Pengguna:Dedhert.Jr/Ruang vektor topologis

Dalam matematika, sebuah ruang vektor topologis (atau disebut ruang topologis linear dan biasanya disingkat TVS atau t.v.s.) merupakan salah satu dari struktur dasar yang diinvestigasi dalam analisis fungsional. Sebuah ruang vektor topologis merupakan sebuah ruang vektor (sebuah struktur aljabar) yang juga merupakan sebuah ruang topologis, ini menyiratkan bahwa operasi ruang vektor menjadi fungsi kontinu. Lebih spesifiknya, ruang topologisnya memiliki sebuah struktur topologis seragam, memungkinkan sebuah gagasan kekonvergenan seragam.

Unsur ruang vektor topologisnya biasanya operator fungsi atau linear bertindak pada ruang vektor topologis, dan topologinya seringkali didefinisikan sehingga menangkap sebuah gagasan khusus mengenai kekonvergenan barisan fungsi.

Ruang Banach, ruang Hilbert, dan ruang Sobolev merupakan contoh yang terkenal.

Kecuali dinyatakan sebaliknya, medan pendasar ruang vektor topologis diasumsikan menjadi baik bilangan kompleks $\mathbb {C}$ atau bilangan real $\mathbb {R}$.

Motivasi

Ruang bernorma

Setiap ruang vektor bernorma memiliki sebuah struktur topologis alami: normanya mengindikasi sebuah metrik dan metriknya mengindukasikan sebuah topologi. Ini adlaha sebuah ruang vektor topologis karena:

1. Penambahan vektor $+:X\times X\to X$ sama-sama kontinu terhadap topologi ini. Ini mengikuti secara langsung dari pertidaksamaan segitiga dipatuhi oleh normanya.
2. Perkalian skalar $\cdot :\mathbb {K} \times X\to X$, dimana $\mathbb {K}$ merupakan medan skalar pendasar $X$, sama-samna kontinu. Ini mengikuti dari pertidaksamaan segitiga dan kehomogenan dari norma.

Demikian semua ruang Banach dan ruang Hilbert merupakan contoh-contoh ruang vektor topologis.

Ruang takbernorma

Terdapat ruang vektor topologis yang topologi tidak disebabkan oleh sebuah norma, tetapi tetap menarik dalam analisis. Contohnya seperti ruang adalah ruang fungsi holomorfik ranah buka, ruang fungsi terdiferensialkan takhingga, ruang Schwartz, dan ruang fungsi uji dan ruang sebaran padanya. Ini merupakan semua contoh ruang Montel. Sebuah ruang Montel berdimensi takhingga tidak pernah ternormakan. Keberadaan norma untuk ruang vektor topologis yang diberikan dicirikan oleh kriteria kenormalan Kolmogorov.

Sebuah medan topologis merupakan sebuah ruang vektor topologi atas masing-masing submedannya.

Definisi

Sebuah keluarga lingkungan asalnya dengan dua sifat di atas menentukan dengan unik sebuah ruang vektor topologis. Sistem lingkungan suatu titik lainnya dalam ruang vektor diperoleh dengan translasi.

Sebuah ruang vektor topologis (TVS) $X$ merupakan sebuah ruang vektor atas medan topologis $\mathbb {K}$ (hampir seringkali bilangan real atau kompleks dengan topologi standarnya) yang diberkahi dengan sebuah topologis sehingga penambahan vektor $\cdot \,+\,\cdot \;:X\times X\to X$ dan perkalian skalar $\cdot :\mathbb {K} \times X\to X$ adalah fungsi kontinu (dimana ranah fungsi ini diberkahi dengan topologi darab). Seperti sebuah topologi disebut ruang vektor atau sebuah topologi ruang vektor topologis pada $X$.

Setiap ruang vektor topologis juga merupakan sebuah grup topologis komutatif terhadap penambahan.

Asumsi Hausdorff

Beberapa penulis (yaitu, Walter Rudin) memerlukan topologi pada $X$ menjadi T₁, ini kemudian mengikuti bahwa ruangnya adalah Hausdorff, dan bahkan Tychonoff. Sebuah ruang vektor topologis dikatakan menjadi terpisah jika ini merupakan ruang Hausdorff, terpenting, "terpisah" tidak berarti terpisahkan. Struktur aljabar topologis dan linear dapat diikat bersama bahkan lebih erat lagi dengan asumsi tambahan, hal-hal paling umum yang didaftarkan di bawah.

Kategori dan morfisme

Kategori ruang vektor topologis atas diberikan medan topologis $\mathbb {K}$ biasanya dilambangkan $\operatorname {TVS} _{\mathbb {K} }$ atau $\operatorname {TVect} _{\mathbb {K} }$. Objeknya adalah ruang vektor topologis atas $\mathbb {K}$ dan morfismenya adalah peta linear-$\mathbb {K}$ kontinu dari satu objek ke objek lainnya.

Sebuah kehomomorgan ruang vektor topologis atau kehomomorfan topologis merupakan sebuah peta linear kontinu $u:X\to Y$ antara ruang vektor topologis sehingga peta terinduksi $u:X\to \operatorname {Im} u$ merupakan sebuah pemetaan buka ketika $\operatorname {Im} (u):=u(X)$, yang kisaran atau citra dari $u$, diberikan topologi subruang yang disebab oleh $Y$.

