorff topological vector space over 𝕂 is locally compact if and only if it is finite-dimensional, that is, isomorphic to 𝕂n for some natural number n.

### Completeness and uniform structure

Definition:[14] The canonical uniformity on a TVS (X, τ) is the unique translation-invariant uniformity that induces the topology τ on X.

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to[clarification needed] about related notions such as completeness, uniform convergence, Cauchy nets, and uniform continuity. etc., which are always assumed to be with respect to this uniformity (unless indicated other). This implies that every Hausdorff topological vector space is Tychonoff.[15] A subset of a TVS is compact if and only if it is complete and totally bounded (for TVSs, a set being totally bounded is equivalent to it being precompact). But if the TVS is not Hausdorff then there exist compact subsets that are not closed. However, the closure of a compact subset of a non-Hausdorff TVS is again compact (so compact subsets are relatively compact).

With respect to this uniformity, a net (or sequence) x = (xi)iI is Cauchy if and only if for every neighborhood V of 0, there exists some index i such that xmxnV whenever ji and ki.

Every Cauchy sequence is bounded, although Cauchy nets and Cauchy filters may not be bounded. A topological vector space where every Cauchy sequence converges is called sequentially complete; in general, it may not be complete (in the sense that all Cauchy filters converge).

The vector space operation of addition is uniformly continuous and an open map. Scalar multiplication is Cauchy continuous but in general, it is almost never uniformly continuous. Because of this, every topological vector space can be completed and is thus a dense linear subspace of a complete topological vector space.

• Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.[4] Every TVS (even those that are Hausdorff and/or complete) have infinitely many non-isomorphic non-Hausdorff completions.
• A compact subset of a TVS (not necessarily Hausdorff) is complete.[16] A complete subset of a Hausdorff TVS is closed.[16]
• If C is a complete subset of a TVS then any subset of C that is closed in C is complete.[16]
• A Cauchy sequence in a Hausdorff TVS X is not necessarily relatively compact (i.e. its closure in X is not necessarily compact).
• If a Cauchy filter in a TVS has an accumulation point x then it converges to x.
• If a series
i=1
xi
converges[note 5] in a TVS X then xi → 0 in X.[17]

## Examples

### Finest and coarsest vector topology

Let X be a real or complex vector space.

Trivial topology

The trivial topology (or the indiscrete topology) { X, ∅ } is always a TVS topology on any vector space X, so it is obviously the coarsest TVS topology possible. This simple observation allows us to conclude that the intersection of any collection of TVS topologies on X always contains a TVS topology. Any vector space (including those that are infinite dimensional) endowed with the trivial topology is a compact (and thus locally compact) complete pseudometrizable seminormable locally convex topological vector space. It is Hausdorff if and only if dim X = 0.

Finest vector topology

There exists a TVS topology τf on X that is finer than every other TVS-topology on X (that is, any TVS-topology on X is necessarily a subset of 𝜏f).[18][19] Every linear map from (X, τf) into another TVS is necessarily continuous. If X has an uncountable Hamel basis then 𝜏f is not locally convex and not metrizable.[19]

### Product vector spaces

A Cartesian product of a family of topological vector spaces, when endowed with the product topology, is a topological vector space. For instance, the set X of all functions f : ℝ → ℝ: this set X can be identified with the product space and carries a natural product topology. With this topology, X becomes a topological vector space, endowed with a topology called the topology of pointwise convergence. The reason for this name is the following: if (fn) is a sequence of elements in X, then fn has limit fX if and only if fn(x) has limit f(x) for every real number x. This space is complete, but not normable: indeed, every neighborhood of 0 in the product topology contains lines, i.e., sets 𝕂 f for f ≠ 0.

### Finite-dimensional spaces

Let 𝔽 denote or and endow 𝔽 with its usual Hausdorff normed Euclidean topology. Let X be a vector space over 𝔽 of finite dimension n := dim X and note that X is vector space isomorphic to 𝔽n. X has a unique Hausdorff vector topology that is TVS-isomorphic to 𝔽n, which has the usual Euclidean (or product) topology, but it has a unique vector topology if and only if dim X = 0. This Hausdorff vector topology is also the finest vector topology on X.

