Topological vector space

In mathematics, a topological vector space (also called a linear topological space) 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.

Hilbert spaces and Banach 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.

DefinitionEdit

 
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.

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

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. The topological and linear algebraic structures can be tied together even more closely with additional assumptions, the most common of which are listed below.

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.

Ways of defining a vector topologyEdit

Definition:[1] 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.

If X is a vector space and τ is a topology on X, then the addition map is continuous at the origin if and only if the set of neighborhoods of 0 in (X, τ) is additive.[1] This is consequently a necessary condition for a topology to form a vector topology.

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[2] — 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. 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.

ExamplesEdit

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.

Discrete and cofinite topologies - Non-vector 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.

Finest and coarsest vector topologyEdit

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 X, so it is obviously the coarsest TVS topology possible. This simple observation allows us to conclude that intersection of any collection of TVS topologies on X always contains a TVS topology.

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).[3] Every linear map from (X, 𝜏f) into another TVS is necessarily continuous.[3] If X has an uncountable Hamel basis then 𝜏f is not locally convex and not metrizable.[3]

Product vector spacesEdit

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.

Topological structureEdit

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.

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.[4] 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.

In particular, topological vector spaces are uniform spaces and one can thus talk about completeness, uniform convergence and uniform continuity. (This implies that every Hausdorff topological vector space is Tychonoff.[5]) The vector space operation of addition is uniformly continuous and the scalar multiplication is Cauchy continuous (it is almost never uniformly continuous, however). Because of this, every topological vector space can be completed and is thus a dense linear subspace of a complete topological vector space.

The Birkhoff–Kakutani theorem gives that the following three conditions on a topological vector space X are equivalent:[6]

A metric linear space means a (real or complex) vector space together with a metric for which addition and scalar multiplication are continuous. By the Birkhoff–Kakutani theorem, it follows that there is an equivalent metric that is translation-invariant.

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.[7]

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.

Let 𝕂 be a non-discrete locally compact topological field, for example the real or complex numbers. A 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.

Local notionsEdit

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

  • symmetric if -EE, or equivalently, if -E = E;
  • balanced if tEE for every scalar {{math||t| ≤ 1
  • convex if tE + (1-t)EE for every real 0 ≤ t ≤ 1
  • bounded if for every neighborhood V of 0, then EtV when t is sufficiently large.[8]

The definition of boundedness can be weakened a bit; E is bounded if and only if every countable subset of it is bounded. 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 topological vector space has a local base of absorbing and balanced sets.

A sequence x = (xi)
i=1
in a TVS X is said to be:

  • Cauchy if for every neighborhood V of 0, the difference xmxn belongs to V when m and n are sufficiently large.
  • Mackey convergent to 0 if there exists a sequence of positive real numbers r = (ri)
    i=1
    diverging to such that (rixi)
    i=1
    converges to 0 in X.[9]

Every Cauchy sequence is bounded, although Cauchy nets or 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). Every compact set is bounded and complete but if the TVS is not Hausdorff then compact subsets are not necessarily closed.

TypesEdit

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 spaceEdit

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*.

PropertiesEdit

Every TVS is a commutative topological group (with addition and negation as the group operations). 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).

Translation invariant topology

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.

Thus for any xX and any subset SX, x + S is a neighborhood (resp. open neighborhood, closed neighborhood) of x if and only if the same is true of S at the origin.

The map XX defined by x ↦ -x is also a homeomorphism.