Sebuah pembenaman ruang vektor topologis atau monomorfisme topologis merupakan sebuah kehomomorfan topologis injektif. Dengan jelas, sebuah pembenaman ruang vektor topologis merupakan sebuah peta linear yang juga merupakan sebuah pembenaman topologis.

Sebuah keisomorfan ruang vektor topologis atau sebuah keisomorfan dalam kategori ruang vektor topologis merupakan sebuah kehomomorfan linear bijektif. Dengan jelas, ini merupakan sebuah pembenaman ruang vektor topologis surjektif.

Banyak sifat-sifat ruang vektor topologis yang dipelajari, seperti kecembungan lokal, ketermetrikkan, kelengkapan, dan kenormalan, adalah invarian terhadap keisomorfan ruang vektor topologis.

Sebuah syarat perlu untuk sebuah ruang vektor

Sebuah kumpulan ${\mathcal {N}}$ dari himpunan bagian ruang vektor disebut aditif jika untuk setiap $N\in {\mathcal {N}}$, terdapat suatu $U\in {\mathcal {N}}$ sehingga $U+U\subseteq N$.

Pencirian kekontinuan penambahan di $0$ — Jika $(X,+)$ adalah sebuah grup (sebagia semua ruang vektor), $\tau$ adalah sebuah topologi pada $X$, dan $X\times X$ diberikan dengan topologi darab, maka peta penambahan $X\times X\to X$ (didefinisikan oleh $(x,y)\mapsto x+y$) adalah kontinu di asalnya dari $X\times X$ jika dan hanya jika himpunan lingkungan dari asalnya di $(X,\tau )$ adalah aditif. Pernyataan ini tetap benar jika kata "lingkungan" diganti oleh "lingkungan buka."

Semua syarat di atas akibatnya sebuah keperluan untuk sebuah topologi untuk membentuk sebuah ruang vektor.

Mendefinisikan topologi menggunakan lingkungan dari asalnya

Karena setiap ruang vektor adalah invarian translasi (yaitu untuk semua $x_{0}\in X$, peta $X\to X$ didefinisikan oleh $x\mapsto x_{0}+x$ merupakan sebuah homeomorfisme), untuk mendefinisikan sebuah topologi vektor, ini mencukupi untuk mendefinisikan sebuah basis lingkungan (atau subbasis) untuknya di asal tersebut.

Teorema (Tapis lingkungan dari asalnya) — Andaikan bahwa $X$ adalah ruang vektor real atau kompleks. Jika ${\mathcal {B}}$ adalah sebuah kumpulan aditif takkosong dari himpunan bagian berimbang dan penyerap $X$, maka ${\mathcal {B}}$ adalah sebuah basis lingkungan di $0$ untuk sebuah topologi vektor pada $X$. Yakni, asumsinya adalah bahwa ${\mathcal {B}}$ adalah sebuah basis tapis yang memenuhi syarat-syarat berikut:

1. Setiap $B\in {\mathcal {B}}$ adalah berimbang dan penyerap,
2. ${\mathcal {B}}$ adalah aditif: Untuk setiap $B\in {\mathcal {B}}$ terdapat $U\in {\mathcal {B}}$ sehingga $U+U\subseteq B$,

Jika ${\mathcal {B}}$ memenuhi dua syarat di atas tetapi bukan sebuah basis tapis maka ini akan membentuk sebuah subbasis lingkungan di $0$ (daripada sebuah basis lingkungan) untuk sebuah topologi vektor pada $X$.

Umumnya, himpunan semua himpunan bagian berimbang dan penyerap mengenai sebuah ruang vektor tidak memenuhi syarat teorema ini dan tidak membentuk sebuah basis lingkungan di asalnya untuk suatu vektor topologi.

Mendefinisikan topologi menggunakan untai

Misalkan $X$ menjadi sebuah ruang vektor dan misalkan $U_{\bullet }=\left(U_{i}\right)_{i=1}^{\infty }$ menjadi sebuah barisan himpunan bagian $X$. Setiap himpunan dalam barisan $U_{\bullet }$ disebut sebuah buhul dari $U_{\bullet }$ dan untuk setiap indeks $i$, $U_{i}$ disebut buhul ke-$i$ dari $U_{\bullet }$. Himpunan $U_{1}$ disebut awal dari $U_{\bullet }$. Barisan $U_{\bullet }$ adalah:

• Sumatif jika $U_{i+1}+U_{i+1}\subseteq U_{i}$ untuk setiap indeks $i$
• Berimbang (masing-masing, penyerap, tertutup, cembung, buka, simetrik, terlaras, cembung/tercakram mutlak, dst.) jika ini adalah benar untuk setiap $U_{i}$.
• Untai jika $U_{\bullet }$ adalah sumatif, penyerap, dan berimbang.
• Untai topologis atau sebuah untai lingkungan dalam sebuah ruang vektor topologis $X$ jika $U_{\bullet }$ adalah sebuah untai dan setiap buhlnya adalah sebuah lingkungan asalnya di $X$.