• If dim X = 0 then X = { 0 } has one vector topology: the trivial topology.
• The trivial topology on a vector space is Hausdorff if and only if the vector space has dimension 0.
• If dim X = 1 then X has two vector topologies: the usual Euclidean topology and the trivial topology.
• Since the field 𝔽 is itself a 1-dimensional topological vector space over 𝔽 and since it plays an important role in the definition of topological vector spaces, this dichotomy plays an important role in the definition of an absorbing set and has consequences that reverberate throughout functional analysis.
Proof outline

The proof of this dichotomy is straightforward an we give only an outline with the important observations. As usual, 𝔽 is assumed have the (normed) Euclidean topology. Let X be a 1-dimensional vector space over 𝔽. Observe that if B ⊆ 𝔽 is a ball centered at 0 and if SX is a subset containing an "unbounded sequence" then BS = X, where an "unbounded sequence" means a sequence of the form (si x)
i=1
where 0 ≠ xX and (si)
i=1
⊆ 𝔽
is unbounded in normed space 𝔽. Any vector topology on X will be translation invariant and invariant under non-zero scalar multiplication, and for every 0 ≠ xX, the map Mx : 𝔽 → X given by Mx (s) := s x is a continuous linear bijection. In particular, for any such x, X = 𝔽 x so every subset of X can be written as F x = Mx(F) for some unique subset F ⊆ 𝔽. And if this vector topology on X has a neighborhood of 0 that is properly contained in X, then the continuity of scalar multiplication 𝔽 × XX at the origin forces the existence of an open neighborhood of the origin in X that doesn't contain any "unbounded sequence". From this, one deduces that if X doesn't carry the trivial topology and if 0 ≠ xX, then for any ball B ⊆ 𝔽 center at 0 in 𝔽, Mx (B) = B x contains an open neighborhood of the origin in X so that Mx is thus a linear homeomorphism. ∎

• If dim X = n ≥ 2 then X has infinitely many distinct vector topologies:
• We now describe some of these topologies: Every linear functional f on X, which is vector space isomorphic to 𝔽n, induces a seminorm |f| : X → ℝ defined by |f|(x) := |f (x)| where ker f = ker |f|. Every seminorm induces a (pseudometrizable locally convex) vector topology on X and seminorms with distinct kernels induce distinct topologies so that in particular, seminorms on X that are induced by linear functionals with distinct kernel will induces distinct vector topologies on X.
• However, while there are infinitely many vector topologies on X when dim X ≥ 2, there are, up to TVS-isomorphism only 1 + dim X vector topologies on X. For instance, if n := dim X = 2 then the vector topologies on X consist of the trivial topology, the Hausdorff Euclidean topology, and then the infinitely many remaining non-trivial non-Euclidean vector topologies on X are all TVS-isomorphic to one another.

### Non-vector topologies

Discrete and cofinite topologies

If X is a non-trivial vector space (i.e. of non-0 dimension) then the discrete topology on X (which is always metrizable) is not a TVS topology because despite making addition and negation continuous (which makes it into a topological group under addition), it fails to make scalar multiplication continuous.

The cofinite topology on X (where a subset is open if and only if its complement is finite) is also not a TVS topology on X.

## Linear maps

A linear operator between two topological vector spaces which is continuous at one point is continuous on the whole domain. Moreover, a linear operator f is continuous if f(X) is bounded (as defined below) for some neighborhood X of 0.

A hyperplane on a topological vector space X is either dense or closed. A linear functional f on a topological vector space X has either dense or closed kernel. Moreover, f is continuous if and only if its kernel is closed.

## Types

Depending on the application additional constraints are usually enforced on the topological structure of the space. In fact, several principal results in functional analysis fail to hold in general for topological vector spaces: the closed graph theorem, the open mapping theorem, and the fact that the dual space of the space separates points in the space.

Below are some common topological vector spaces, roughly ordered by their niceness.

