Topological vector space

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{{Citation needed|date=April 2024}}:

1. The vector addition map $\backslash cdot\backslash ,\; +\; \backslash ,\backslash cdot\backslash ;\; :\; X\; \backslash times\; X\; \backslash to\; X$ defined by $(x,\; y)\; \backslash mapsto\; x\; +\; y$ is (jointly) continuous with respect to this topology. This follows directly from the triangle inequality obeyed by the norm.
2. The scalar multiplication map $\backslash cdot\; :\; \backslash mathbb\{K\}\; \backslash times\; X\; \backslash to\; X$ defined by $(s,\; x)\; \backslash mapsto\; s\; \backslash cdot\; x,$ where $\backslash mathbb\{K\}$ 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.

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.{{sfn|Rudin|1991|p=4-5 §1.3}} 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.

Image:Topological vector space illust.svg.]]

A topological vector space (TVS) $X$ is a vector space over a topological field $\backslash mathbb\{K\}$ (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that vector addition $\backslash cdot\backslash ,\; +\; \backslash ,\backslash cdot\backslash ;\; :\; X\; \backslash times\; X\; \backslash to\; X$ and scalar multiplication $\backslash cdot\; :\; \backslash mathbb\{K\}\; \backslash times\; X\; \backslash to\; X$ are continuous functions (where the domains of these functions are endowed with product topologies). Such a topology is called a {{visible anchor|vector topology}} or a {{visible anchor|TVS topology}} on $X.$

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

Hausdorff assumption

Many authors (for example, Walter Rudin), but not this page, 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 {{em|{{visible anchor|separated}}}} if it is Hausdorff; importantly, "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 $\backslash mathbb\{K\}$ is commonly denoted $\backslash mathrm\{TVS\}\_\backslash mathbb\{K\}$ or $\backslash mathrm\{TVect\}\_\backslash mathbb\{K\}.$ The objects are the topological vector spaces over $\backslash mathbb\{K\}$ and the morphisms are the continuous $\backslash mathbb\{K\}$-linear maps from one object to another.

A {{em|{{visible anchor|topological vector space homomorphism}}}} (abbreviated {{em|{{visible anchor|TVS homomorphism|TVS-homomorphism}}}}), also called a {{em|{{visible anchor|topological homomorphism|text=topological homomorphism}}}},{{sfn|Köthe|1983|p=91}}{{sfn|Schaefer|Wolff|1999|pp=74–78}} is a continuous linear map $u\; :\; X\; \backslash to\; Y$ between topological vector spaces (TVSs) such that the induced map $u\; :\; X\; \backslash to\; \backslash operatorname\{Im\}\; u$ is an open mapping when $\backslash operatorname\{Im\}\; u\; :=\; u(X),$ which is the range or image of $u,$ is given the subspace topology induced by $Y.$

A {{em|{{visible anchor|topological vector space embedding}}}} (abbreviated {{em|{{visible anchor|TVS embedding|TVS-embedding}}}}), also called a {{em|{{visible anchor|topological monomorphism|text=topological monomorphism}}}}, is an injective topological homomorphism. Equivalently, a TVS-embedding is a linear map that is also a topological embedding.{{sfn|Köthe|1983|p=91}}

A {{em|{{visible anchor|topological vector space isomorphism}}}} (abbreviated {{em|{{visible anchor|TVS isomorphism|TVS-isomorphism}}}}), also called a {{em|{{visible anchor|topological vector isomorphism}}}}{{sfn|Grothendieck|1973|pp=34-36}} or an {{em|{{visible anchor|isomorphism in the category of TVSs|isomorphism in the category of topological vector spaces}}}}, is a bijective linear homeomorphism. Equivalently, it is a surjective TVS embedding{{sfn|Köthe|1983|p=91}}

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

A collection $\backslash mathcal\{N\}$ of subsets of a vector space is called {{em|additive}}{{sfn|Wilansky|2013|pp=40-47}} if for every $N\; \backslash in\; \backslash mathcal\{N\},$ there exists some $U\; \backslash in\; \backslash mathcal\{N\}$ such that $U\; +\; U\; \backslash subseteq\; N.$

{{Math theorem|name=Characterization of continuity of addition at $0${{sfn|Wilansky|2013|pp=40-47}}|note=|math_statement=

If $(X,\; +)$ is a group (as all vector spaces are), $\backslash tau$ is a topology on $X,$ and $X\; \backslash times\; X$ is endowed with the product topology, then the addition map $X\; \backslash times\; X\; \backslash to\; X$ (defined by $(x,\; y)\; \backslash mapsto\; x\; +\; y$) is continuous at the origin of $X\; \backslash times\; X$ if and only if the set of neighborhoods of the origin in $(X,\; \backslash tau)$ 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.

Since every vector topology is translation invariant (which means that for all $x\_0\; \backslash in\; X,$ the map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; x\_0\; +\; x$ is a homeomorphism), to define a vector topology it suffices to define a neighborhood basis (or subbasis) for it at the origin.

{{Math theorem|name=Theorem{{sfn|Narici|Beckenstein|2011|pp=67-113}}|note=Neighborhood filter of the origin|math_statement=

Suppose that $X$ is a real or complex vector space. If $\backslash mathcal\{B\}$ is a non-empty additive collection of balanced and absorbing subsets of $X$ then $\backslash mathcal\{B\}$ is a neighborhood base at $0$ for a vector topology on $X.$ That is, the assumptions are that $\backslash mathcal\{B\}$ is a filter base that satisfies the following conditions:

1. Every $B\; \backslash in\; \backslash mathcal\{B\}$ is balanced and absorbing,
2. $\backslash mathcal\{B\}$ is additive: For every $B\; \backslash in\; \backslash mathcal\{B\}$ there exists a $U\; \backslash in\; \backslash mathcal\{B\}$ such that $U\; +\; U\; \backslash subseteq\; B,$

If $\backslash mathcal\{B\}$ satisfies the above two conditions but is {{em|not}} a filter base then it will form a neighborhood {{em|sub}}basis at $0$ (rather than a neighborhood basis) for a vector topology on $X.$

}}

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.{{sfn|Wilansky|2013|pp=40-47}}

{{anchor|String|Strings}}

Let $X$ be a vector space and let $U\_\{\backslash bull\}\; =\; \backslash left(U\_i\backslash right)\_\{i\; =\; 1\}^\{\backslash infty\}$ be a sequence of subsets of $X.$ Each set in the sequence $U\_\{\backslash bull\}$ is called a {{visible anchor|knot}} of $U\_\{\backslash bull\}$ and for every index $i,$ $U\_i$ is called the $i$-th knot of $U\_\{\backslash bull\}.$ The set $U\_1$ is called the beginning of $U\_\{\backslash bull\}.$ The sequence $U\_\{\backslash bull\}$ is/is a:{{sfn|Adasch|Ernst|Keim|1978|pp=5-9}}{{sfn|Schechter|1996|pp=721-751}}{{sfn|Narici|Beckenstein|2011|pp=371-423}}

• {{visible anchor|Summative}} if $U\_\{i+1\}\; +\; U\_\{i+1\}\; \backslash subseteq\; U\_i$ for every index $i.$
• Balanced (resp. absorbing, closed,The topological properties of course also require that $X$ be a TVS. convex, open, symmetric, barrelled, absolutely convex/disked, etc.) if this is true of every $U\_i.$
• {{visible anchor|String}} if $U\_\{\backslash bull\}$ is summative, absorbing, and balanced.
• {{visible anchor|Topological string}} or a {{visible anchor|neighborhood string}} in a TVS $X$ if $U\_\{\backslash bull\}$ is a string and each of its knots is a neighborhood of the origin in $X.$

If $U$ is an absorbing disk in a vector space $X$ then the sequence defined by $U\_i\; :=\; 2^\{1-i\}\; U$ forms a string beginning with $U\_1\; =\; U.$ This is called the natural string of $U${{sfn|Adasch|Ernst|Keim|1978|pp=5-9}} 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.