Jika $U$ adalah sebuah cakram penyerap dalam sebuah ruang vektor $X$ maka barisannya didefinisikan oleh $U_{i}:=2^{1-i}U$ membentuk sebuah untai diawali dengan $U_{1}=U$. Ini disebut untai alami dari $U$. Selain itu, jika sebuah ruang vektor $X$ memiliki dimensi tercacahkan maka setiap untai berisi sebuah untai cembung mutlak.

Barisan sumatif himpunan khususnya memiliki sifat yang baik bahwa ini mendefinisikan fungsi subaditif bernilai real kontinu taknegatif. Fungsi-fungsi ini dapat kemudian digunakan untuk membuktikan banyak sifat-sifat dasar ruang vektor topologis.

Teorema (Fungsi bernilai-$\mathbb {R}$ disebabkan oleh sebuah untai) — Misalkan $U_{\bullet }=\left(U_{i}\right)_{i=0}^{\infty }$ menjadui sebuah kumpulan himpunan bagian ruang vektor sehingga $0\in U_{i}$ dan $U_{i+1}+U_{i+1}\subseteq U_{i}$ untuk semua $i\geq 0.$ Untuk semua $u\in U_{0},$ misalkan

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

Mendefinisikan $f:X\to [0,1]$ oleh $f(x)=1$ if $x\not \in U_{0}$ dan jika tidak, misalkan

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

Maka $f$ adalah subaditif (berarti $f(x+y)\leq f(x)+f(y)$ untuk semua $x,y\in X$) dan $f=0$ pada $\bigcap _{i\geq 0}U_{i}$, jadi khususnya $f(0)=0$. Jika semua $U_{i}$ merupakan himpunan simetrik, maka $f(-x)=f(x)$ dan jika semua $U_{i}$ adalah berimbang, maka $f(sx)\leq f(x)$ untuk semua skalar $s$ sehingga $|s|\leq 1$ dan semua $x\in X.$ Jika $X$ adalah sebuah ruang vektor topologis dan jika semua $U_{i}$ adalah lingkungan dari asalnya, maka $f$ adalah kontinu, dimana jika sebagai tambahan $X$ is Hausdorff and $U_{\bullet }$ membentuk sebuau basis lingkungan berimbang dari asalnya di $X$, maka $d(x,y):=f(x-y)$ adalah sebuah metrik menentukan topologi vektor pada $X$.

Sebuah bukti dari teorema di atas diberikan dalam artikel pada ruang vektor topologis termetrikkan.

Jika $U_{\bullet }=\left(U_{i}\right)_{i\in \mathbb {N} }$ dan $V_{\bullet }=\left(V_{i}\right)_{i\in \mathbb {N} }$ adalah dua kumpulan himpunan bagian ruang vektor $X$ dan jika $s$ adalah sebuah skalar, maka oleh definisi:

• $V_{\bullet }$ berisi $U_{\bullet }$: $U_{\bullet }\subseteq V_{\bullet }$ jika dan hanya jika $U_{i}\subseteq V_{i}$ untuk setiap indeks $i$.
• Himpunan buhul: $\operatorname {Knots} U_{\bullet }:=\left\{U_{i}:i\in \mathbb {N} \right\}$.
• Kernel: $\operatorname {ker} U_{\bullet }:=\bigcap _{i\in \mathbb {N} }U_{i}$.
• Kelipatan skalar: $sU_{\bullet }:=\left(sU_{i}\right)_{i\in \mathbb {N} }$.
• Jumlah: $U_{\bullet }+V_{\bullet }:=\left(U_{i}+V_{i}\right)_{i\in \mathbb {N} }$.
• Irisan: $U_{\bullet }\cap V_{\bullet }:=\left(U_{i}\cap V_{i}\right)_{i\in \mathbb {N} }$.

Jika $\mathbb {S}$ adalah sebuah barisan kumpulan himpunan bagian $X$, maka $\mathbb {S}$ dikatakan menjadi berarah (ke bawah) terhadap inklusi atau berarah sederhana jika $\mathbb {S}$ tidak kosong dan untuk semua $U_{\bullet },V_{\bullet }\in \mathbb {S}$, terdapat $W_{\bullet }\in \mathbb {S}$ sehingga $W_{\bullet }\subseteq U_{\bullet }$ dan $W_{\bullet }\subseteq V_{\bullet }$ (dikatakan dengan berbeda, jika dan hanya jika $\mathbb {S}$ adalah sebuah pratapis terhadap kurungan $\subseteq$ yang didefinisikan di atas).

Notasi: Misalkan $\operatorname {Knots} \mathbb {S} :=\bigcup _{U_{\bullet }\in \mathbb {S} }\operatorname {Knots} U_{\bullet }$ menjadi himpunan semua buhul mengenai semua untai di $\mathbb {S}$.

Mendefinisikan topologi vektor menggunakan kumpulan untai khususnya berguna untuk mendefinisikan kelas ruang vektor topologis yang tidak memerlukan cembung lokal.