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.[10]
  • If SX and U is an open subset of X then S + U is an open set in X.[2]
  • Every neighborhood of 0 is an absorbing set and contains an open balanced neighborhood of 0.[2]
  • The origin 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.
  • If SX has non-empty interior then S - S is a neighborhood of 0.[2]
  • If K is an absorbing disk in a TVS X and if p := pK is the Minkowski functional of K then[11]
    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).
  • A disk in a TVS is not nowhere dense if and only if its closure is a neighborhood of the origin.[12]
  • 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).[13]
  • 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.[14]
    • 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.
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.[15]
    • However, a closed balanced subset of X with non-empty interior may fail to contain 0 in its interior.[15]
  • 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.[2]
    • 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 it this interior can not be a balanced set.
  • If x belongs to the interior of SX and y ∈ cl S, then the half-open line segment [x, y) := {tx + (1 - t)y : 0 < t ≤ 1} ⊆ Int S.
  • If C is convex and 0 < t ≤ 1, then t Int C + (1 - t) cl C ⊆ Int C.[16]
Completeness
  • Every TVS has a completion and every Hausdorff TVS has a Hausdorff completion.[2]
  • A compact subset of a TVS (not necessarily Hausdorff) is complete.[17]
  • If a TVS has a complete neighborhood of the origin then it is complete.[17]
  • A complete subset of a Hausdorff TVS is closed.[17]
  • If C is a complete subset of a TVS then any subset of C that is closed in C is complete.[17]
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).
    • Thus, every non-zero scalar multiple of a closed set is closed. If X is Hausdorff then every scalar multiple of a closed set is closed.
  • If SX and S + S ⊆ 2 cl S then cl S is convex.[18]
  • If R, SX then cl(R) + cl(S) ⊆ cl(R+S) and cl[cl(R) + cl(S)] = cl(R+S).[2] Thus if R + S is closed then so is cl(R) + cl(S).[18]
  • If SX and x ∈ X, cl(x + S) = x + cl(S).[2]
  • 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).[18]
  • The closure of a vector subspace of a TVS is a vector subspace (so in particular, the closure of { 0 } is a vector subspace).
  • If SX then cl S = N ∈ 𝒩 (S + N) where 𝒩 is any neighborhood basis at the origin for X.[19]
    • In particular, cl { 0 } = N ∈ 𝒩 N and every neighborhood of the origin contains the closure of { 0 }.
    • It follows that every subset S of cl({ 0 }) is compact and thus complete (see footnote for proof)[20]. In particular, if X is not Hausdorff then there exist compact complete sets that are not closed.
    • Note however, that there are sets S in TVSs X such that cl S { U : SU, U open in X}.[21]
    • It also follows that cl UU + U for every neighborhood U of the origin in X.[22]
  • If X is a real TVS and SX, then r > 1 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.[2]
  • 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.[23]
  • A topologically complemented vector subspace of a Hausdorff TVS is closed.[2]
  • 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.[2]
  • A vector subspace of a TVS that is closed but not open is nowhere dense.[12]
Closed hulls
  • 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))).[2]
  • 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))).[2]
  • 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))).[2]
  • 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)).[2]
  • 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))].
Compact and totally bounded sets
  • A subset of a TVS is compact if and only if it is complete and totally bounded.[24]
    • Thus, in a complete TVS, a closed and totally bounded subset is compact.[24]
  • In any non-Hausdorff TVS there is a compact (and thus complete) set that is not closed (see footnote for example).[25][24]
  • The closure of a totally bounded set is totally bounded.
  • Every complete totally bounded set relatively compact.[24]
  • Every relatively compact set is totally bounded.[24]
  • The image of a totally bounded set under a uniformly continuous map (e.g. a continuous linear map) is totally bounded.[24]
  • In a locally convex space, any linear combination of totally bounded sets is totally bounded.[26]
  • If K is a compact subset of a TVS X and U is an open subset of X containg K, then there exists a neighborhood N of 0 such that K + NU.[23]
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.[2]
  • In a Fréchet space, the closed convex hull of a compact set is compact.[27]
  • The convex hull of a finite union of compact convex sets is again compact and convex.[2]
  • In a locally convex space, the convex hull and the disked hull of a totally bounded set is totally bounded.[2]
  • In a complete locally convex space, the convex hull and the disked hull of a compact set are both compact.[2]
    • More generally, if K is a compact subset of a locally convex space, then the convex hull co K (resp. the disked hull cobal K) is compact if and only if it is complete.[2]
Boundedness
  • Every totally bounded set is bounded.[26]
  • A set is bounded if and only if each of its subsequences is a bounded set.[26]
  • A vector subspace of a TVS is bounded if and only if it is contained in the closure of { 0 }.[26]
  • In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.[26]
  • 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.[26]
Series and sequences
  • A series
    i=1
    xi
    is said to converge in a TVS X if the sequence of partial sums converges.
  • If a series
    i=1
    xi
    converges in a TVS X then xi → 0 in X.[28]
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.[29]
  • 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)).
  • A balanced subset S is absorbing if and only if for every xX, there exists some scalar a such that axS.[2]
  • If R, SX and a is a scalar then co(R + S) = co R + co S and co(aS) = aco S.[2]
  • 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 is completely regular. However, a TVS need not be normal.[30]
  • Every TVS is connected.[2]
  • 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.
  • Every TVS topology can be generated by a family of F-seminorms.[31]
  • Suppose X is a TVS that does not carry the indiscrete topology. Then X is a Baire space if and only if X has no balanced absorbing nowhere dense subset.[12]
  • A TVS X is a Baire space if and only if X is nonmeager, which happens if and only if there does not exist a nowhere dense set D such that X = n ∈ ℕ nD.[12]
  • If X is a TVS then every vector subspace of X that is an algebraic complement of cl { 0 } is a topological complement of cl { 0 }. Thus every TVS is the product of a Hausdorff TVS and a TVS with the indiscrete topology (i.e cl { 0 }).[32]