• F-spaces are complete topological vector spaces with a translation-invariant metric. These include Lp spaces for all p > 0.
• Locally convex topological vector spaces: here each point has a local base consisting of convex sets. By a technique known as Minkowski functionals it can be shown that a space is locally convex if and only if its topology can be defined by a family of semi-norms. Local convexity is the minimum requirement for "geometrical" arguments like the Hahn–Banach theorem. The Lp spaces are locally convex (in fact, Banach spaces) for all p ≥ 1, but not for 0 < p < 1.
• Barrelled spaces: locally convex spaces where the Banach–Steinhaus theorem holds.
• Bornological space: a locally convex space where the continuous linear operators to any locally convex space are exactly the bounded linear operators.
• Stereotype space: a locally convex space satisfying a variant of reflexivity condition, where the dual space is endowed with the topology of uniform convergence on totally bounded sets.
• Montel space: a barrelled space where every closed and bounded set is compact
• Fréchet spaces: these are complete locally convex spaces where the topology comes from a translation-invariant metric, or equivalently: from a countable family of semi-norms. Many interesting spaces of functions fall into this class. A locally convex F-space is a Fréchet space.
• LF-spaces are limits of Fréchet spaces. ILH spaces are inverse limits of Hilbert spaces.
• Nuclear spaces: these are locally convex spaces with the property that every bounded map from the nuclear space to an arbitrary Banach space is a nuclear operator.
• Normed spaces and semi-normed spaces: locally convex spaces where the topology can be described by a single norm or semi-norm. In normed spaces a linear operator is continuous if and only if it is bounded.
• Banach spaces: Complete normed vector spaces. Most of functional analysis is formulated for Banach spaces.
• Reflexive Banach spaces: Banach spaces naturally isomorphic to their double dual (see below), which ensures that some geometrical arguments can be carried out. An important example which is not reflexive is L1, whose dual is L but is strictly contained in the dual of L.
• Hilbert spaces: these have an inner product; even though these spaces may be infinite-dimensional, most geometrical reasoning familiar from finite dimensions can be carried out in them. These include L2 spaces.
• Euclidean spaces: n or n with the topology induced by the standard inner product. As pointed out in the preceding section, for a given finite n, there is only one n-dimensional topological vector space, up to isomorphism. It follows from this that any finite-dimensional subspace of a TVS is closed. A characterization of finite dimensionality is that a Hausdorff TVS is locally compact if and only if it is finite-dimensional (therefore isomorphic to some Euclidean space).

## Dual space

Every topological vector space has a continuous dual space—the set X* of all continuous linear functionals, i.e. continuous linear maps from the space into the base field 𝕂. A topology on the dual can be defined to be the coarsest topology such that the dual pairing each point evaluation X* → 𝕂 is continuous. This turns the dual into a locally convex topological vector space. This topology is called the weak-* topology. This may not be the only natural topology on the dual space; for instance, the dual of a normed space has a natural norm defined on it. However, it is very important in applications because of its compactness properties (see Banach–Alaoglu theorem). Caution: Whenever X is a not-normable locally convex space, then the pairing map X* × X → 𝕂 is never continuous, no matter which vector space topology one chooses on V*.

## Properties

Let X be a TVS (not necessarily Hausdorff or locally convex).

Definition: For any SX, the convex (resp. balanced, disked, closed convex, closed balanced, closed disked) hull of S is the smallest subset of X that has this property and contains S.

We denote the closure (resp. interior, convex hull, balanced hull, disked hull) of a set S by cl S (resp. Int S, co S, bal S, cobal S).

### Neighborhoods and open sets

Properties of neighborhoods and open sets
• The open convex subsets of a TVS X (not necessarily Hausdorff or locally convex) are exactly those that are of the form z + { xX : p(x) < 1 } = { xX : p(x - z) < 1 } for some zX and some positive continuous sublinear functional p on X.[20]
• If SX and U is an open subset of X then S + U is an open set in X.[4]
• If SX has non-empty interior then S - S is a neighborhood of 0.[4]
• If K is an absorbing disk in a TVS X and if p := pK is the Minkowski functional of K then[21]
Int K    { xX : p(x) < 1 }    K    { xX : p(x) ≤ 1 }    cl K
• Note that we did not assume that K had any topological properties nor that p was continuous (which happens if and only if K is a neighborhood of 0).
• Every TVS is connected[4] and locally connected.[22] Any connected open subset of a TVS is arcwise connected.
• Let 𝜏 and 𝜐 be two vector topologies on X. Then 𝜏 ⊆ 𝜐 if and only if whenever a net x = (xi)iI in X converges 0 in (X, 𝜐) then x → 0 in (X, 𝜏).[23]
• Let 𝒩 be a neighborhood basis of the origin in X, let SX, and let xX. Then x ∈ cl S if and only if there exists a net s = (sN)N ∈ 𝒩 in S (indexed by 𝒩) such that sx in X.[24][note 6]
Interior
• If S has non-empty interior then Int S = Int(cl S) and cl S = cl(Int S).
• If R, SX and S has non-empty interior then Int(R) + Int(S) ⊆ R + Int(S) ⊆ Int(R+S).
• If S is a disk in X that has non-empty interior then 0 belongs to the interior of S.[25]
• However, a closed balanced subset of X with non-empty interior may fail to contain 0 in its interior.[25]
• If S is a balanced subset of X with non-empty interior then { 0 } ∪ Int S is balanced; in particular, if the interior of a balanced set contains the origin then Int S is balanced.[4][note 7]
• If x belongs to the interior of a convex set SX and y ∈ clX S, then the half-open line segment [x, y) := {tx + (1 - t)y : 0 < t ≤ 1} ⊆ Int S.[26] If N is a balanced neighborhood of 0 in X then by considering intersections of the form N ∩ ℝ x (which are convex symmetric neighborhoods of 0 in the real TVS x) it follows that:
• Int N  =  [0, 1) Int N  =  [0, 1) N  =  (-1, 1) N  =  B1 N, where B1 := { a ∈ 𝕂 : |a| < 1 }.
• if x ∈ Int N and r := sup { r > 0 : [0, r) x ⊆ Int N} then r > 1, [0, r) x ⊆ Int N, and if r ≠ ∞ then r x ∈ cl N ∖ Int N.
• If C is convex and 0 < t ≤ 1, then t Int C + (1 - t) cl C ⊆ Int C.[27]