Theorem (Topology induced by strings) — If $(X,\tau )$ is a topological vector space then there exists a set $\mathbb {S}$ of neighborhood strings in $X$ that is directed downward and such that the set of all knots of all strings in $\mathbb {S}$ is a neighborhood basis at the origin for $(X,\tau ).$ Such a collection of strings is said to be $\tau$ fundamental.

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

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

Topological structure

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

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

Invariance of vector topologies

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

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

If $x\in X$ and any subset $S\subseteq X$, then $\operatorname {cl} _{X}(x+S)=x+\operatorname {cl} _{X}S$ and moreover, if $0\in S$ 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.

Local notions

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

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

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

Bounded subsets

A subset $E$ of a topological vector space $X$ is bounded if for every neighborhood $V$ of the origin, then $E\subseteq tV$ when $t$ is sufficiently large.

The definition of boundedness can be weakened a bit; $E$ is bounded if and only if every countable subset of it is bounded. A set is bounded if and only if each of its subsequences is a bounded set. Also, $E$ is bounded if and only if for every balanced neighborhood $V$ of 0, there exists $t$ such that $E\subseteq tV$. 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 $E$.

Every totally bounded set is bounded. 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 $X$.

Metrizability

Birkhoff–Kakutani theorem — If $(X,\tau )$ is a topological vector space then the following three conditions are equivalent:

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

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

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

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

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

Completeness and uniform structure

Artikel utama: Complete topological vector space

The canonical uniformity on a TVS $(X,\tau )$ is the unique translation-invariant uniformity that induces the topology $\tau$ on $X$.

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

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. 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. A complete subset of a Hausdorff TVS is closed.
• If $C$ is a complete subset of a TVS then any subset of $C$ that is closed in $C$ is complete.
• 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 $\sum _{i=1}^{\infty }x_{i}$ converges in a TVS $X$ then $x_{\bullet }\to 0$ in $X$.

Examples

Finest and coarsest vector topology

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

Trivial topology

The trivial topology or indiscrete topology $\{X,\varnothing \}$ 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 $\operatorname {dim} X=0$.

Finest vector topology

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

Product vector spaces

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 $f:\mathbb {R} \to \mathbb {R}$ where $\mathbb {R}$ 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 $\mathbb {R} ^{\mathbb {R} },$, which carries the natural product topology. With this product topology, $X:=\mathbb {R} ^{\mathbb {R} }$ becomes a topological vector space whose topology is called the topology of pointwise convergence on $\mathbb {R}$. The reason for this name is the following: if $\left(f_{n}\right)_{n=1}^{\infty }$ is a sequence (or more generally, a net) of elements in $X$ and if $f\in X$ 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 $\mathbb {R}$. 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 (i.e. 1-dimensional vector subspaces, which are subsets of the form $\mathbb {R} f:=\{rf:r\in \mathbb {R} \}$ with $f\neq 0$).

Finite-dimensional spaces

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

• If $\operatorname {dim} X=0$ then $X=\{0\}$ 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 $0$.
• If $\operatorname {dim} X=1$ then $X$ has two vector topologies: the usual Euclidean topology and the (non-Hausdorff) trivial topology.
  • Since the field $\mathbb {K}$ is itself a 1-dimensional topological vector space over $\mathbb {K}$ 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.

Proof outline

The proof of this dichotomy is straightforward so only an outline with the important observations is given. As usual, $\mathbb {K}$ is assumed have the (normed) Euclidean topology. Let $X$ be a 1-dimensional vector space over $\mathbb {K}$. Observe that if $B\subseteq \mathbb {K}$ is a ball centered at 0 and if $S\subseteq X$ is a subset containing an "unbounded sequence" then $B\cdot S=X$, where an "unbounded sequence" means a sequence of the form $s_{\bullet }=\left(s_{i}\right)_{i=1}^{\infty }$ where $0\neq x\in X$ and $\left(s_{i}\right)_{i=1}^{\infty }\subseteq \mathbb {K}$ is unbounded in normed space $\mathbb {K}$. Any vector topology on $X$ will be translation invariant and invariant under non-zero scalar multiplication, and for every $0\neq x\in X$, the map $M_{x}:\mathbb {K} \to X$ given by $M_{x}(s):=sx$ is a continuous linear bijection. In particular, for any such $x$, $X=\mathbb {K} x$ so every subset of $X$ can be written as $Fx=M_{x}(F)$ for some unique subset $F\subseteq \mathbb {K} .$ And if this vector topology on $X$ has a neighborhood of 0 that is properly contained in $X$, then the continuity of scalar multiplication $\mathbb {K} \times X\to X$ at the origin forces the existence of an open neighborhood of the origin in $X$ that does not contain any "unbounded sequence". From this, one deduces that if $X$ doesn't carry the trivial topology and if $0\neq x\in X$, then for any ball $B\subseteq \mathbb {K}$ center at 0 in $\mathbb {K}$, $M_{x}(B)=Bx$ contains an open neighborhood of the origin in $X$ so that $M_{x}$ is thus a linear homeomorphism. ∎

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

Non-vector topologies

Discrete and cofinite topologies

If $X$ is a non-trivial vector space (i.e. of non-zero dimension) then the discrete topology on $X$ (which is always metrizable) is not a TVS topology because despite making addition and negation continuous (which makes it into a topological group under addition), it fails to make scalar multiplication continuous. The cofinite topology on $X$ (where a subset is open if and only if its complement is finite) is also not a TVS topology on $X$.