{{Math theorem|name=Theorem|note=$\backslash R$-valued function induced by a string|math_statement=

Let $U\_\{\backslash bull\}\; =\; \backslash left(U\_i\backslash right)\_\{i=0\}^\{\backslash infty\}$ be a collection of subsets of a vector space such that $0\; \backslash in\; U\_i$ and $U\_\{i+1\}\; +\; U\_\{i+1\}\; \backslash subseteq\; U\_i$ for all $i\; \backslash geq\; 0.$ For all $u\; \backslash in\; U\_0,$ let $$\backslash mathbb\{S\}(u)\; :=\; \backslash left\backslash \{n\_\{\backslash bull\}\; =\; \backslash left(n\_1,\; \backslash ldots,\; n\_k\backslash right)\; ~:~\; k\; \backslash geq\; 1,\; n\_i\; \backslash geq\; 0\; \backslash text\{\; for\; all\; \}\; i,\; \backslash text\{\; and\; \}\; u\; \backslash in\; U\_\{n\_1\}\; +\; \backslash cdots\; +\; U\_\{n\_k\}\backslash right\backslash \}.$$

Define $f\; :\; X\; \backslash to\; [0,\; 1]$ by $f(x)\; =\; 1$ if $x\; \backslash not\backslash in\; U\_0$ and otherwise let $$f(x)\; :=\; \backslash inf\_\{\}\; \backslash left\backslash \{2^\{-\; n\_1\}\; +\; \backslash cdots\; 2^\{-\; n\_k\}\; ~:~\; n\_\{\backslash bull\}\; =\; \backslash left(n\_1,\; \backslash ldots,\; n\_k\backslash right)\; \backslash in\; \backslash mathbb\{S\}(x)\backslash right\backslash \}.$$

Then $f$ is subadditive (meaning $f(x\; +\; y)\; \backslash leq\; f(x)\; +\; f(y)$ for all $x,\; y\; \backslash in\; X$) and $f\; =\; 0$ on $\backslash bigcap\_\{i\; \backslash geq\; 0\}\; U\_i;$ so in particular, $f(0)\; =\; 0.$ If all $U\_i$ are symmetric sets then $f(-x)\; =\; f(x)$ and if all $U\_i$ are balanced then $f(s\; x)\; \backslash leq\; f(x)$ for all scalars $s$ such that $|s|\; \backslash leq\; 1$ and all $x\; \backslash in\; X.$ If $X$ is a topological vector space and if all $U\_i$ are neighborhoods of the origin then $f$ is continuous, where if in addition $X$ is Hausdorff and $U\_\{\backslash bull\}$ 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 $X.$

}}

A proof of the above theorem is given in the article on metrizable topological vector spaces.

If $U\_\{\backslash bull\}\; =\; \backslash left(U\_i\backslash right)\_\{i\; \backslash in\; \backslash N\}$ and $V\_\{\backslash bull\}\; =\; \backslash left(V\_i\backslash right)\_\{i\; \backslash in\; \backslash N\}$ are two collections of subsets of a vector space $X$ and if $s$ is a scalar, then by definition:{{sfn|Adasch|Ernst|Keim|1978|pp=5-9}}

• $V\_\{\backslash bull\}$ contains $U\_\{\backslash bull\}$: $\backslash \; U\_\{\backslash bull\}\; \backslash subseteq\; V\_\{\backslash bull\}$ if and only if $U\_i\; \backslash subseteq\; V\_i$ for every index $i.$
• Set of knots: $\backslash \; \backslash operatorname\{Knots\}\; U\_\{\backslash bull\}\; :=\; \backslash left\backslash \{U\_i\; :\; i\; \backslash in\; \backslash N\backslash right\backslash \}.$
• Kernel: $\backslash \; \backslash ker\; U\_\{\backslash bull\}\; :=\; \backslash bigcap\_\{i\; \backslash in\; \backslash N\}\; U\_i.$
• Scalar multiple: $\backslash \; s\; U\_\{\backslash bull\}\; :=\; \backslash left(s\; U\_i\backslash right)\_\{i\; \backslash in\; \backslash N\}.$
• Sum: $\backslash \; U\_\{\backslash bull\}\; +\; V\_\{\backslash bull\}\; :=\; \backslash left(U\_i\; +\; V\_i\backslash right)\_\{i\; \backslash in\; \backslash N\}.$
• Intersection: $\backslash \; U\_\{\backslash bull\}\; \backslash cap\; V\_\{\backslash bull\}\; :=\; \backslash left(U\_i\; \backslash cap\; V\_i\backslash right)\_\{i\; \backslash in\; \backslash N\}.$

If $\backslash mathbb\{S\}$ is a collection sequences of subsets of $X,$ then $\backslash mathbb\{S\}$ is said to be directed (downwards) under inclusion or simply directed downward if $\backslash mathbb\{S\}$ is not empty and for all $U\_\{\backslash bull\},\; V\_\{\backslash bull\}\; \backslash in\; \backslash mathbb\{S\},$ there exists some $W\_\{\backslash bull\}\; \backslash in\; \backslash mathbb\{S\}$ such that $W\_\{\backslash bull\}\; \backslash subseteq\; U\_\{\backslash bull\}$ and $W\_\{\backslash bull\}\; \backslash subseteq\; V\_\{\backslash bull\}$ (said differently, if and only if $\backslash mathbb\{S\}$ is a prefilter with respect to the containment $\backslash ,\backslash subseteq\backslash ,$ defined above).

Notation: Let $\backslash operatorname\{Knots\}\; \backslash mathbb\{S\}\; :=\; \backslash bigcup\_\{U\_\{\backslash bull\}\; \backslash in\; \backslash mathbb\{S\}\}\; \backslash operatorname\{Knots\}\; U\_\{\backslash bull\}$ be the set of all knots of all strings in $\backslash mathbb\{S\}.$

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

{{Math theorem|name=Theorem{{sfn|Adasch|Ernst|Keim|1978|pp=5-9}}|note=Topology induced by strings|math_statement=If $(X,\; \backslash tau)$ is a topological vector space then there exists a set $\backslash mathbb\{S\}$This condition is satisfied if $\backslash mathbb\{S\}$ denotes the set of all topological strings in $(X,\; \backslash tau).$ of neighborhood strings in $X$ that is directed downward and such that the set of all knots of all strings in $\backslash mathbb\{S\}$ is a neighborhood basis at the origin for $(X,\; \backslash tau).$ Such a collection of strings is said to be {{em|$\backslash tau$ fundamental}}.

Conversely, if $X$ is a vector space and if $\backslash mathbb\{S\}$ is a collection of strings in $X$ that is directed downward, then the set $\backslash operatorname\{Knots\}\; \backslash mathbb\{S\}$ of all knots of all strings in $\backslash mathbb\{S\}$ forms a neighborhood basis at the origin for a vector topology on $X.$ In this case, this topology is denoted by $\backslash tau\_\backslash mathbb\{S\}$ and it is called the topology generated by $\backslash mathbb\{S\}.$

}}

If $\backslash mathbb\{S\}$ is the set of all topological strings in a TVS $(X,\; \backslash tau)$ then $\backslash tau\_\{\backslash mathbb\{S\}\}\; =\; \backslash tau.${{sfn|Adasch|Ernst|Keim|1978|pp=5-9}} A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.{{sfn|Adasch|Ernst|Keim|1978|pp=10-15}}

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.{{sfn|Wilansky|2013|p=53}}

Let $X$ be a topological vector space. Given a subspace $M\; \backslash subseteq\; X,$ the quotient space $X\; /\; M$ with the usual quotient topology is a Hausdorff topological vector space if and only if $M$ is closed.In particular, $X$ is Hausdorff if and only if the set $\backslash \{0\backslash \}$ is closed (that is, $X$ is a T₁ space). 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 $\backslash \{0\backslash \}.$ $X\; /\; M$ is then a Hausdorff topological vector space that can be studied instead of $X.$

One of the most used properties of vector topologies is that every vector topology is {{em|{{visible anchor|translation invariant}}}}:

:for all $x\_0\; \backslash in\; X,$ the map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; x\_0\; +\; x$ is a homeomorphism, but if $x\_0\; \backslash neq\; 0$ 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 $s\; \backslash neq\; 0$ then the linear map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; s\; x$ is a homeomorphism. Using $s\; =\; -1$ produces the negation map $X\; \backslash to\; X$ defined by $x\; \backslash mapsto\; -\; x,$ which is consequently a linear homeomorphism and thus a TVS-isomorphism.

If $x\; \backslash in\; X$ and any subset $S\; \backslash subseteq\; X,$ then $\backslash operatorname\{cl\}\_X\; (x\; +\; S)\; =\; x\; +\; \backslash operatorname\{cl\}\_X\; S${{sfn|Narici|Beckenstein|2011|pp=67-113}} and moreover, if $0\; \backslash in\; S$ 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.