### Non-Hausdorff spaces and the closure of the origin

• X is Hausdorff if and only if { 0 } is closed in X.
• clX { 0 } = N so every neighborhood of the origin contains the closure of { 0 }.
• clX { 0 } is a vector subspace of X and its subspace topology is the trivial topology (which makes clX { 0 } compact).
• Every subset of clX { 0 } is compact and thus complete (see footnote for proof).[proof 2] In particular, if X is not Hausdorff then there exist compact complete subsets that are not closed.[28]
• S + clX { 0 }  ⊆  clX S for every subset SX.[proof 3]
• So if SX is open or closed in X then S + clX { 0 }  =  S (so S is a "tube" with vertical side clX { 0  }).
• The quotient map q : XX / clX { 0 } is a closed map onto a Hausdorff TVS.[29]
• A subset S of a TVS X is totally bounded if and only if S + cl { 0 } is totally bounded,[30] if and only if clX S is totally bounded,[31][32] if and only if its image under the canonical quotient map XX / clX ({ 0 }) is totally bounded.[30]
• If SX is compact, then clX S  = S + clX { 0 } and this set is compact. Thus the closure of a compact set is compact[note 8] (i.e. all compact sets are relatively compact).[33]
• A vector subspace of a TVS is bounded if and only if it is contained in the closure of { 0  }.[11]
• If M is a vector subspace of a TVS X then X / M is Hausdorff if and only if M is closed in X.
• Every vector subspace of X that is an algebraic complement of cl { 0 } is a topological complement of cl { 0  }. Thus if H is an algebraic complement of clX { 0 } in X then the addition map H × clX { 0 } → X, defined by (h, n) ↦ h + n is a TVS-isomorphsim, where H is Hausdorff and cl { 0 } has the indiscrete topology.[34] Moreover, if C is a Hausdorff completion of H then C × clX { 0 } is a completion of XH × clX { 0 }.[30]