Linear maps

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

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

Types

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

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

• F-spaces are complete topological vector spaces with a translation-invariant metric. These include $L^{p}$ spaces for all $p>0$.
• Locally convex topological vector spaces: here each point has a local base consisting of convex sets. By a technique known as Minkowski functionals it can be shown that a space is locally convex if and only if its topology can be defined by a family of seminorms. 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 $p\geq 1$, but not for $0<p<1$.
• Barrelled spaces: locally convex spaces where the Banach–Steinhaus theorem holds.
• Bornological space: a locally convex space where the continuous linear operators to any locally convex space are exactly the bounded linear operators.
• Stereotype space: a locally convex space satisfying a variant of reflexivity condition, where the dual space is endowed with the topology of uniform convergence on totally bounded sets.
• Montel space: a barrelled space where every closed and bounded set is compact
• Fréchet spaces: these are complete locally convex spaces where the topology comes from a translation-invariant metric, or equivalently: from a countable family of seminorms. Many interesting spaces of functions fall into this class. A locally convex F-space is a Fréchet space.
• LF-spaces are limits of Fréchet spaces. ILH spaces are inverse limits of Hilbert spaces.
• Nuclear spaces: these are locally convex spaces with the property that every bounded map from the nuclear space to an arbitrary Banach space is a nuclear operator.
• Normed spaces and 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.
• Reflexive Banach spaces: Banach spaces naturally isomorphic to their double dual (see below), which ensures that some geometrical arguments can be carried out. An important example which is not reflexive is $L^{1}$, whose dual is $L^{\infty }$ but is strictly contained in the dual of $L^{\infty }$.
• 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.
• Euclidean spaces: $\mathbb {R} ^{n}$ or $\mathbb {C} ^{n}$ with the topology induced by the standard inner product. As pointed out in the preceding section, for a given finite $n$, there is only one $n$-dimensional topological vector space, up to isomorphism. It follows from this that any finite-dimensional subspace of a TVS is closed. A characterization of finite dimensionality is that a Hausdorff TVS is locally compact if and only if it is finite-dimensional (therefore isomorphic to some Euclidean space).

Dual space

Artikel utama: Algebraic dual space, Continuous dual space, dan Strong dual space

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

Properties

Lihat pula: Locally convex topological vector space § Properties

For any $S\subseteq X$ 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 (resp. interior, convex hull, balanced hull, disked hull) of a set $S$ is sometimes denoted by $\operatorname {cl} _{X}S$ (resp. $\operatorname {Int} _{X}S$, $\operatorname {co} S$, $\operatorname {bal} S$, $\operatorname {cobal} S$).

Neighborhoods and open sets

Properties of neighborhoods and open sets

• The open convex subsets of a TVS $X$ (not necessarily Hausdorff or locally convex) are exactly those that are of the form $z+\{x\in X:p(x)<1\}=\{x\in X:p(x-z)<1\}$ for some $z\in X$ and some positive continuous sublinear functional $p$ on $X$.
• If $S\subseteq X$ and $U$ is an open subset of $X$ then $S+U$ is an open set in $X$.
• If $S\subseteq X$ has non-empty interior then $S-S$ is a neighborhood of the origin.
• If $K$ is an absorbing disk in a TVS $X$ and if $p:=p_{K}$ is the Minkowski functional of $K$ then

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

  • It was not assumed that $K$ had any topological properties nor that $p$ was continuous (which happens if and only if $K$ is a neighborhood of 0).
• Every TVS is connected and locally connected. Any connected open subset of a TVS is arcwise connected.
• Let $\tau$ and $\nu$ be two vector topologies on $X$. Then $\tau \subseteq \nu$ if and only if whenever a net $x_{\bullet }=\left(x_{i}\right)_{i\in I}$ in $X$ converges 0 in $(X,\nu )$ then $x_{\bullet }\to 0$ in $(X,\tau )$.
• Let ${\mathcal {N}}$ be a neighborhood basis of the origin in $X$, let $S\subseteq X$, and let $x\in X$. Then $x\ in\operatorname {cl} _{X}S$ if and only if there exists a net $s_{\bullet }=\left(s_{N}\right)_{N\in {\mathcal {N}}}$ in $S$ (indexed by ${\mathcal {N}}$) such that $s_{\bullet }\to x$ in $X$.