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

• absorbing (in $X$): if for every $x\; \backslash in\; X,$ there exists a real $r\; >\; 0$ such that $c\; x\; \backslash in\; E$ for any scalar $c$ satisfying $|c|\; \backslash leq\; r.${{sfn|Rudin|1991|p=6 §1.4}}
• balanced or circled: if $t\; E\; \backslash subseteq\; E$ for every scalar $|t|\; \backslash leq\; 1.${{sfn|Rudin|1991|p=6 §1.4}}
• convex: if $t\; E\; +\; (1\; -\; t)\; E\; \backslash subseteq\; E$ for every real $0\; \backslash leq\; t\; \backslash leq\; 1.${{sfn|Rudin|1991|p=6 §1.4}}
• a disk or absolutely convex: if $E$ is convex and balanced.
• symmetric: if $-\; E\; \backslash subseteq\; E,$ or equivalently, if $-\; E\; =\; E.$

Every neighborhood of the origin is an absorbing set and contains an open balanced neighborhood of $0${{sfn|Narici|Beckenstein|2011|pp=67-113}} 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 the origin.

Bounded subsets

A subset $E$ of a topological vector space $X$ is bounded{{sfn|Rudin|1991|p=8}} if for every neighborhood $V$ of the origin there exists $t$ such that $E\; \backslash subseteq\; t\; V$.

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.{{sfn|Narici|Beckenstein|2011|pp=155-176}} Also, $E$ is bounded if and only if for every balanced neighborhood $V$ of the origin, there exists $t$ such that $E\; \backslash subseteq\; t\; V.$ Moreover, when $X$ is locally convex, the boundedness can be characterized by seminorms: the subset $E$ is bounded if and only if every continuous seminorm $p$ is bounded on $E.${{sfn|Rudin|1991|p=27-28 Theorem 1.37}}

Every totally bounded set is bounded.{{sfn|Narici|Beckenstein|2011|pp=155-176}} 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 $X.${{sfn|Narici|Beckenstein|2011|pp=155-176}}

{{Math theorem|name=Birkhoff–Kakutani theorem|math_statement=

If $(X,\; \backslash tau)$ is a topological vector space then the following four conditions are equivalent:{{sfn|Köthe|1983|loc=section 15.11}}In fact, this is true for topological group, since the proof does not use the scalar multiplications.

1. The origin $\backslash \{0\backslash \}$ is closed in $X$ and there is a countable basis of neighborhoods at the origin in $X.$
2. $(X,\; \backslash tau)$ is metrizable (as a topological space).
3. There is a translation-invariant metric on $X$ that induces on $X$ the topology $\backslash tau,$ which is the given topology on $X.$
4. $(X,\; \backslash tau)$ is a metrizable topological vector space.Also called a metric linear space, which means that it is a real or complex vector space together with a translation-invariant 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.

}}

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 the origin.{{SpringerEOM|title=Topological vector space|access-date=26 February 2021}}

Let $\backslash mathbb\{K\}$ be a non-discrete locally compact topological field, for example the real or complex numbers. A Hausdorff topological vector space over $\backslash mathbb\{K\}$ is locally compact if and only if it is finite-dimensional, that is, isomorphic to $\backslash mathbb\{K\}^n$ for some natural number $n.${{sfn|Rudin|1991|p=17 Theorem 1.22}}

{{Main|Complete topological vector space}}

The canonical uniformity{{sfn|Schaefer|Wolff|1999|pp=12-19}} on a TVS $(X,\; \backslash tau)$ is the unique translation-invariant uniformity that induces the topology $\backslash tau$ on $X.$

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to talk{{clarify|date=September 2020}} 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.{{sfn|Schaefer|Wolff|1999|p=16}} A subset of a TVS is compact if and only if it is complete and totally bounded (for Hausdorff 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) $x\_\{\backslash bull\}\; =\; \backslash left(x\_i\backslash right)\_\{i\; \backslash in\; I\}$ is Cauchy if and only if for every neighborhood $V$ of $0,$ there exists some index $n$ such that $x\_i\; -\; x\_j\; \backslash in\; V$ whenever $i\; \backslash geq\; n$ and $j\; \backslash geq\; n.$

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.{{sfn|Narici|Beckenstein|2011|pp=67-113}} Every TVS (even those that are Hausdorff and/or complete) has infinitely many non-isomorphic non-Hausdorff completions.
• A compact subset of a TVS (not necessarily Hausdorff) is complete.{{sfn|Narici|Beckenstein|2011|pp=115-154}} A complete subset of a Hausdorff TVS is closed.{{sfn|Narici|Beckenstein|2011|pp=115-154}}
• If $C$ is a complete subset of a TVS then any subset of $C$ that is closed in $C$ is complete.{{sfn|Narici|Beckenstein|2011|pp=115-154}}
• A Cauchy sequence in a Hausdorff TVS $X$ is not necessarily relatively compact (that is, 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 $\backslash sum\_\{i=1\}^\{\backslash infty\}\; x\_i$ convergesA series $\backslash sum\_\{i=1\}^\{\backslash infty\}\; x\_i$ is said to converge in a TVS $X$ if the sequence of partial sums converges. in a TVS $X$ then $x\_\{\backslash bull\}\; \backslash to\; 0$ in $X.${{sfn|Swartz|1992|pp=27-29}}

Let $X$ be a real or complex vector space.

Trivial topology

The trivial topology or indiscrete topology $\backslash \{X,\; \backslash varnothing\backslash \}$ is always a TVS topology on any vector space $X$ and it is the coarsest TVS topology possible. An important consequence of this is 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 $\backslash dim\; X\; =\; 0.$

Finest vector topology

There exists a TVS topology $\backslash tau\_f$ on $X,$ called the {{visible anchor|finest vector topology}} on $X,$ that is finer than every other TVS-topology on $X$ (that is, any TVS-topology on $X$ is necessarily a subset of $\backslash tau\_f$).{{Cite web|date=2016-04-22|title=A quick application of the closed graph theorem|url=https://terrytao.wordpress.com/2016/04/22/a-quick-application-of-the-closed-graph-theorem/|access-date=2020-10-07| website=What's new| language=en}}{{sfn|Narici|Beckenstein|2011|p=111}} Every linear map from $\backslash left(X,\; \backslash tau\_f\backslash right)$ into another TVS is necessarily continuous. If $X$ has an uncountable Hamel basis then $\backslash tau\_f$ is {{em|not}} locally convex and {{em|not}} metrizable.{{sfn|Narici|Beckenstein|2011|p=111}}

A Cartesian product of a family of topological vector spaces, when endowed with the product topology, is a topological vector space. Consider for instance the set $X$ of all functions $f:\; \backslash R\; \backslash to\; \backslash R$ where $\backslash R$ carries its usual Euclidean topology. This set $X$ is a real vector space (where addition and scalar multiplication are defined pointwise, as usual) that can be identified with (and indeed, is often defined to be) the Cartesian product $\backslash R^\backslash R,,$ which carries the natural product topology. With this product topology, $X\; :=\; \backslash R^\{\backslash R\}$ becomes a topological vector space whose topology is called {{em|the topology of pointwise convergence on $\backslash R.$}} The reason for this name is the following: if $\backslash left(f\_n\backslash right)\_\{n=1\}^\{\backslash infty\}$ is a sequence (or more generally, a net) of elements in $X$ and if $f\; \backslash in\; X$ then $f\_n$ converges to $f$ in $X$ if and only if for every real number $x,$ $f\_n(x)$ converges to $f(x)$ in $\backslash R.$ This TVS is complete, Hausdorff, and locally convex but not metrizable and consequently not normable; indeed, every neighborhood of the origin in the product topology contains lines (that is, 1-dimensional vector subspaces, which are subsets of the form $\backslash R\; f\; :=\; \backslash \{r\; f\; :\; r\; \backslash in\; \backslash R\backslash \}$ with $f\; \backslash neq\; 0$).

By F. Riesz's theorem, a Hausdorff topological vector space is finite-dimensional if and only if it is locally compact, which happens if and only if it has a compact neighborhood of the origin.

