Inverse-producing extensions of Topological Algebras/topological algebra

A topological vector space ${\textstyle V}$ over ${\displaystyle \mathbb {K} }$ is a vector space over the field ${\displaystyle \mathbb {K} }$ that has a topology with which scalar multiplication and addition are continuous mappings.

${\displaystyle {\begin{array}{rcl}\cdot :\mathbb {K} \times V&\longrightarrow &V\quad (\lambda ,v)\longmapsto \lambda \cdot v\\+:V\times V&\longrightarrow &V\quad (v,w)\longmapsto v+w\end{array}}}$

In the following, for all topological vector spaces, we shall use the Hausdorff property be assumed.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space with a topology ${\displaystyle {\mathcal {T}}}$ as a system of open sets ${\displaystyle {\mathcal {T}}\subset \wp (X)}$ and ${\displaystyle a\in X}$, then denote

• ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a):=\left\{U\subseteq X\,:\,\exists _{U_{o}\in {\mathcal {T}}}:\,a\in U_{o}\subseteq U\right\}}$ the set of all neighbourhoods from the point ${\displaystyle a}$,
• ${\displaystyle {\stackrel {o}{\mathfrak {U}}}_{\mathcal {T}}(a):={\mathfrak {U}}_{\mathcal {T}}(a)\cap {\mathcal {T}}}$ the set of all open Neighbourhoods from the point ${\displaystyle a}$,
• ${\displaystyle {\overline {\mathfrak {U}}}_{\mathcal {T}}(a):=\left\{{\overline {U}}\,:\,U\in {\mathfrak {U}}_{\mathcal {T}}(a)\right\}}$ the set of all closed neighbourhoods of point ${\displaystyle a}$.

If no misunderstanding about the underlying topological space can occur, the index ${\displaystyle {\mathcal {T}}}$ is not included as a designation of the topology used.

In convergence statements in the real numbers one usually considers only ${\displaystyle \varepsilon }$ neighbourhood. In doing so, one would actually have to consider in topological spaces for arbitrary neighbourhoods from ${\displaystyle U\in {\mathfrak {U}}_{\mathcal {T}}(a)}$ find an index bound ${\displaystyle i_{U}\in I}$ of a net ${\displaystyle (x_{i})_{i\in I}}$ above which all ${\displaystyle x_{i}\in U}$ lie with ${\displaystyle i\geq i_{U}}$. However, since the ${\displaystyle \varepsilon }$ neighbourhoods are an neighbourhood basis, by the convergence definition one only needs to show the property for all neighbourhoods with ${\displaystyle \varepsilon >0}$.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space, ${\displaystyle a\in X}$, ${\displaystyle I}$ an index set (partial order) and ${\displaystyle (x_{i})_{i\in I}\in X^{I}}$ a mesh. The convergence of ${\displaystyle (x_{i})_{i\in I}\in X^{I}}$ against ${\displaystyle a\in X}$ is then defined as follows:

$${\displaystyle {\begin{array}{rcl}\displaystyle {{\stackrel {\mathcal {T}}{\lim _{i\in I}}}\,x_{i}=a}&:\longleftrightarrow &\forall _{U\in {\mathfrak {U}}_{\mathcal {T}}(a)}\exists _{i_{U}\in I}\forall _{i\geq i_{U}}:\,x_{i}\in U\\\end{array}}}$$.

(where "${\displaystyle \leq }$" for ${\displaystyle I}$ is the partial order on the index set).

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space, ${\displaystyle a\in X}$ and ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a)}$ the set of all neighbourhoods of ${\displaystyle a\in X}$. ${\displaystyle {\mathfrak {B}}_{\mathcal {T}}(a)}$ is called the neighbourhood basis of ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a)}$ if for every :${\displaystyle {\mathfrak {B}}_{\mathcal {T}}(a)\subseteq {\mathfrak {U}}_{\mathcal {T}}(a)\wedge \forall _{U\in {\mathfrak {U}}_{\mathcal {T}}(a)}\exists _{B\in {\mathfrak {B}}_{\mathcal {T}}(a)}:\,B\subseteq U}$.

Let ${\displaystyle (V,\|\cdot \|)}$ be a normed space, then the ${\displaystyle \varepsilon }$ spheres form

${\displaystyle B_{\varepsilon }^{\|\cdot \|}(a):=\left\{v\in V\,;\,\|v-a\|<\varepsilon \right\}}$

an ambient basis of ${\displaystyle {\mathfrak {B}}_{\mathcal {T}}(a)}$ the set of all environments of ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a)}$ of ${\displaystyle a\in V}$.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a toplogic space with chaotic topology ${\displaystyle {\mathcal {T}}:=\{\emptyset ,X\}}$.