Properties preserved by set operatorsEdit

  • The balanced hull of a compact (resp. totally bounded, open) set has that same property.[2]
  • The interior of a convex set is convex.[2]
  • The closure of a convex (resp. balanced, bounded, absorbing) set is has that same property.[2]
  • The closure of a vector subspace is a vector subspace.[2]
  • The (Minkowski) sum of two compact (resp. bounded, balanced, convex) sets has that same property.[2]
    • The sum of two closed sets need not be closed.
  • Finite sums, finite unions, scalar multiples, subsets, closures, interiors, balanced hulls of bounded sets are bounded.[26]
  • Any scalar multiple of a convex (resp. balanced) set has that same property.
  • Every non-zero scalar multiple of an open set (resp. a neighborhood of 0, an absorbing set) is an open set (resp. a neighborhood of 0, an absorbing set).
  • The convex hull of a balanced (resp. open) set is balanced (resp. open).
    • The convex hull of a closed set need not be closed.[2]
    • 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.

See alsoEdit

NotesEdit

  1. ^ a b Wilansky 2013, pp. 40-46.
  2. ^ a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab ac Narici 2011, pp. 67-113.
  3. ^ a b c Narici 2011, p. 111.
  4. ^ In particular, X is Hausdorff if and only if the set {0} is closed (i.e., X is a T1 space).
  5. ^ Schaefer (1999), 16.
  6. ^ Kŏthe (1983), section 15.11.
  7. ^ http://eom.springer.de/T/t093180.htm
  8. ^ Rudin (1991), p. 8.
  9. ^ Swartz 1992, pp. 15-16.
  10. ^ Narici 2011, pp. 177-220.
  11. ^ Narici 2011, p. 119-120.
  12. ^ a b c d e Narici 2011, pp. 371-423.
  13. ^ Wilansky 2013, p. 43.
  14. ^ Wilansky 2013, p. 42.
  15. ^ a b Narici 2011, p. 108.
  16. ^ Jarchow 1981, pp. 101-104.
  17. ^ a b c d Narici 2011, pp. 115-154.
  18. ^ a b c Wilansky 2013, pp. 43-44.
  19. ^ Narici 2011, pp. 80.
  20. ^ Given any open cover of S, pick any open set U from that cover that contains the origin. Since U is a neighborhood of the origin, U contains cl({ 0 }) and thus contains S.
  21. ^ Narici 2011, pp. 108-109.
  22. ^ Jarchow 1981, pp. 30-32.
  23. ^ a b Narici 2011, pp. 19-45.
  24. ^ a b c d e f Narici 2011, pp. 47-66.
  25. ^ Let X be a non-Hausdorff TVS, let C denote the closure of { 0 }, and let x be a non-0 element of C. Then C ∖ { x } is compact but not closed.
  26. ^ a b c d e f g Narici 2011, pp. 155-176.
  27. ^ Rudin 1991, p. 7.
  28. ^ Swartz 1992, pp. 27-29.
  29. ^ Rudin 1991, p. 38.
  30. ^ Wilansky 2013, p. 53.
  31. ^ Swartz 1992, p. 35.
  32. ^ Wilansky 2013, p. 63.

ReferencesEdit

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