Let $\backslash mathbb\{K\}$ denote $\backslash R$ or $\backslash Complex$ and endow $\backslash mathbb\{K\}$ with its usual Hausdorff normed Euclidean topology. Let $X$ be a vector space over $\backslash mathbb\{K\}$ of finite dimension $n\; :=\; \backslash dim\; X$ and so that $X$ is vector space isomorphic to $\backslash mathbb\{K\}^n$ (explicitly, this means that there exists a linear isomorphism between the vector spaces $X$ and $\backslash mathbb\{K\}^n$). This finite-dimensional vector space $X$ always has a unique {{em|Hausdorff}} vector topology, which makes it TVS-isomorphic to $\backslash mathbb\{K\}^n,$ where $\backslash mathbb\{K\}^n$ is endowed with the usual Euclidean topology (which is the same as the product topology). This Hausdorff vector topology is also the (unique) finest vector topology on $X.$ $X$ has a unique vector topology if and only if $\backslash dim\; X\; =\; 0.$ If $\backslash dim\; X\; \backslash neq\; 0$ then although $X$ does not have a unique vector topology, it does have a unique {{em|Hausdorff}} vector topology.

• If $\backslash dim\; X\; =\; 0$ then $X\; =\; \backslash \{0\backslash \}$ has exactly one vector topology: the trivial topology, which in this case (and {{em|only}} in this case) is Hausdorff. The trivial topology on a vector space is Hausdorff if and only if the vector space has dimension $0.$
• If $\backslash dim\; X\; =\; 1$ then $X$ has two vector topologies: the usual Euclidean topology and the (non-Hausdorff) trivial topology.
• Since the field $\backslash mathbb\{K\}$ is itself a $1$-dimensional topological vector space over $\backslash mathbb\{K\}$ 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.

{{math proof | title=Proof outline| proof =

The proof of this dichotomy (i.e. that a vector topology is either trivial or isomorphic to $\backslash mathbb\{K\}$) is straightforward so only an outline with the important observations is given. As usual, $\backslash mathbb\{K\}$ is assumed have the (normed) Euclidean topology. Let for all $r\; >\; 0.$ Let $X$ be a $1$-dimensional vector space over $\backslash mathbb\{K\}.$ If $S\; \backslash subseteq\; X$ and $B\; \backslash subseteq\; \backslash mathbb\{K\}$ is a ball centered at $0$ then $B\; \backslash cdot\; S\; =\; X$ whenever $S$ contains an "unbounded sequence", by which it is meant a sequence of the form $\backslash left(a\_i\; x\backslash right)\_\{i=1\}^\{\backslash infty\}$ where $0\; \backslash neq\; x\; \backslash in\; X$ and $\backslash left(a\_i\backslash right)\_\{i=1\}^\{\backslash infty\}\; \backslash subseteq\; \backslash mathbb\{K\}$ is unbounded in normed space $\backslash mathbb\{K\}$ (in the usual sense). Any vector topology on $X$ will be translation invariant and invariant under non-zero scalar multiplication, and for every $0\; \backslash neq\; x\; \backslash in\; X,$ the map $M\_x\; :\; \backslash mathbb\{K\}\; \backslash to\; X$ given by $M\_x(a)\; :=\; a\; x$ is a continuous linear bijection. Because $X\; =\; \backslash mathbb\{K\}\; x$ for any such $x,$ every subset of $X$ can be written as $F\; x\; =\; M\_x(F)$ for some unique subset $F\; \backslash subseteq\; \backslash mathbb\{K\}.$ And if this vector topology on $X$ has a neighborhood $W$ of the origin that is not equal to all of $X,$ then the continuity of scalar multiplication $\backslash mathbb\{K\}\; \backslash times\; X\; \backslash to\; X$ at the origin guarantees the existence of an open ball $B\_r\; \backslash subseteq\; \backslash mathbb\{K\}$ centered at $0$ and an open neighborhood $S$ of the origin in $X$ such that $B\_r\; \backslash cdot\; S\; \backslash subseteq\; W\; \backslash neq\; X,$ which implies that $S$ does {{em|not}} contain any "unbounded sequence". This implies that for every $0\; \backslash neq\; x\; \backslash in\; X,$ there exists some positive integer $n$ such that $S\; \backslash subseteq\; B\_n\; x.$ From this, it can be deduced that if $X$ does not carry the trivial topology and if $0\; \backslash neq\; x\; \backslash in\; X,$ then for any ball $B\; \backslash subseteq\; \backslash mathbb\{K\}$ center at 0 in $\backslash mathbb\{K\},$ $M\_x(B)\; =\; B\; x$ contains an open neighborhood of the origin in $X,$ which then proves that $M\_x$ is a linear homeomorphism. Q.E.D. $\backslash blacksquare$

}}

• If $\backslash dim\; X\; =\; n\; \backslash geq\; 2$ then $X$ has {{em|infinitely many}} distinct vector topologies:
• Some of these topologies are now described: Every linear functional $f$ on $X,$ which is vector space isomorphic to $\backslash mathbb\{K\}^n,$ induces a seminorm $|f|\; :\; X\; \backslash to\; \backslash R$ defined by $|f|(x)\; =\; |f(x)|$ where $\backslash ker\; f\; =\; \backslash ker\; |f|.$ 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 kernels will induce distinct vector topologies on $X.$
• However, while there are infinitely many vector topologies on $X$ when $\backslash dim\; X\; \backslash geq\; 2,$ there are, {{em|up to TVS-isomorphism}}, only $1\; +\; \backslash dim\; X$ vector topologies on $X.$ For instance, if $n\; :=\; \backslash dim\; X\; =\; 2$ 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.

Discrete and cofinite topologies

If $X$ is a non-trivial vector space (that is, of non-zero dimension) then the discrete topology on $X$ (which is always metrizable) is {{em|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 {{em|not}} a TVS topology on $X.$

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 the origin.

A hyperplane in 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.

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 in order of increasing "niceness."

• F-spaces are complete topological vector spaces with a translation-invariant metric.{{sfn|Rudin|1991|p=9 §1.8}} These include $L^p$ spaces for all $p\; >\; 0.$
• Locally convex topological vector spaces: here each point has a local base consisting of convex sets.{{sfn|Rudin|1991|p=9 §1.8}} 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 seminorms.{{sfn|Rudin|1991|p=27 Theorem 1.36}} Local convexity is the minimum requirement for "geometrical" arguments like the Hahn–Banach theorem. The $L^p$ spaces are locally convex (in fact, Banach spaces) for all $p\; \backslash geq\; 1,$ but not for
• 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 seminorms. Many interesting spaces of functions fall into this class -- $C^\backslash infty(\backslash R)$ is a Fréchet space under the seminorms $\backslash |f\backslash |\_\{k,\backslash ell\}\; =\; \backslash sup\_\{x\backslash in[-k,k]\}\; |f^\{(\backslash ell)\}(x)|.$ A locally convex F-space is a Fréchet space.{{sfn|Rudin|1991|p=9 §1.8}}
• 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 seminormed spaces: locally convex spaces where the topology can be described by a single norm or seminorm. 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. This class includes the $L^p$ spaces with $1\backslash leq\; p\; \backslash leq\; \backslash infty,$ the space $BV$ of functions of bounded variation, and certain spaces of measures.
• 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 {{em|not}} reflexive is $L^1$, whose dual is $L^\{\backslash infty\}$ but is strictly contained in the dual of $L^\{\backslash infty\}.$
• 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 $L^2$ spaces, the $L^2$ Sobolev spaces $W^\{2,k\},$ and Hardy spaces.
• Euclidean spaces: $\backslash R^n$ or $\backslash Complex^n$ 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).

{{Main|Algebraic dual space|Continuous dual space|Strong dual space}}

Every topological vector space has a continuous dual space—the set $X\text{'}$ of all continuous linear functionals, that is, continuous linear maps from the space into the base field $\backslash mathbb\{K\}.$ A topology on the dual can be defined to be the coarsest topology such that the dual pairing each point evaluation $X\text{'}\; \backslash to\; \backslash mathbb\{K\}$ is continuous. This turns the dual into a locally convex topological vector space. This topology is called the weak-* topology.{{sfn|Rudin|1991|p=62-68 §3.8-3.14}} 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 non-normable locally convex space, then the pairing map $X\text{'}\; \backslash times\; X\; \backslash to\; \backslash mathbb\{K\}$ is never continuous, no matter which vector space topology one chooses on $X\text{'}.$ A topological vector space has a non-trivial continuous dual space if and only if it has a proper convex neighborhood of the origin.{{sfn|Narici|Beckenstein|2011|pp=177-220}}

{{See also|Locally convex topological vector space#Properties}}

For any $S\; \backslash subseteq\; X$ of a TVS $X,$ 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 $S.$ The closure (respectively, interior, convex hull, balanced hull, disked hull) of a set $S$ is sometimes denoted by $\backslash operatorname\{cl\}\_X\; S$ (respectively, $\backslash operatorname\{Int\}\_X\; S,$ $\backslash operatorname\{co\}\; S,$ $\backslash operatorname\{bal\}\; S,$ $\backslash operatorname\{cobal\}\; S$).