### Closed and compact sets

Compact and totally bounded sets
• A subset of a TVS is compact if and only if it is complete and totally bounded.[28]
• Thus, in a complete TVS, a closed and totally bounded subset is compact.[28]
• A subset S of a TVS X is totally bounded if and only if clX S is totally bounded,[31][32] if and only if its image under the canonical quotient map XX / clX ({ 0 }) is totally bounded.[30]
• Every relatively compact set is totally bounded.[28] The closure of a totally bounded set is totally bounded.[28]
• The image of a totally bounded set under a uniformly continuous map (e.g. a continuous linear map) is totally bounded.[28]
• If K is a compact subset of a TVS X and U is an open subset of X containing K, then there exists a neighborhood N of 0 such that K + NU.[35]
• If S is a subset of a TVS X such that every sequence in S has a cluster point in S then S is totally bounded.[30]
Closure and closed set
• If SX and a is a scalar then acl(S) ⊆ cl(aS); if X is Hausdorff, a ≠ 0, or S = ∅ then equality holds: cl(aS) = acl(S).
• In particular, every non-zero scalar multiple of a closed set is closed.
• If SX and S + S ⊆ 2 cl S then cl S is convex.[36]
• If R, SX then cl(R) + cl(S) ⊆ cl(R+S) and cl[cl(R) + cl(S)] = cl(R+S).[4] Thus if R + S is closed then so is cl(R) + cl(S).[36]
• If SX and if R is a set of scalars such that neither cl S nor cl R contain zero then (cl R) (cl S) = cl(RS).[36]
• The closure of a vector subspace of a TVS is a vector subspace.
• If SX then cl S = (S + N) where 𝒩 is any neighborhood basis at the origin for X.[37]
• But we only have cl S { U : SU, U open in X } and this containment can be proper[38] (e.g. X = ℝ and S is the rational numbers).
• It follows that cl UU + U for every neighborhood U of the origin in X.[39]
• If X is a real TVS and SX, then rS ⊆ cl S (observe that the left hand side is independent of the topology on X); if S is a convex neighborhood of the origin then equality holds.
• The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed[4] (see this footnote[note 9] for examples).
• If M is a vector subspace of X and N is a closed neighborhood of 0 in X such that UN is closed in X then M is closed in X.[35]
• Every finite dimensional vector subspace of a Hausdorff TVS is closed. The sum of a closed vector subspace and a finite-dimensional vector subspace is closed.[4]
Closed hulls
• In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.[11]
• The closed convex hull of a set is equal to the closure of the convex hull of that set (i.e. to cl(co(S))).[4]
• The closed balanced hull of a set is equal to the closure of the balanced hull of that set (i.e. to cl(bal(S))).[4]
• The closed disked hull of a set is equal to the closure of the disked hull of that set (i.e. to cl(cobal(S))).[4]
• If R, SX and the closed convex hull of one of the sets S or R is compact then cl(co(R)) + cl(co(S)) = cl(co(R+S)).[4]
• If R, SX each have a closed convex hull that is compact (i.e. cl(co(R)) and cl(co(S)) are compact) then cl(co(RS))  =  co[cl(co(R)) ∪ cl(co(S))].
Hulls and compactness
• In a general TVS, the closed convex hull of a compact set may fail to be compact.
• The balanced hull of a compact (resp. totally bounded) set has that same property.[4]
• The convex hull of a finite union of compact convex sets is again compact and convex.[4]

### Other properties

Meager, nowhere dense, and Baire
Important algebraic facts and common misconceptions
• If SX then 2SS + S; if S is convex then equality holds.
• For an example where equality does not hold, let x be non-zero and set S = { -x, x }; S = { x, 2x } also works.
• A subset C is convex if and only if (s + t)C = sC + tC for all positive real s and t.[40]
• The disked hull of a set SX is equal to the convex hull of the balanced hull of S (i.e. to co(bal(S))).
• However, in general co(bal(S)) ≠ bal(co(S)).
• If R, SX and a is a scalar then co(R + S) = co R + co S and co(aS) = aco S.[4]
• If R, SX are convex non-empty disjoint sets and xRS, then S ∩ co(R ∪ { x }) = ∅ or R ∩ co(S ∪ { x }) = ∅.
• In any non-trivial vector space X, there exist two disjoint non-empty convex subsets whose union is X.
Other properties
• Every TVS topology can be generated by a family of F-seminorms.[41]

### Properties preserved by set operators

• The balanced hull of a compact (resp. totally bounded, open) set has that same property.[4]
• The (Minkowski) sum of two compact (resp. bounded, balanced, convex) sets has that same property.[4] But the sum of two closed sets need not be closed.
• The convex hull of a balanced (resp. open) set is balanced (resp. open).
• The convex hull of a closed set need not be closed.[4]
• The convex hull of a bounded set need not be bounded.

The following table, the color of each cell indicates whether or not a given property of subsets of X (indicated by the column name e.g. "convex") is preserved under the set operator (indicated by the row's name e.g. "closure"). If in every TVS, a property is preserved under the indicated set operator then that cell will be colored green; otherwise, it will be colored red.

So for instance, since the union of two absorbing sets is again absorbing, the cell in row "RS" and column "Absorbing" is colored green. But since the arbitrary intersection of absorbing sets need not be absorbing, the cell in row "Arbitrary intersections (of at least 1 set)" and column "Absorbing" is colored red. If a cell is not colored then that information has yet to be filled in.

