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

Defining topologies using neighborhoods of the origin

Since every vector topology is translation invariant (which means that for all ${\displaystyle x_{0}\in X,}$ the map ${\displaystyle X\to X}$ defined by ${\displaystyle x\mapsto x_{0}+x}$ is a homeomorphism), to define a vector topology it suffices to define a neighborhood basis (or subbasis) for it at the origin.

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.^[5]

Defining topologies using strings

Let ${\displaystyle X}$ be a vector space and let ${\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=1}^{\infty }}$ be a sequence of subsets of ${\displaystyle X.}$ Each set in the sequence ${\displaystyle U_{\bullet }}$ is called a knot of ${\displaystyle U_{\bullet }}$ and for every index ${\displaystyle i,}$ ${\displaystyle U_{i}}$ is called the ${\displaystyle i}$-th knot of ${\displaystyle U_{\bullet }.}$ The set ${\displaystyle U_{1}}$ is called the beginning of ${\displaystyle U_{\bullet }.}$ The sequence ${\displaystyle U_{\bullet }}$ is/is a:^[7][8][9]

• Summative if ${\displaystyle U_{i+1}+U_{i+1}\subseteq U_{i}}$ for every index ${\displaystyle i.}$
• Balanced (resp. absorbing, closed,^{[note 1]} convex, open, symmetric, barrelled, absolutely convex/disked, etc.) if this is true of every ${\displaystyle U_{i}.}$
• String if ${\displaystyle U_{\bullet }}$ is summative, absorbing, and balanced.
• Topological string or a neighborhood string in a TVS ${\displaystyle X}$ if ${\displaystyle U_{\bullet }}$ is a string and each of its knots is a neighborhood of the origin in ${\displaystyle X.}$

If ${\displaystyle U}$is an absorbing disk in a vector space ${\displaystyle X}$ then the sequence defined by ${\displaystyle U_{i}:=2^{1-i}U}$ forms a string beginning with ${\displaystyle U_{1}=U.}$ This is called the natural string of ${\displaystyle U}$^[7] Moreover, if a vector space ${\displaystyle X}$ has countable dimension then every string contains an absolutely convex string.

Summative sequences of sets have the particularly nice property that they define non-negative continuous real-valued subadditive functions. These functions can then be used to prove many of the basic properties of topological vector spaces.

Theorem (${\displaystyle \mathbb {R} }$-valued function induced by a string) — Let ${\displaystyle U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty }}$ be a collection of subsets of a vector space such that ${\displaystyle 0\in U_{i}}$ and ${\displaystyle U_{i+1}+U_{i+1}\subseteq U_{i}}$ for all ${\displaystyle i\geq 0.}$ For all ${\displaystyle u\in U_{0},}$ let

$${\displaystyle \mathbb {S} (u):=\left\{n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)~:~k\geq 1,n_{i}\geq 0{\text{ for all }}i,{\text{ and }}u\in U_{n_{1}}+\cdots +U_{n_{k}}\right\}.}$$

Define ${\displaystyle f:X\to [0,1]}$ by ${\displaystyle f(x)=1}$ if ${\displaystyle x\not \in U_{0}}$ and otherwise let

$${\displaystyle f(x):=\inf _{}\left\{2^{-n_{1}}+\cdots 2^{-n_{k}}~:~n_{\bullet }=\left(n_{1},\ldots ,n_{k}\right)\in \mathbb {S} (x)\right\}.}$$

Then ${\displaystyle f}$ is subadditive (meaning ${\displaystyle f(x+y)\leq f(x)+f(y)}$ for all ${\displaystyle x,y\in X}$) and ${\displaystyle f=0}$ on ${\textstyle \bigcap _{i\geq 0}U_{i};}$ so in particular, ${\displaystyle f(0)=0.}$ If all ${\displaystyle U_{i}}$ are symmetric sets then ${\displaystyle f(-x)=f(x)}$ and if all ${\displaystyle U_{i}}$ are balanced then ${\displaystyle f(sx)\leq f(x)}$ for all scalars ${\displaystyle s}$ such that ${\displaystyle |s|\leq 1}$ and all ${\displaystyle x\in X.}$ If ${\displaystyle X}$ is a topological vector space and if all ${\displaystyle U_{i}}$ are neighborhoods of the origin then ${\displaystyle f}$ is continuous, where if in addition ${\displaystyle X}$ is Hausdorff and ${\displaystyle U_{\bullet }}$ forms a basis of balanced neighborhoods of the origin in ${\displaystyle X}$ then ${\displaystyle d(x,y):=f(x-y)}$ is a metric defining the vector topology on ${\displaystyle X.}$