The convex hull $\backslash operatorname\{co\}\; S$ of a subset $S$ is equal to the set of all {{em|convex combinations}} of elements in $S,$ which are finite linear combinations of the form $t\_1\; s\_1\; +\; \backslash cdots\; +\; t\_n\; s\_n$ where $n\; \backslash geq\; 1$ is an integer, $s\_1,\; \backslash ldots,\; s\_n\; \backslash in\; S$ and $t\_1,\; \backslash ldots,\; t\_n\; \backslash in\; [0,\; 1]$ sum to $1.${{sfn|Rudin|1991|p=38}} The intersection of any family of convex sets is convex and the convex hull of a subset is equal to the intersection of all convex sets that contain it.{{sfn|Rudin|1991|p=38}}

Properties of neighborhoods and open sets

Every TVS is connected{{sfn|Narici|Beckenstein|2011|pp=67-113}} and locally connected{{sfn|Schaefer|Wolff|1999|p=35}} and any connected open subset of a TVS is arcwise connected. If $S\; \backslash subseteq\; X$ and $U$ is an open subset of $X$ then $S\; +\; U$ is an open set in $X${{sfn|Narici|Beckenstein|2011|pp=67-113}} and if $S\; \backslash subseteq\; X$ has non-empty interior then $S\; -\; S$ is a neighborhood of the origin.{{sfn|Narici|Beckenstein|2011|pp=67-113}}

The open convex subsets of a TVS $X$ (not necessarily Hausdorff or locally convex) are exactly those that are of the form for some $z\; \backslash in\; X$ and some positive continuous sublinear functional $p$ on $X.${{sfn|Narici|Beckenstein|2011|pp=177-220}}

If $K$ is an absorbing disk in a TVS $X$ and if $p\; :=\; p\_K$ is the Minkowski functional of $K$ then{{sfn|Narici|Beckenstein|2011|p=119-120}} where importantly, it was {{em|not}} assumed that $K$ had any topological properties nor that $p$ was continuous (which happens if and only if $K$ is a neighborhood of the origin).

Let $\backslash tau$ and $\backslash nu$ be two vector topologies on $X.$ Then $\backslash tau\; \backslash subseteq\; \backslash nu$ if and only if whenever a net $x\_\{\backslash bull\}\; =\; \backslash left(x\_i\backslash right)\_\{i\; \backslash in\; I\}$ in $X$ converges $0$ in $(X,\; \backslash nu)$ then $x\_\{\backslash bull\}\; \backslash to\; 0$ in $(X,\; \backslash tau).${{sfn|Wilansky|2013|p=43}}

Let $\backslash mathcal\{N\}$ be a neighborhood basis of the origin in $X,$ let $S\; \backslash subseteq\; X,$ and let $x\; \backslash in\; X.$ Then $x\; \backslash in\; \backslash operatorname\{cl\}\_X\; S$ if and only if there exists a net $s\_\{\backslash bull\}\; =\; \backslash left(s\_N\backslash right)\_\{N\; \backslash in\; \backslash mathcal\{N\}\}$ in $S$ (indexed by $\backslash mathcal\{N\}$) such that $s\_\{\backslash bull\}\; \backslash to\; x$ in $X.${{sfn|Wilansky|2013|p=42}} 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.

If $X$ is a TVS that is of the second category in itself (that is, a nonmeager space) then any closed convex absorbing subset of $X$ is a neighborhood of the origin.{{sfn|Rudin|1991|p=55}} This is no longer guaranteed if the set is not convex (a counter-example exists even in $X\; =\; \backslash R^2$) or if $X$ is not of the second category in itself.{{sfn|Rudin|1991|p=55}}

Interior

If $R,\; S\; \backslash subseteq\; X$ and $S$ has non-empty interior then

$$\backslash operatorname\{Int\}\_X\; S\; ~=~\; \backslash operatorname\{Int\}\_X\; \backslash left(\backslash operatorname\{cl\}\_X\; S\backslash right)~\; \backslash text\{\; and\; \}\; ~\backslash operatorname\{cl\}\_X\; S\; ~=~\; \backslash operatorname\{cl\}\_X\; \backslash left(\backslash operatorname\{Int\}\_X\; S\backslash right)$$

and

$$\backslash operatorname\{Int\}\_X\; (R)\; +\; \backslash operatorname\{Int\}\_X\; (S)\; ~\backslash subseteq~\; R\; +\; \backslash operatorname\{Int\}\_X\; S\; \backslash subseteq\; \backslash operatorname\{Int\}\_X\; (R\; +\; S).$$

The topological interior of a disk is not empty if and only if this interior contains the origin.{{sfn|Narici|Beckenstein|2011|p=108}}

More generally, if $S$ is a balanced set with non-empty interior $\backslash operatorname\{Int\}\_X\; S\; \backslash neq\; \backslash varnothing$ in a TVS $X$ then $\backslash \{0\backslash \}\; \backslash cup\; \backslash operatorname\{Int\}\_X\; S$ will necessarily be balanced;{{sfn|Narici|Beckenstein|2011|pp=67-113}} consequently, $\backslash operatorname\{Int\}\_X\; S$ will be balanced if and only if it contains the origin.This is because every non-empty balanced set must contain the origin and because $0\; \backslash in\; \backslash operatorname\{Int\}\_X\; S$ if and only if $\backslash operatorname\{Int\}\_X\; S\; =\; \backslash \{0\backslash \}\; \backslash cup\; \backslash operatorname\{Int\}\_X\; S.$ For this (i.e. $0\; \backslash in\; \backslash operatorname\{Int\}\_X\; S$) to be true, it suffices for $S$ to also be convex (in addition to being balanced and having non-empty interior).;{{sfn|Narici|Beckenstein|2011|pp=67-113}}

The conclusion $0\; \backslash in\; \backslash operatorname\{Int\}\_X\; S$ could be false if $S$ is not also convex;{{sfn|Narici|Beckenstein|2011|p=108}} for example, in $X\; :=\; \backslash R^2,$ the interior of the closed and balanced set $S\; :=\; \backslash \{(x,\; y)\; :\; x\; y\; \backslash geq\; 0\backslash \}$ is $\backslash \{(x,\; y)\; :\; x\; y\; >\; 0\backslash \}.$

If $C$ is convex and then{{sfn|Jarchow|1981|pp=101-104}} $t\; \backslash operatorname\{Int\}\; C\; +\; (1\; -\; t)\; \backslash operatorname\{cl\}\; C\; ~\backslash subseteq~\; \backslash operatorname\{Int\}\; C.$

Explicitly, this means that if $C$ is a convex subset of a TVS $X$ (not necessarily Hausdorff or locally convex), $y\; \backslash in\; \backslash operatorname\{int\}\_X\; C,$ and $x\; \backslash in\; \backslash operatorname\{cl\}\_X\; C$ then the open line segment joining $x$ and $y$ belongs to the interior of $C;$ that is, {{sfn|Schaefer|Wolff|1999|p=38}}{{sfn|Conway|1990|p=102}}Fix so it remains to show that $w\_0\; ~\backslash stackrel\{\backslash scriptscriptstyle\backslash text\{def\}\}\{=\}~\; r\; x\; +\; (1\; -\; r)\; y$ belongs to $\backslash operatorname\{int\}\_X\; C.$ By replacing $C,\; x,\; y$ with $C\; -\; w\_0,\; x\; -\; w\_0,\; y\; -\; w\_0$ if necessary, we may assume without loss of generality that $r\; x\; +\; (1\; -\; r)\; y\; =\; 0,$ and so it remains to show that $C$ is a neighborhood of the origin. Let so that $y\; =\; \backslash tfrac\{r\}\{r\; -\; 1\}\; x\; =\; s\; x.$ Since scalar multiplication by $s\; \backslash neq\; 0$ is a linear homeomorphism $X\; \backslash to\; X,$ $\backslash operatorname\{cl\}\_X\; \backslash left(\backslash tfrac\{1\}\{s\}\; C\backslash right)\; =\; \backslash tfrac\{1\}\{s\}\; \backslash operatorname\{cl\}\_X\; C.$ Since $x\; \backslash in\; \backslash operatorname\{int\}\; C$ and $y\; \backslash in\; \backslash operatorname\{cl\}\; C,$ it follows that $x\; =\; \backslash tfrac\{1\}\{s\}\; y\; \backslash in\; \backslash operatorname\{cl\}\; \backslash left(\backslash tfrac\{1\}\{s\}\; C\backslash right)\; \backslash cap\; \backslash operatorname\{int\}\; C$ where because $\backslash operatorname\{int\}\; C$ is open, there exists some $c\_0\; \backslash in\; \backslash left(\backslash tfrac\{1\}\{s\}\; C\backslash right)\; \backslash cap\; \backslash operatorname\{int\}\; C,$ which satisfies $s\; c\_0\; \backslash in\; C.$ Define $h\; :\; X\; \backslash to\; X$ by $x\; \backslash mapsto\; r\; x\; +\; (1\; -\; r)\; s\; c\_0\; =\; r\; x\; -\; r\; c\_0,$ which is a homeomorphism because The set $h\backslash left(\backslash operatorname\{int\}\; C\backslash right)$ is thus an open subset of $X$ that moreover contains $h(c\_0)\; =\; r\; c\_0\; -\; r\; c\_0\; =\; 0.$ If $c\; \backslash in\; \backslash operatorname\{int\}\; C$ then $h(c)\; =\; r\; c\; +\; (1\; -\; r)\; s\; c\_0\; \backslash in\; C$ since $C$ is convex, and $s\; c\_0,\; c\; \backslash in\; C,$ which proves that $h\backslash left(\backslash operatorname\{int\}\; C\backslash right)\; \backslash subseteq\; C.$ Thus $h\backslash left(\backslash operatorname\{int\}\; C\backslash right)$ is an open subset of $X$ that contains the origin and is contained in $C.$ Q.E.D.