Interior

• If $S$ has non-empty interior then $\operatorname {Int} _{X}S=\operatorname {Int} _{X}\left(\operatorname {cl} _{X}S\right)$ and $\operatorname {cl} _{X}S=\operatorname {cl} _{X}\left(\operatorname {Int} _{X}S\right)$.
• If $R,S\subseteq X$ and $S$ has non-empty interior then $\operatorname {Int} _{X}(R)+\operatorname {Int} _{X}(S)\subseteq R+\operatorname {Int} _{X}S\subseteq \operatorname {Int} _{X}(R+S)$.
• If $S$ is a disk in $X$ that has non-empty interior then 0 belongs to the interior of $S$.
  • However, a closed balanced subset of $X$ with non-empty interior may fail to contain 0 in its interior.
• If $S$ is a balanced subset of $X$ with non-empty interior then $\{0\}\cup \operatorname {Int} _{X}S$ is balanced; in particular, if the interior of a balanced set contains the origin then $\operatorname {Int} _{X}S$ is balanced.
• If $x$ belongs to the interior of a convex set $S\subseteq X$ and $y\in \operatorname {cl} _{X}S$, then the half-open line segment $[x,y):=\left\{tx+(1-t)y:0<t\leq 1\right\}$ if $x\neq y$ and $[x,x)=\varnothing$ if $x=y$. If $N$ is a balanced neighborhood of $0$ in $X$ then by considering intersections of the form $N\cap \mathbb {R} x$ (which are convex symmetric neighborhoods of $0$ in the real TVS $\mathbb {R} x$) it follows that:
  • $\operatorname {Int} N=[0,1)\operatorname {Int} N=(-1,1)N=B_{1}N$, where $B_{1}:=\left\{a\in \mathbb {K} :|a|<1\right\}$.
  • if $x\in \operatorname {Int} N$ and $r:=\sup _{}\left\{r>0:[0,r)x\subseteq N\right\}$ then $r>1$, $[0,r)x\subseteq \operatorname {Int} N$, and if $r\neq \infty$ then $rx\in \operatorname {cl} N\setminus \operatorname {Int} N$.
• If $C$ is convex and $0<t\leq 1$, then $t\operatorname {Int} C+(1-t)\operatorname {cl} C\subseteq \operatorname {Int} C$.

Non-Hausdorff spaces and the closure of the origin

• $X$ is Hausdorff if and only if $\{0\}$ is closed in $X$.
• $\operatorname {cl} _{X}\{0\}=\bigcap _{N\in {\mathcal {N}}(0)}N$ so every neighborhood of the origin contains the closure of $\{0\}$.
• $\operatorname {cl} _{X}\{0\}$ is a vector subspace of $X$ and its subspace topology is the trivial topology (which makes $\operatorname {cl} _{X}\{0\}$ compact).
• Every subset of $\operatorname {cl} _{X}\{0\}$ is compact and thus complete (see footnote for a proof). In particular, if $X$ is not Hausdorff then there exist compact complete subsets that are not closed.
• $S+\operatorname {cl} _{X}\{0\}\subseteq \operatorname {cl} _{X}S$ for every subset $S\subseteq X$.
  • So if $S\subseteq X$ is open or closed in $X$ then $S+\operatorname {cl} _{X}\{0\}=S$ (so $S$ is a "tube" with vertical side $\operatorname {cl} _{X}\{0\}$).
  • The quotient map $q:X\to X/\operatorname {cl} _{X}\{0\}$ is a closed map onto a Hausdorff TVS.
• A subset $S$ of a TVS $X$ is totally bounded if and only if $S+\operatorname {cl} _{X}\{0\}$ is totally bounded, if and only if $\operatorname {cl} _{X}S$ is totally bounded, if and only if its image under the canonical quotient map $X\to X/\operatorname {cl} _{X}\left(\{0\}\right)$ is totally bounded.
• If $S\subseteq X$ is compact, then $\operatorname {cl} _{X}S=S+\operatorname {cl} _{X}\{0\}$ and this set is compact. Thus the closure of a compact set is compact (i.e. all compact sets are relatively compact).
• A vector subspace of a TVS is bounded if and only if it is contained in the closure of $\{0\}$.
• If $M$ is a vector subspace of a TVS $X$ then $X/M$ is Hausdorff if and only if $M$ is closed in $X$.
• Every vector subspace of $X$ that is an algebraic complement of $\operatorname {cl} _{X}\{0\}$ is a topological complement of $\operatorname {cl} _{X}\{0\}$. Thus if $H$ is an algebraic complement of $\operatorname {cl} _{X}\{0\}$ in $X$ then the addition map $H\times \operatorname {cl} _{X}\{0\}\to X$, defined by $(h,n)\mapsto h+n$ is a TVS-isomorphism, where $H$ is Hausdorff and $\operatorname {cl} _{X}\{0\}$ has the indiscrete topology. Moreover, if $C$ is a Hausdorff completion of $H$ then $C\times \operatorname {cl} _{X}\{0\}$ is a completion of $X\cong H\times \operatorname {cl} _{X}\{0\}$.

Closed and compact sets

Compact and totally bounded sets

• A subset of a TVS is compact if and only if it is complete and totally bounded.
  • Thus, in a complete TVS, a closed and totally bounded subset is compact.
• A subset $S$ of a TVS $X$ is totally bounded if and only if $\operatorname {cl} _{X}S$ is totally bounded, if and only if its image under the canonical quotient map $X\to X/\operatorname {cl} _{X}\left(\{0\}\right)$ is totally bounded.
• Every relatively compact set is totally bounded. The closure of a totally bounded set is totally bounded.
• The image of a totally bounded set under a uniformly continuous map (e.g. a continuous linear map) is totally bounded.
• 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 $K+N\subseteq U$.
• 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.