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

If ${\displaystyle U_{\bullet }=\left(U_{i}\right)_{i\in \mathbb {N} }}$ and ${\displaystyle V_{\bullet }=\left(V_{i}\right)_{i\in \mathbb {N} }}$ are two collections of subsets of a vector space ${\displaystyle X}$ and if ${\displaystyle s}$ is a scalar, then by definition:^[7]

• ${\displaystyle V_{\bullet }}$ contains ${\displaystyle U_{\bullet }}$: ${\displaystyle \ U_{\bullet }\subseteq V_{\bullet }}$ if and only if ${\displaystyle U_{i}\subseteq V_{i}}$ for every index ${\displaystyle i.}$
• Set of knots: ${\displaystyle \ \operatorname {Knots} U_{\bullet }:=\left\{U_{i}:i\in \mathbb {N} \right\}.}$
• Kernel: ${\textstyle \ \ker U_{\bullet }:=\bigcap _{i\in \mathbb {N} }U_{i}.}$
• Scalar multiple: ${\displaystyle \ sU_{\bullet }:=\left(sU_{i}\right)_{i\in \mathbb {N} }.}$
• Sum: ${\displaystyle \ U_{\bullet }+V_{\bullet }:=\left(U_{i}+V_{i}\right)_{i\in \mathbb {N} }.}$
• Intersection: ${\displaystyle \ U_{\bullet }\cap V_{\bullet }:=\left(U_{i}\cap V_{i}\right)_{i\in \mathbb {N} }.}$

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

Notation: Let ${\textstyle \operatorname {Knots} \mathbb {S}$ :=\bigcup _{U_{\bullet }\in \mathbb {S} }\operatorname {Knots} U_{\bullet }} be the set of all knots of all strings in ${\displaystyle \mathbb {S} .}$

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

If ${\displaystyle \mathbb {S} }$ is the set of all topological strings in a TVS ${\displaystyle (X,\tau )}$ then ${\displaystyle \tau _{\mathbb {S} }=\tau .}$^[7] A Hausdorff TVS is metrizable if and only if its topology can be induced by a single topological string.^[10]

Invariance of vector topologies

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

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

If ${\displaystyle x\in X}$ and any subset ${\displaystyle S\subseteq X,}$ then ${\displaystyle \operatorname {cl} _{X}(x+S)=x+\operatorname {cl} _{X}S}$^[6] and moreover, if ${\displaystyle 0\in S}$ then ${\displaystyle x+S}$ is a neighborhood (resp. open neighborhood, closed neighborhood) of ${\displaystyle x}$ in ${\displaystyle X}$ if and only if the same is true of ${\displaystyle S}$ at the origin.

Local notions

A subset ${\displaystyle E}$ of a vector space ${\displaystyle X}$ is said to be

• absorbing (in ${\displaystyle X}$): if for every ${\displaystyle x\in X,}$ there exists a real ${\displaystyle r>0}$ such that ${\displaystyle cx\in E}$ for any scalar ${\displaystyle c}$ satisfying ${\displaystyle |c|\leq r.}$^[12]
• balanced or circled: if ${\displaystyle tE\subseteq E}$ for every scalar ${\displaystyle |t|\leq 1.}$^[12]
• convex: if ${\displaystyle tE+(1-t)E\subseteq E}$ for every real ${\displaystyle 0\leq t\leq 1.}$^[12]
• a disk or absolutely convex: if ${\displaystyle E}$ is convex and balanced.
• symmetric: if ${\displaystyle -E\subseteq E,}$ or equivalently, if ${\displaystyle -E=E.}$

Every neighborhood of the origin is an absorbing set and contains an open balanced neighborhood of ${\displaystyle 0}$^[6] 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 ${\displaystyle 0;}$ 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 ${\displaystyle E}$ of a topological vector space ${\displaystyle X}$ is bounded^[13] if for every neighborhood ${\displaystyle V}$ of the origin there exists ${\displaystyle t}$ such that ${\displaystyle E\subseteq tV}$.