If $N\; \backslash subseteq\; X$ is any balanced neighborhood of the origin in $X$ then where $B\_1$ is the set of all scalars $a$ such that

If $x$ belongs to the interior of a convex set $S\; \backslash subseteq\; X$ and $y\; \backslash in\; \backslash operatorname\{cl\}\_X\; S,$ then the half-open line segment and{{sfn|Schaefer|Wolff|1999|p=38}} $$[x,\; x)\; =\; \backslash varnothing\; \backslash text\{\; if\; \}\; x\; =\; y.$$

If $N$ is a balanced neighborhood of $0$ in $X$ and then by considering intersections of the form $N\; \backslash cap\; \backslash R\; x$ (which are convex symmetric neighborhoods of $0$ in the real TVS $\backslash R\; x$) it follows that: $\backslash operatorname\{Int\}\; N\; =\; [0,\; 1)\; \backslash operatorname\{Int\}\; N\; =\; (-1,\; 1)\; N\; =\; B\_1\; N,$ and furthermore, if $x\; \backslash in\; \backslash operatorname\{Int\}\; N\; \backslash text\{\; and\; \}\; r\; :=\; \backslash sup\; \backslash \{r\; >\; 0\; :\; [0,\; r)\; x\; \backslash subseteq\; N\backslash \}$ then $r\; >\; 1\; \backslash text\{\; and\; \}\; [0,\; r)\; x\; \backslash subseteq\; \backslash operatorname\{Int\}\; N,$ and if $r\; \backslash neq\; \backslash infty$ then $r\; x\; \backslash in\; \backslash operatorname\{cl\}\; N\; \backslash setminus\; \backslash operatorname\{Int\}\; N.$

A topological vector space $X$ is Hausdorff if and only if $\backslash \{0\backslash \}$ is a closed subset of $X,$ or equivalently, if and only if $\backslash \{0\backslash \}\; =\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}.$ Because $\backslash \{0\backslash \}$ is a vector subspace of $X,$ the same is true of its closure $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \},$ which is referred to as {{em|the closure of the origin}} in $X.$ This vector space satisfies $$\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; =\; \backslash bigcap\_\{N\; \backslash in\; \backslash mathcal\{N\}(0)\}\; N$$ so that in particular, every neighborhood of the origin in $X$ contains the vector space $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ as a subset.

The subspace topology on $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is always the trivial topology, which in particular implies that the topological vector space $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ a compact space (even if its dimension is non-zero or even infinite) and consequently also a bounded subset of $X.$ In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of $\backslash \{0\backslash \}.${{sfn|Narici|Beckenstein|2011|pp=155-176}}

Every subset of $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ also carries the trivial topology and so is itself a compact, and thus also complete, subspace (see footnote for a proof).Since $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ 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. In particular, if $X$ is not Hausdorff then there exist subsets that are both {{em|compact and complete}} but {{em|not closed}} in $X$;{{sfn|Narici|Beckenstein|2011|pp=47-66}} for instance, this will be true of any non-empty proper subset of $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}.$

If $S\; \backslash subseteq\; X$ is compact, then $\backslash operatorname\{cl\}\_X\; S\; =\; S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ and this set is compact. Thus the closure of a compact subset of a TVS is compact (said differently, all compact sets are relatively compact),{{sfn|Narici|Beckenstein|2011|p=156}} which is not guaranteed for arbitrary non-Hausdorff topological spaces.In general topology, the closure of a compact subset of a non-Hausdorff space may fail to be compact (for example, the particular point topology on an infinite set). This result shows that this does not happen in non-Hausdorff TVSs. $S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is compact because it is the image of the compact set $S\; \backslash times\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ under the continuous addition map $\backslash cdot\backslash ,\; +\; \backslash ,\backslash cdot\backslash ;\; :\; X\; \backslash times\; X\; \backslash to\; X.$ Recall also that the sum of a compact set (that is, $S$) and a closed set is closed so $S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is closed in $X.$

For every subset $S\; \backslash subseteq\; X,$ $$S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; \backslash subseteq\; \backslash operatorname\{cl\}\_X\; S$$ and consequently, if $S\; \backslash subseteq\; X$ is open or closed in $X$ then $S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; =\; S$If $s\; \backslash in\; S$ then $s\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; =\; \backslash operatorname\{cl\}\_X\; (s\; +\; \backslash \{0\backslash \})\; =\; \backslash operatorname\{cl\}\_X\; \backslash \{s\backslash \}\; \backslash subseteq\; \backslash operatorname\{cl\}\_X\; S.$ Because $S\; \backslash subseteq\; S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; \backslash subseteq\; \backslash operatorname\{cl\}\_X\; S,$ if $S$ is closed then equality holds. Using the fact that $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is a vector space, it is readily verified that the complement in $X$ of any set $S$ satisfying the equality $S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; =\; S$ must also satisfy this equality (when $X\; \backslash setminus\; S$ is substituted for $S$). (so that this {{em|arbitrary}} open {{em|or}} closed subsets $S$ can be described as a "tube" whose vertical side is the vector space $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$).

For any subset $S\; \backslash subseteq\; X$ of this TVS $X,$ the following are equivalent:

• $S$ is totally bounded.
• $S\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}
• $\backslash operatorname\{cl\}\_X\; S$ is totally bounded.{{sfn|Schaefer|Wolff|1999|p=25}}{{sfn|Jarchow|1981|pp=56-73}}
• The image if $S$ under the canonical quotient map $X\; \backslash to\; X\; /\; \backslash operatorname\{cl\}\_X\; (\backslash \{0\backslash \})$ is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

If $M$ is a vector subspace of a TVS $X$ then $X\; /\; M$ is Hausdorff if and only if $M$ is closed in $X.$

Moreover, the quotient map $q\; :\; X\; \backslash to\; X\; /\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is always a closed map onto the (necessarily) Hausdorff TVS.{{sfn|Narici|Beckenstein|2011|pp=107-112}}

Every vector subspace of $X$ that is an algebraic complement of $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ (that is, a vector subspace $H$ that satisfies $\backslash \{0\backslash \}\; =\; H\; \backslash cap\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ and $X\; =\; H\; +\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$) is a topological complement of $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}.$

Consequently, if $H$ is an algebraic complement of $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ in $X$ then the addition map $H\; \backslash times\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}\; \backslash to\; X,$ defined by $(h,\; n)\; \backslash mapsto\; h\; +\; n$ is a TVS-isomorphism, where $H$ is necessarily Hausdorff and $\backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ has the indiscrete topology.{{sfn|Wilansky|2013|p=63}} Moreover, if $C$ is a Hausdorff completion of $H$ then $C\; \backslash times\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}$ is a completion of $X\; \backslash cong\; H\; \backslash times\; \backslash operatorname\{cl\}\_X\; \backslash \{0\backslash \}.${{sfn|Schaefer|Wolff|1999|pp=12-35}}