## Notes

1. ^ The topological properties of course also require that X be a TVS.
2. ^ In particular, X is Hausdorff if and only if the set {0} is closed (i.e., X is a T1 space).
3. ^ In fact, this is true for topological group, since the proof doesn't use the scalar multiplications.
4. ^ Also called a metric linear space, which means that it's a real or complex vector space together with a translation-invariant metric for which addition and scalar multiplication are continuous.
5. ^ A series
i=1
xi
is said to converge in a TVS X if the sequence of partial sums converges.
6. ^ This shows, in particular, that it will often suffice to consider nets indexed by a neighborhood basis of the origin rather than nets on arbitrary directed sets.
7. ^ If the interior of a balanced set is non-empty but does not contain the origin (such sets exists even in 2 and 2) then the interior of this set can not be a balanced set.
8. ^ In general topology, the closure of a compact subset of a non-Hausdorff space may fail to be compact (e.g. the particular point topology on an infinite set). This result shows that this does not happen in non-Hausdorff TVSs. S + clX { 0 } is compact because it is the image of the compact set S × clX { 0 } under the continuous addition map ⋅ + ⋅  :  X × XX. Recall also that the sum of a compact set (i.e. S) and a closed set is closed so S + clX { 0 } is closed in X.
9. ^ In the 2, the sum of the y-axis and the graph of y = 1/x, which is the complement of the y-axis, is open in 2. In , the sum of and 2 is a countable dense subset of so not closed in .
1. ^ This condition is satisfied if we let 𝕊 be the set of all topological strings in (X, 𝜏).
2. ^ Since clX { 0 } has the trivial topology, so does each of its subsets, which makes them all compact. It is known that a subset of any uniform space is compact if and only if it is complete and totally bounded.
3. ^ If sS then s + clX { 0 } = clX (s + { 0 }) = clX { s } ⊆ clX S. Since SS + clX { 0 } ⊆ clX S, if S is closed then equality holds. Clearly, the complement of any set S satisfying the equality S + clX { 0 } = S must also satisfy this equality.

## References

1. ^ a b c Köthe 1969, p. 91.
2. ^ Schaefer & Wolff 1999, pp. 74–78.
3. ^ a b c Wilansky 2013, pp. 40-47.
4. Narici & Beckenstein 2011, pp. 67-113.
5. Adasch, Ernst & Keim 1978, pp. 5-9.
6. ^ Schechter 1996, pp. 721-751.
7. Narici & Beckenstein 2011, pp. 371-423.
8. ^ Adasch, Ernst & Keim 1978, pp. 10-15.
9. ^ Wilansky 2013, p. 53.
10. ^ Rudin 1991, p. 8.
11. Narici & Beckenstein 2011, pp. 155-176.
12. ^ Köthe 1983, section 15.11.
13. ^ http://eom.springer.de/T/t093180.htm
14. ^ Schaefer & Wolff 1999, pp. 12-19.
15. ^ Schaefer & Wolff 1999, p. 16.
16. ^ a b c Narici & Beckenstein 2011, pp. 115-154.
17. ^ Swartz 1992, pp. 27-29.
18. ^ "A quick application of the closed graph theorem". What's new. 2016-04-22. Retrieved 2020-10-07.
19. ^ a b Narici & Beckenstein 2011, p. 111.
20. ^ Narici & Beckenstein 2011, pp. 177-220.
21. ^ Narici & Beckenstein 2011, p. 119-120.
22. ^ Schaefer & Wolff 1999, p. 35.
23. ^ Wilansky 2013, p. 43.
24. ^ Wilansky 2013, p. 42.
25. ^ a b Narici & Beckenstein 2011, p. 108.
26. ^ Schaefer & Wolff 1999, p. 38.
27. ^ Jarchow 1981, pp. 101-104.
28. Narici & Beckenstein 2011, pp. 47-66.
29. ^ Narici & Beckenstein 2011, pp. 107-112.
30. Schaefer & Wolff 1999, pp. 12-35.
31. ^ a b Schaefer & Wolff 1999, p. 25.
32. ^ a b Jarchow 1981, pp. 56-73.
33. ^ Narici & Beckenstein 2011, p. 156.
34. ^ Wilansky 2013, p. 63.
35. ^ a b Narici & Beckenstein 2011, pp. 19-45.
36. ^ a b c Wilansky 2013, pp. 43-44.
37. ^ Narici & Beckenstein 2011, pp. 80.
38. ^ Narici & Beckenstein 2011, pp. 108-109.
39. ^ Jarchow 1981, pp. 30-32.
40. ^ Rudin 1991, p. 38.
41. ^ Swartz 1992, p. 35.