The definition of boundedness can be weakened a bit; ${\displaystyle E}$ 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.^[14] Also, ${\displaystyle E}$ is bounded if and only if for every balanced neighborhood ${\displaystyle V}$ of the origin, there exists ${\displaystyle t}$ such that ${\displaystyle E\subseteq tV.}$ Moreover, when ${\displaystyle X}$ is locally convex, the boundedness can be characterized by seminorms: the subset ${\displaystyle E}$ is bounded if and only if every continuous seminorm ${\displaystyle p}$ is bounded on ${\displaystyle E.}$^[15]

Every totally bounded set is bounded.^[14] If ${\displaystyle M}$ is a vector subspace of a TVS ${\displaystyle X,}$ then a subset of ${\displaystyle M}$ is bounded in ${\displaystyle M}$ if and only if it is bounded in ${\displaystyle X.}$^[14]

Metrizability

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.^[17]

Let ${\displaystyle \mathbb {K} }$ be a non-discrete locally compact topological field, for example the real or complex numbers. A Hausdorff topological vector space over ${\displaystyle \mathbb {K} }$ is locally compact if and only if it is finite-dimensional, that is, isomorphic to ${\displaystyle \mathbb {K} ^{n}}$ for some natural number ${\displaystyle n.}$^[18]

Completeness and uniform structure

The canonical uniformity^[19] on a TVS ${\displaystyle (X,\tau )}$ is the unique translation-invariant uniformity that induces the topology ${\displaystyle \tau }$ on ${\displaystyle X.}$

Every TVS is assumed to be endowed with this canonical uniformity, which makes all TVSs into uniform spaces. This allows one to talk 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.^[20] 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) ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ is Cauchy if and only if for every neighborhood ${\displaystyle V}$ of ${\displaystyle 0,}$ there exists some index ${\displaystyle n}$ such that ${\displaystyle x_{i}-x_{j}\in V}$ whenever ${\displaystyle i\geq n}$ and ${\displaystyle j\geq n.}$

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.^[6] 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.^[21] A complete subset of a Hausdorff TVS is closed.^[21]
• If ${\displaystyle C}$ is a complete subset of a TVS then any subset of ${\displaystyle C}$ that is closed in ${\displaystyle C}$ is complete.^[21]
• A Cauchy sequence in a Hausdorff TVS ${\displaystyle X}$ is not necessarily relatively compact (that is, its closure in ${\displaystyle X}$ is not necessarily compact).
• If a Cauchy filter in a TVS has an accumulation point ${\displaystyle x}$ then it converges to ${\displaystyle x.}$
• If a series ${\textstyle \sum _{i=1}^{\infty }x_{i}}$ converges^{[note 5]} in a TVS ${\displaystyle X}$ then ${\displaystyle x_{\bullet }\to 0}$ in ${\displaystyle X.}$^[22]

Finest and coarsest vector topology

Let ${\displaystyle X}$ be a real or complex vector space.

Trivial topology

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

Finest vector topology

There exists a TVS topology ${\displaystyle \tau _{f}}$ on ${\displaystyle X,}$ called the finest vector topology on ${\displaystyle X,}$ that is finer than every other TVS-topology on ${\displaystyle X}$ (that is, any TVS-topology on ${\displaystyle X}$ is necessarily a subset of ${\displaystyle \tau _{f}}$).^[23][24] Every linear map from ${\displaystyle \left(X,\tau _{f}\right)}$ into another TVS is necessarily continuous. If ${\displaystyle X}$ has an uncountable Hamel basis then ${\displaystyle \tau _{f}}$ is not locally convex and not metrizable.^[24]