Closure and closed set

• If $S\subseteq X$ and $a$ is a scalar then $a\operatorname {cl} _{X}S\subseteq \operatorname {cl} _{X}(aS)$; if $X$ is Hausdorff, $a\neq 0$, or $S=\varnothing$ then equality holds: $\operatorname {cl} _{X}(aS)=a\operatorname {cl} _{X}S$.
  • In particular, every non-zero scalar multiple of a closed set is closed.
• If $S\subseteq X$ and $S+S\subseteq 2\operatorname {cl} _{X}S$ then $\operatorname {cl} _{X}S$ is convex.
• If $R,S\subseteq X$ then $\operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S)\subseteq \operatorname {cl} _{X}(R+S)$ and $\operatorname {cl} _{X}\left[\operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S)\right]=\operatorname {cl} _{X}(R+S)$. Thus if $R+S$ is closed then so is $\operatorname {cl} _{X}(R)+\operatorname {cl} _{X}(S)$.
• If $S\subseteq X$ and if $R$ is a set of scalars such that neither $\operatorname {cl} _{X}S$ nor $\operatorname {cl} _{X}R$ contain zero then $\left(\operatorname {cl} R\right)\left(\operatorname {cl} _{X}S\right)=\operatorname {cl} _{X}(RS)$.
• The closure of a vector subspace of a TVS is a vector subspace.
• If $S\subseteq X$ then $\operatorname {cl} _{X}S=\bigcap _{N\in {\mathcal {N}}}(S+N)$ where ${\mathcal {N}}$ is any neighborhood basis at the origin for $X$.
  • However, $\operatorname {cl} _{X}U\supseteq \bigcap \left\{U:S\subseteq U,U{\text{ is open in }}X\right\}$ and it's possible for this containment to be proper (e.g. if $X=\mathbb {R}$ and $S$ is the rational numbers).
  • It follows that $\operatorname {cl} _{X}U\subseteq U+U$ for every neighborhood $U$ of the origin in $X$.
• If $X$ is a real TVS and $S\subseteq X$, then $\bigcap _{r>1}rS\subseteq \operatorname {cl} _{X}S$ (observe that the left hand side is independent of the topology on $X$); if $S$ is a convex neighborhood of the origin then equality holds.
• The sum of a compact set and a closed set is closed. However, the sum of two closed subsets may fail to be closed (see this footnote for examples).
• If $M$ is a vector subspace of $X$ and $N$ is a closed neighborhood of the origin in $X$ such that $U\cap N$ is closed in $X$ then $M$ is closed in $X$.
• 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.

Closed hulls

• In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.
• The closed convex hull of a set is equal to the closure of the convex hull of that set; that is, equal to $\operatorname {cl} _{X}(\operatorname {co} S)$.
• The closed balanced hull of a set is equal to the closure of the balanced hull of that set; that is, equal to $\operatorname {cl} _{X}(\operatorname {bal} S)$.
• The closed disked hull of a set is equal to the closure of the disked hull of that set; that is, equal to $\operatorname {cl} _{X}(\operatorname {cobal} S)$.
• If $R,S\subseteq X$ and the closed convex hull of one of the sets $S$ or $R$ is compact then $\operatorname {cl} _{X}(\operatorname {co} (R+S))=\operatorname {cl} _{X}(\operatorname {co} R)+\operatorname {cl} _{X}(\operatorname {co} S)$.
• If $R,S\subseteq X$ each have a closed convex hull that is compact (that is, $\operatorname {cl} _{X}(\operatorname {co} R)$ and $\operatorname {cl} _{X}(\operatorname {co} S)$ are compact) then $\operatorname {cl} _{X}(\operatorname {co} (R\cup S))=\operatorname {co} \left[\operatorname {cl} _{X}(\operatorname {co} R)\cup \operatorname {cl} _{X}(\operatorname {co} S)\right]$.

Hulls and compactness

• In a general TVS, the closed convex hull of a compact set may fail to be compact.
• The balanced hull of a compact (resp. totally bounded) set has that same property.
• The convex hull of a finite union of compact convex sets is again compact and convex.

Other properties

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.
• A vector subspace of a TVS that is closed but not open is nowhere dense.
• 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.
• 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 $X=\bigcup _{n\in \mathbb {N} }nD$.
  • Every nonmeager locally convex TVS is a barrelled space.

Important algebraic facts and common misconceptions

• If $S\subseteq X$ then $2S\subseteq S+S$; if $S$ is convex then equality holds.
  • For an example where equality does not hold, let $x$ be non-zero and set $S=\{-x,x\}$; $S=\{x,2x\}$ also works.
• A subset $C$ is convex if and only if $(s+t)C=sC+tC$ for all positive real $s$ and $t.$
• The disked hull of a set $S\subseteq X$ is equal to the convex hull of the balanced hull of $S$; that is, equal to $\operatorname {co} (\operatorname {bal} S)$. However, in general $\operatorname {co} (\operatorname {bal} S)\neq \operatorname {bal} (\operatorname {co} S).$
• If $R,S\subseteq X$ and $a$ is a scalar then $a(R+S)=aR+aS,$ and $\operatorname {co} (R+S)=\operatorname {co} R+\operatorname {co} S,$ and $\operatorname {co} (aS)=a\operatorname {co} S,$
• If $R,S\subseteq X$ are convex non-empty disjoint sets and $x\not \in R\cup S,$ then $S\cap \operatorname {co} \left(R\cup \{x\}\right)=\varnothing$ or $R\cap \operatorname {co} \left(S\cup \{x\}\right)=\varnothing .$
• 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 family of F-seminorms.