Compact and totally bounded sets

A subset of a TVS is compact if and only if it is complete and totally bounded.{{sfn|Narici|Beckenstein|2011|pp=47-66}} Thus, in a complete topological vector space, a closed and totally bounded subset is compact.{{sfn|Narici|Beckenstein|2011|pp=47-66}}

A subset $S$ of a TVS $X$ is totally bounded if and only if $\backslash operatorname\{cl\}\_X\; S$ is totally bounded,{{sfn|Schaefer|Wolff|1999|p=25}}{{sfn|Jarchow|1981|pp=56-73}} if and only if its image under the canonical quotient map $$X\; \backslash to\; X\; /\; \backslash operatorname\{cl\}\_X\; (\backslash \{0\backslash \})$$ is totally bounded.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

Every relatively compact set is totally bounded{{sfn|Narici|Beckenstein|2011|pp=47-66}} and the closure of a totally bounded set is totally bounded.{{sfn|Narici|Beckenstein|2011|pp=47-66}}

The image of a totally bounded set under a uniformly continuous map (such as a continuous linear map for instance) is totally bounded.{{sfn|Narici|Beckenstein|2011|pp=47-66}}

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.{{sfn|Schaefer|Wolff|1999|pp=12-35}}

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 $K\; +\; N\; \backslash subseteq\; U.${{sfn|Narici|Beckenstein|2011|pp=19-45}}

Closure and closed set

The closure of any convex (respectively, any balanced, any absorbing) subset of any TVS has this same property. In particular, the closure of any convex, balanced, and absorbing subset is a barrel.

The closure of a vector subspace of a TVS is a vector subspace. 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.{{sfn|Narici|Beckenstein|2011|pp=67-113}}

If $M$ is a vector subspace of $X$ and $N$ is a closed neighborhood of the origin in $X$ such that $U\; \backslash cap\; N$ is closed in $X$ then $M$ is closed in $X.${{sfn|Narici|Beckenstein|2011|pp=19-45}}

The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed{{sfn|Narici|Beckenstein|2011|pp=67-113}} (see this footnoteIn $\backslash R^2,$ the sum of the $y$-axis and the graph of $y\; =\; \backslash frac\{1\}\{x\},$ which is the complement of the $y$-axis, is open in $\backslash R^2.$ In $\backslash R,$ the Minkowski sum $\backslash Z\; +\; \backslash sqrt\{2\}\backslash Z$ is a countable dense subset of $\backslash R$ so not closed in $\backslash R.$ for examples).

If $S\; \backslash subseteq\; X$ and $a$ is a scalar then $$a\; \backslash operatorname\{cl\}\_X\; S\; \backslash subseteq\; \backslash operatorname\{cl\}\_X\; (a\; S),$$ where if $X$ is Hausdorff, $a\; \backslash neq\; 0,\; \backslash text\{\; or\; \}\; S\; =\; \backslash varnothing$ then equality holds: $\backslash operatorname\{cl\}\_X\; (a\; S)\; =\; a\; \backslash operatorname\{cl\}\_X\; S.$ In particular, every non-zero scalar multiple of a closed set is closed.

If $S\; \backslash subseteq\; X$ and if $A$ is a set of scalars such that neither $\backslash operatorname\{cl\}\; S\; \backslash text\{\; nor\; \}\; \backslash operatorname\{cl\}\; A$ contain zero then{{sfn|Wilansky|2013|pp=43-44}} $\backslash left(\backslash operatorname\{cl\}\; A\backslash right)\; \backslash left(\backslash operatorname\{cl\}\_X\; S\backslash right)\; =\; \backslash operatorname\{cl\}\_X\; (A\; S).$

If $S\; \backslash subseteq\; X\; \backslash text\{\; and\; \}\; S\; +\; S\; \backslash subseteq\; 2\; \backslash operatorname\{cl\}\_X\; S$ then $\backslash operatorname\{cl\}\_X\; S$ is convex.{{sfn|Wilansky|2013|pp=43-44}}

If $R,\; S\; \backslash subseteq\; X$ then{{sfn|Narici|Beckenstein|2011|pp=67-113}} $$\backslash operatorname\{cl\}\_X\; (R)\; +\; \backslash operatorname\{cl\}\_X\; (S)\; ~\backslash subseteq~\; \backslash operatorname\{cl\}\_X\; (R\; +\; S)~\; \backslash text\{\; and\; \}\; ~\backslash operatorname\{cl\}\_X\; \backslash left[\; \backslash operatorname\{cl\}\_X\; (R)\; +\; \backslash operatorname\{cl\}\_X\; (S)\; \backslash right]\; ~=~\; \backslash operatorname\{cl\}\_X\; (R\; +\; S)$$ and so consequently, if $R\; +\; S$ is closed then so is $\backslash operatorname\{cl\}\_X\; (R)\; +\; \backslash operatorname\{cl\}\_X\; (S).${{sfn|Wilansky|2013|pp=43-44}}

If $X$ is a real TVS and $S\; \backslash subseteq\; X,$ then $$\backslash bigcap\_\{r\; >\; 1\}\; r\; S\; \backslash subseteq\; \backslash operatorname\{cl\}\_X\; S$$ where the left hand side is independent of the topology on $X;$ moreover, if $S$ is a convex neighborhood of the origin then equality holds.

For any subset $S\; \backslash subseteq\; X,$ $$\backslash operatorname\{cl\}\_X\; S\; ~=~\; \backslash bigcap\_\{N\; \backslash in\; \backslash mathcal\{N\}\}\; (S\; +\; N)$$ where $\backslash mathcal\{N\}$ is any neighborhood basis at the origin for $X.${{sfn|Narici|Beckenstein|2011|pp=80}}

However, $$\backslash operatorname\{cl\}\_X\; U\; ~\backslash supseteq~\; \backslash bigcap\; \backslash \{U\; :\; S\; \backslash subseteq\; U,\; U\; \backslash text\{\; is\; open\; in\; \}\; X\backslash \}$$ and it is possible for this containment to be proper{{sfn|Narici|Beckenstein|2011|pp=108-109}} (for example, if $X\; =\; \backslash R$ and $S$ is the rational numbers). It follows that $\backslash operatorname\{cl\}\_X\; U\; \backslash subseteq\; U\; +\; U$ for every neighborhood $U$ of the origin in $X.${{sfn|Jarchow|1981|pp=30-32}}

Closed hulls

In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.{{sfn|Narici|Beckenstein|2011|pp=155-176}}

• The closed convex hull of a set is equal to the closure of the convex hull of that set; that is, equal to $\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; S).${{sfn|Narici|Beckenstein|2011|pp=67-113}}
• The closed balanced hull of a set is equal to the closure of the balanced hull of that set; that is, equal to $\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{bal\}\; S).${{sfn|Narici|Beckenstein|2011|pp=67-113}}
• The closed disked hull of a set is equal to the closure of the disked hull of that set; that is, equal to $\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{cobal\}\; S).${{sfn|Narici|Beckenstein|2011|p=109}}

If $R,\; S\; \backslash subseteq\; X$ and the closed convex hull of one of the sets $S$ or $R$ is compact then{{sfn|Narici|Beckenstein|2011|p=109}} $$\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; (R\; +\; S))\; ~=~\; \backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; R)\; +\; \backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; S).$$

If $R,\; S\; \backslash subseteq\; X$ each have a closed convex hull that is compact (that is, $\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; R)$ and $\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; S)$ are compact) then{{sfn|Narici|Beckenstein|2011|p=109}} $$\backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; (R\; \backslash cup\; S))\; ~=~\; \backslash operatorname\{co\}\; \backslash left[\; \backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; R)\; \backslash cup\; \backslash operatorname\{cl\}\_X\; (\backslash operatorname\{co\}\; S)\; \backslash right].$$

Hulls and compactness

In a general TVS, the closed convex hull of a compact set may {{em|fail}} to be compact.