Cartesian products

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 ${\displaystyle X}$ of all functions ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ where ${\displaystyle \mathbb {R} }$ carries its usual Euclidean topology. This set ${\displaystyle X}$ 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 ${\displaystyle \mathbb {R} ^{\mathbb {R} },,}$ which carries the natural product topology. With this product topology, ${\displaystyle X:=\mathbb {R} ^{\mathbb {R} }}$ becomes a topological vector space whose topology is called the topology of pointwise convergence on ${\displaystyle \mathbb {R} .}$ The reason for this name is the following: if ${\displaystyle \left(f_{n}\right)_{n=1}^{\infty }}$ is a sequence (or more generally, a net) of elements in ${\displaystyle X}$ and if ${\displaystyle f\in X}$ then ${\displaystyle f_{n}}$ converges to ${\displaystyle f}$ in ${\displaystyle X}$ if and only if for every real number ${\displaystyle x,}$ ${\displaystyle f_{n}(x)}$ converges to ${\displaystyle f(x)}$ in ${\displaystyle \mathbb {R} .}$ 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 ${\displaystyle \mathbb {R} f:=\{rf:r\in \mathbb {R} \}}$ with ${\displaystyle f\neq 0}$).

Finite-dimensional spaces

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

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

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

Non-vector topologies

Discrete and cofinite topologies

If ${\displaystyle X}$ is a non-trivial vector space (that is, of non-zero dimension) then the discrete topology on ${\displaystyle X}$ (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 ${\displaystyle X}$ (where a subset is open if and only if its complement is finite) is also not a TVS topology on ${\displaystyle X.}$

Neighborhoods and open sets

Properties of neighborhoods and open sets

Every TVS is connected^[6] and locally connected^[30] and any connected open subset of a TVS is arcwise connected. If ${\displaystyle S\subseteq X}$ and ${\displaystyle U}$ is an open subset of ${\displaystyle X}$ then ${\displaystyle S+U}$ is an open set in ${\displaystyle X}$^[6] and if ${\displaystyle S\subseteq X}$ has non-empty interior then ${\displaystyle S-S}$ is a neighborhood of the origin.^[6]

The open convex subsets of a TVS ${\displaystyle X}$ (not necessarily Hausdorff or locally convex) are exactly those that are of the form

$${\displaystyle z+\{x\in X:p(x)<1\}~=~\{x\in X:p(x-z)<1\}}$$

for some ${\displaystyle z\in X}$ and some positive continuous sublinear functional ${\displaystyle p}$ on ${\displaystyle X.}$^[28] If ${\displaystyle K}$ is an absorbing disk in a TVS ${\displaystyle X}$ and if ${\displaystyle p:=p_{K}}$ is the Minkowski functional of ${\displaystyle K}$ then^[31]

$${\displaystyle \operatorname {Int} _{X}K~\subseteq ~\{x\in X:p(x)<1\}~\subseteq ~K~\subseteq ~\{x\in X:p(x)\leq 1\}~\subseteq ~\operatorname {cl} _{X}K}$$

where importantly, it was not assumed that ${\displaystyle K}$ had any topological properties nor that ${\displaystyle p}$ was continuous (which happens if and only if ${\displaystyle K}$ is a neighborhood of the origin).

Let ${\displaystyle \tau }$ and ${\displaystyle \nu }$ be two vector topologies on ${\displaystyle X.}$ Then ${\displaystyle \tau \subseteq \nu }$ if and only if whenever a net ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ in ${\displaystyle X}$ converges ${\displaystyle 0}$ in ${\displaystyle (X,\nu )}$ then ${\displaystyle x_{\bullet }\to 0}$ in ${\displaystyle (X,\tau ).}$^[32]

Let ${\displaystyle {\mathcal {N}}}$ be a neighborhood basis of the origin in ${\displaystyle X,}$ let ${\displaystyle S\subseteq X,}$ and let ${\displaystyle x\in X.}$ Then ${\displaystyle x\in \operatorname {cl} _{X}S}$ if and only if there exists a net ${\displaystyle s_{\bullet }=\left(s_{N}\right)_{N\in {\mathcal {N}}}}$ in ${\displaystyle S}$ (indexed by ${\displaystyle {\mathcal {N}}}$) such that ${\displaystyle s_{\bullet }\to x}$ in ${\displaystyle X.}$^[33] 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 ${\displaystyle X}$ is a TVS that is of the second category in itself (that is, a nonmeager space) then any closed convex absorbing subset of ${\displaystyle X}$ is a neighborhood of the origin.^[34] This is no longer guaranteed if the set is not convex (a counter-example exists even in ${\displaystyle X=\mathbb {R} ^{2}}$) or if ${\displaystyle X}$ is not of the second category in itself.^[34]