Properties preserved by set operators

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

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

So for instance, since the union of two absorbing sets is again absorbing, the cell in row "$R\cup S$" 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.

Properties preserved by set operators

Operation	Property of $R$, $S$, and any other subsets of $X$ that is considered
	Absorbing	Balanced	Convex	Symmetric	Convex Balanced	Vector subspace	Open	Neighborhood of 0	Closed	Closed Balanced	Closed Convex	Closed Convex Balanced	Barrel	Closed Vector subspace	Totally bounded	Compact	Compact Convex	Relatively compact	Complete	Sequentially Complete	Banach disk	Bounded	Bornivorous	Infrabornivorous	Nowhere dense (in $X$)	Meager	Separable	Pseudometrizable	Operation
$R\cup S$																													$R\cup S$
$\cup$ of increasing non-∅ chain																													$\cup$ of increasing non-∅ chain
Arbitrary unions (of at least 1 set)																													Arbitrary unions (of at least 1 set)
$R\cap S$																													$R\cap S$
$\cap$ of decreasing non-$\varnothing$ chain																													$\cap$ of decreasing non-$\varnothing$ chain
Arbitrary intersections (of at least 1 set)																													Arbitrary intersections (of at least 1 set)
$R+S$																													$R+S$
Scalar multiple																													Scalar multiple
Non-0 scalar multiple																													Non-0 scalar multiple
Positive scalar multiple																													Positive scalar multiple
Closure																													Closure
Interior																													Interior
Balanced core																													Balanced core
Balanced hull																													Balanced hull
Convex hull																													Convex hull
Convex balanced hull																													Convex balanced hull
Closed balanced hull																													Closed balanced hull
Closed convex hull																													Closed convex hull
Closed convex balanced hull																													Closed convex balanced hull
Linear span																													Linear span
Pre-image under a continuous linear map																													Pre-image under a continuous linear map
Image under a continuous linear map																													Image under a continuous linear map
Image under a continuous linear surjection																													Image under a continuous linear surjection
Non-empty subset of $R$																													Non-empty subset of $R$
Operation	Absorbing	Balanced	Convex	Symmetric	Convex Balanced	Vector subspace	Open	Neighborhood of 0	Closed	Closed Balanced	Closed Convex	Closed Convex Balanced	Barrel	Closed Vector subspace	Totally bounded	Compact	Compact Convex	Relatively compact	Complete	Sequentially Complete	Banach disk	Bounded	Bornivorous	Infrabornivorous	Nowhere dense (in $X$)	Meager	Separable	Pseudometrizable	Operation

See also

• Banach space
• Hilbert space
• Normed space
• Locally convex topological vector space
• Topological group
• Vector space

References

• Templat:Adasch Topological Vector Spaces
• Templat:Bierstedt An Introduction to Locally Convex Inductive Limits
• Templat:Bourbaki Topological Vector Spaces Part 1 Chapters 1–5
• Templat:Conway A Course in Functional Analysis
• Templat:Dunford Schwartz Linear Operators Part 1 General Theory
• Templat:Edwards Functional Analysis Theory and Applications
• Templat:Grothendieck Topological Vector Spaces
• Templat:Horváth Topological Vector Spaces and Distributions Volume 1 1966
• Templat:Jarchow Locally Convex Spaces
• Templat:Köthe Topological Vector Spaces I
• Templat:Köthe Topological Vector Spaces II
• Köthe, Gottfried (1983) [1969]. Topological vector spaces I. Grundlehren der mathematischen Wissenschaften. 159. New York: Springer-Verlag. ISBN 978-3-642-64990-5. 
• Lang, Serge (1972). Differential manifolds. Reading, Mass.–London–Don Mills, Ont.: Addison-Wesley Publishing Co., Inc. ISBN 0-201-04166-9. 
• Templat:Narici Beckenstein Topological Vector Spaces
• Robertson, A.P.; Robertson, W.J. (1964). Topological vector spaces. Cambridge Tracts in Mathematics. 53. Cambridge University Press. 
• Templat:Robertson Topological Vector Spaces
• Templat:Rudin Walter Functional Analysis
• Templat:Schaefer Wolff Topological Vector Spaces
• Templat:Schechter Handbook of Analysis and Its Foundations
• Templat:Swartz An Introduction to Functional Analysis
• Templat:Trèves François Topological vector spaces, distributions and kernels
• Templat:Valdivia Topics in Locally Convex Spaces
• Templat:Voigt A Course on Topological Vector Spaces
• Templat:Wilansky Modern Methods in Topological Vector Spaces

External links

• Media terkait Topological vector spaces di Wikimedia Commons