The balanced hull of a compact (respectively, totally bounded) set has that same property.{{sfn|Narici|Beckenstein|2011|pp=67-113}}

The convex hull of a finite union of compact {{em|convex}} sets is again compact and convex.{{sfn|Narici|Beckenstein|2011|pp=67-113}}

Meager, nowhere dense, and Baire

A disk in a TVS is not nowhere dense if and only if its closure is a neighborhood of the origin.{{sfn|Narici|Beckenstein|2011|pp=371-423}}

A vector subspace of a TVS that is closed but not open is nowhere dense.{{sfn|Narici|Beckenstein|2011|pp=371-423}}

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.{{sfn|Narici|Beckenstein|2011|pp=371-423}}

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 $X\; =\; \backslash bigcup\_\{n\; \backslash in\; \backslash N\}\; n\; D.${{sfn|Narici|Beckenstein|2011|pp=371-423}}

Every nonmeager locally convex TVS is a barrelled space.{{sfn|Narici|Beckenstein|2011|pp=371-423}}

Important algebraic facts and common misconceptions

If $S\; \backslash subseteq\; X$ then $2\; S\; \backslash subseteq\; S\; +\; S$; if $S$ is convex then equality holds. For an example where equality does {{em|not}} hold, let $x$ be non-zero and set $S\; =\; \backslash \{-\; x,\; x\backslash \};$ $S\; =\; \backslash \{x,\; 2\; x\backslash \}$ also works.

A subset $C$ is convex if and only if $(s\; +\; t)\; C\; =\; s\; C\; +\; t\; C$ for all positive real $s\; >\; 0\; \backslash text\{\; and\; \}\; t\; >\; 0,${{sfn|Rudin|1991|p=38}} or equivalently, if and only if $t\; C\; +\; (1\; -\; t)\; C\; \backslash subseteq\; C$ for all $0\; \backslash leq\; t\; \backslash leq\; 1.${{sfn|Rudin|1991|p=6}}

The convex balanced hull of a set $S\; \backslash subseteq\; X$ is equal to the convex hull of the balanced hull of $S;$ that is, it is equal to $\backslash operatorname\{co\}\; (\backslash operatorname\{bal\}\; S).$ But in general, $$\backslash operatorname\{bal\}\; (\backslash operatorname\{co\}\; S)\; ~\backslash subseteq~\; \backslash operatorname\{cobal\}\; S\; ~=~\; \backslash operatorname\{co\}\; (\backslash operatorname\{bal\}\; S),$$ where the inclusion might be strict since the balanced hull of a convex set need not be convex (counter-examples exist even in $\backslash R^2$).

If $R,\; S\; \backslash subseteq\; X$ and $a$ is a scalar then{{sfn|Narici|Beckenstein|2011|pp=67-113}} $$a(R\; +\; S)\; =\; aR\; +\; a\; S,~\; \backslash text\{\; and\; \}\; ~\backslash operatorname\{co\}\; (R\; +\; S)\; =\; \backslash operatorname\{co\}\; R\; +\; \backslash operatorname\{co\}\; S,~\; \backslash text\{\; and\; \}\; ~\backslash operatorname\{co\}\; (a\; S)\; =\; a\; \backslash operatorname\{co\}\; S.$$

If $R,\; S\; \backslash subseteq\; X$ are convex non-empty disjoint sets and $x\; \backslash not\backslash in\; R\; \backslash cup\; S,$ then $S\; \backslash cap\; \backslash operatorname\{co\}\; (R\; \backslash cup\; \backslash \{x\backslash \})\; =\; \backslash varnothing$ or $R\; \backslash cap\; \backslash operatorname\{co\}\; (S\; \backslash cup\; \backslash \{x\backslash \})\; =\; \backslash varnothing.$

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 {{em|family}} of F-seminorms.{{sfn|Swartz|1992|p=35}}

If $P(x)$ is some unary predicate (a true or false statement dependent on $x\; \backslash in\; X$) then for any $z\; \backslash in\; X,$ $z\; +\; \backslash \{x\; \backslash in\; X\; :\; P(x)\backslash \}\; =\; \backslash \{x\; \backslash in\; X\; :\; P(x\; -\; z)\backslash \}.$$$z\; +\; \backslash \{x\; \backslash in\; X\; :\; P(x)\backslash \}\; =\; \backslash \{z\; +\; x\; :\; x\; \backslash in\; X,\; P(x)\backslash \}\; =\; \backslash \{z\; +\; x\; :\; x\; \backslash in\; X,\; P((z\; +\; x)\; -\; z)\backslash \}$$ and so using $y\; =\; z\; +\; x$ and the fact that $z\; +\; X\; =\; X,$ this is equal to $$\backslash \{y\; :\; y\; -\; z\; \backslash in\; X,\; P(y\; -\; z)\backslash \}\; =\; \backslash \{y\; :\; y\; \backslash in\; X,\; P(y\; -\; z)\backslash \}\; =\; \backslash \{y\; \backslash in\; X\; :\; P(y\; -\; z)\backslash \}.$$ Q.E.D. $\backslash blacksquare$

So for example, if $P(x)$ denotes "" then for any $z\; \backslash in\; X,$ Similarly, if $s\; \backslash neq\; 0$ is a scalar then $s\; \backslash \{x\; \backslash in\; X\; :\; P(x)\backslash \}\; =\; \backslash left\backslash \{x\; \backslash in\; X\; :\; P\backslash left(\backslash tfrac\{1\}\{s\}\; x\backslash right)\backslash right\backslash \}.$ The elements $x\; \backslash in\; X$ of these sets must range over a vector space (that is, over $X$) rather than not just a subset or else these equalities are no longer guaranteed; similarly, $z$ must belong to this vector space (that is, $z\; \backslash in\; X$).

• The balanced hull of a compact (respectively, totally bounded, open) set has that same property.{{sfn|Narici|Beckenstein|2011|pp=67-113}}
• The (Minkowski) sum of two compact (respectively, bounded, balanced, convex) sets has that same property.{{sfn|Narici|Beckenstein|2011|pp=67-113}} But the sum of two closed sets need {{em|not}} be closed.
• The convex hull of a balanced (resp. open) set is balanced (respectively, open). However, the convex hull of a closed set need {{em|not}} be closed.{{sfn|Narici|Beckenstein|2011|pp=67-113}} And the convex hull of a bounded set need {{em|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, "convex" for instance) is preserved under the set operator (indicated by the row's name, "closure" for instance). 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 "$R\; \backslash cup\; S$" 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.

• {{annotated link|Banach space}}
• {{annotated link|Complete field}}
• {{annotated link|Hilbert space}}
• {{annotated link|Liquid vector space}}
• {{annotated link|Normed space}}
• {{annotated link|Locally compact field}}
• {{annotated link|Locally compact group}}
• {{annotated link|Locally compact quantum group}}
• {{annotated link|Locally convex topological vector space}}
• {{annotated link|Ordered topological vector space}}
• {{annotated link|Topological abelian group}}
• {{annotated link|Topological field}}
• {{annotated link|Topological group}}
• {{annotated link|Topological module}}
• {{annotated link|Topological ring}}
• {{annotated link|Topological semigroup}}
• {{annotated link|Topological vector lattice}}
• {{annotated link|Measure theory in topological vector spaces}}

{{reflist|group=note}}

{{reflist|group=proof}}

{{reflist}}

{{refbegin|2}}

• {{Adasch Topological Vector Spaces|edition=2}}
• {{Jarchow Locally Convex Spaces}}
• {{Köthe Topological Vector Spaces I}}
• {{Narici Beckenstein Topological Vector Spaces|edition=2}}
• {{Rudin Walter Functional Analysis|edition=2}}
• {{Schaefer Wolff Topological Vector Spaces|edition=2}}
• {{Schechter Handbook of Analysis and Its Foundations}}
• {{Swartz An Introduction to Functional Analysis}}
• {{Wilansky Modern Methods in Topological Vector Spaces|edition=1}}

{{refend}}

• {{Bierstedt An Introduction to Locally Convex Inductive Limits}}
• {{Bourbaki Topological Vector Spaces Part 1 Chapters 1–5}}
• {{Conway A Course in Functional Analysis|edition=2}}
• {{Dunford Schwartz Linear Operators Part 1 General Theory}}
• {{Edwards Functional Analysis Theory and Applications}}
• {{Grothendieck Topological Vector Spaces}}
• {{Horváth Topological Vector Spaces and Distributions Volume 1 1966}}
• {{Köthe Topological Vector Spaces II}}
• {{cite book|last=Lang|first=Serge|author-link=Serge Lang|title=Differential manifolds|publisher=Addison-Wesley Publishing Co., Inc.|location=Reading, Mass.–London–Don Mills, Ont.|year=1972|isbn=0-201-04166-9}}
• {{Robertson Topological Vector Spaces}}
• {{Trèves François Topological vector spaces, distributions and kernels}}
• {{Valdivia Topics in Locally Convex Spaces|edition=1}}
• {{Voigt A Course on Topological Vector Spaces|edition=1}}

• {{Commons category-inline|Topological vector spaces}}

{{Functional Analysis}}

{{TopologicalVectorSpaces}}

{{Authority control}}

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