Interior

If ${\displaystyle R,S\subseteq X}$ and ${\displaystyle S}$ has non-empty interior then

$${\displaystyle \operatorname {Int} _{X}S~=~\operatorname {Int} _{X}\left(\operatorname {cl} _{X}S\right)~{\text{ and }}~\operatorname {cl} _{X}S~=~\operatorname {cl} _{X}\left(\operatorname {Int} _{X}S\right)}$$

and

$${\displaystyle \operatorname {Int} _{X}(R)+\operatorname {Int} _{X}(S)~\subseteq ~R+\operatorname {Int} _{X}S\subseteq \operatorname {Int} _{X}(R+S).}$$

The topological interior of a disk is not empty if and only if this interior contains the origin.^[35] More generally, if ${\displaystyle S}$ is a balanced set with non-empty interior ${\displaystyle \operatorname {Int} _{X}S\neq \varnothing }$ in a TVS ${\displaystyle X}$ then ${\displaystyle \{0\}\cup \operatorname {Int} _{X}S}$ will necessarily be balanced;^[6] consequently, ${\displaystyle \operatorname {Int} _{X}S}$ will be balanced if and only if it contains the origin.^{[proof 2]} For this (i.e. ${\displaystyle 0\in \operatorname {Int} _{X}S}$) to be true, it suffices for ${\displaystyle S}$ to also be convex (in addition to being balanced and having non-empty interior).;^[6] The conclusion ${\displaystyle 0\in \operatorname {Int} _{X}S}$ could be false if ${\displaystyle S}$ is not also convex;^[35] for example, in ${\displaystyle X:=\mathbb {R} ^{2},}$ the interior of the closed and balanced set ${\displaystyle S:=\{(x,y):xy\geq 0\}}$ is ${\displaystyle \{(x,y):xy>0\}.}$

If ${\displaystyle C}$ is convex and ${\displaystyle 0<t\leq 1,}$ then^[36] ${\displaystyle t\operatorname {Int} C+(1-t)\operatorname {cl} C~\subseteq ~\operatorname {Int} C.}$ Explicitly, this means that if ${\displaystyle C}$ is a convex subset of a TVS ${\displaystyle X}$ (not necessarily Hausdorff or locally convex), ${\displaystyle y\in \operatorname {int} _{X}C,}$ and ${\displaystyle x\in \operatorname {cl} _{X}C}$ then the open line segment joining ${\displaystyle x}$ and ${\displaystyle y}$ belongs to the interior of ${\displaystyle C;}$ that is, ${\displaystyle \{tx+(1-t)y:0<t<1\}\subseteq \operatorname {int} _{X}C.}$^{[37][38][proof 3]}

If ${\displaystyle N\subseteq X}$ is any balanced neighborhood of the origin in ${\displaystyle X}$ then ${\textstyle \operatorname {Int} _{X}N\subseteq B_{1}N=\bigcup _{0<|a|<1}aN\subseteq N}$ where ${\displaystyle B_{1}}$ is the set of all scalars ${\displaystyle a}$ such that ${\displaystyle |a|<1.}$

If ${\displaystyle x}$ belongs to the interior of a convex set ${\displaystyle S\subseteq X}$ and ${\displaystyle y\in \operatorname {cl} _{X}S,}$ then the half-open line segment

$${\displaystyle [x,y):=\{tx+(1-t)y:0<t\leq 1\}\subseteq \operatorname {Int} _{X}{\text{ if }}x\neq y}$$

and^[37]

$${\displaystyle [x,x)=\varnothing {\text{ if }}x=y.}$$

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

Non-Hausdorff spaces and the closure of the origin

A topological vector space ${\displaystyle X}$ is Hausdorff if and only if ${\displaystyle \{0\}}$ is a closed subset of ${\displaystyle X,}$ or equivalently, if and only if ${\displaystyle \{0\}=\operatorname {cl} _{X}\{0\}.}$ Because ${\displaystyle \{0\}}$ is a vector subspace of ${\displaystyle X,}$ the same is true of its closure ${\displaystyle \operatorname {cl} _{X}\{0\},}$ which is referred to as the closure of the origin in ${\displaystyle X.}$ This vector space satisfies