• Determine ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a)}$ for any ${\displaystyle a\in X}$.
• Show that any sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }\in X^{\mathbb {N} }}$ converges in ${\displaystyle (X,{\mathcal {T}})}$ against any limit ${\displaystyle a\in X}$.

Let ${\displaystyle (X,d)}$ be a metric space with the discrete topology given by the metric:

${\displaystyle d(x,y):=\left\{{\begin{array}{lcl}0&{\mbox{ for }}&x=y\\1&{\mbox{ for }}&x\not =y\\\end{array}}\right.}$.

• Determine ${\displaystyle {\mathfrak {U}}_{\mathcal {T}}(a)}$ for any ${\displaystyle a\in X}$.
• How many sets make up ${\displaystyle {\mathfrak {B}}_{\mathcal {T}}(a)}$ minimal for any ${\displaystyle a\in X}$?
• Formally state all sequences ${\displaystyle (x_{n})_{n\in \mathbb {N} }\in X^{\mathbb {N} }}$ in ${\displaystyle (X,d)}$ that converge to a limit ${\displaystyle a\in X}$!

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space and ${\displaystyle {\mathcal {T}}\subseteq \wp (X)}$ be the system of open sets, that is:

${\displaystyle U\subseteq X{\mbox{ open }}:\Longleftrightarrow U\in {\mathcal {T}}}$.

Let ${\displaystyle (\mathbb {R} ,{\mathcal {T}})}$ be a topological space on the basic set of real numbers. However, the topology does not correspond to the Euclic topology over the set ${\displaystyle |\cdot |}$, but the open sets are defined as follows.

${\displaystyle U\in {\mathcal {T}}{\mbox{ open }}:\Longleftrightarrow U=\emptyset {\mbox{ or }}U\subseteq \mathbb {R} \,{\mbox{ with }}\,U^{c}{\mbox{ countable}}}$

• Show that ${\displaystyle (\mathbb {R} ,{\mathcal {T}})}$ is a topological space.
• Show that the sequence ${\displaystyle \left({\frac {1}{n}}\right)_{n\in \mathbb {N} }}$ does not converge to ${\displaystyle 0}$ in the topological space ${\displaystyle (\mathbb {R} ,{\mathcal {T}})}$.

Here ${\displaystyle U^{c}:=\mathbb {R} \setminus U}$ is the complement of ${\displaystyle U}$ in ${\displaystyle \mathbb {R} }$.

By the system of open sets in a topology ${\displaystyle {\mathcal {T}}\subseteq \wp (X)}$ the closed sets of the topology are also defined at the same time as their complements.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space and ${\displaystyle {\mathcal {T}}\subseteq \wp (X)}$ be the system of open sets.

${\displaystyle M\subseteq X{\mbox{ completed }}:\Longleftrightarrow \exists _{U\in {\mathcal {T}}}:\,M=U^{c}:=X\setminus U}$

Let ${\displaystyle (V,{\mathcal {T}})}$ be a topological space and ${\displaystyle M\subset V}$, then the open kernel ${\displaystyle {\stackrel {\circ }{M}}}$ of ${\displaystyle M}$ is the union of all open subsets of ${\displaystyle M}$.

$${\displaystyle {\stackrel {\circ }{M}}:=\bigcup _{U\in {\mathcal {T}},U\subseteq M}U}$$.

Let ${\displaystyle (X,{\mathcal {T}})}$ be a topological space. The closed hull ${\displaystyle {\overline {M}}}$ of ${\displaystyle M}$ is the intersection over all closed subsets of ${\displaystyle W=U^{c}}$ containing ${\displaystyle M}$ and ${\displaystyle U}$ is open.

$${\displaystyle {\overline {M}}:=\bigcap _{U\in {\mathcal {T}},U^{c}:=X\setminus U\supseteq M}U^{c}}$$

The topological edge ${\displaystyle \partial M}$ of ${\displaystyle M}$ is defined as follows:

$${\displaystyle \partial M:={\overline {M}}\backslash {\stackrel {\circ }{M}}}$$

In metric spaces, one can still work with the natural numbers as countable index sets. In arbitrary topological spaces one has to generalize the notion of sequences to the notion of nets.

Let ${\displaystyle T}$ be a topological space and ${\displaystyle I}$ an index set (with partial order), then ${\displaystyle T^{I}}$ denotes the set of all families indexed by ${\displaystyle I}$ in ${\displaystyle T}$:

$${\displaystyle T^{I}:=\{(t_{i})_{i\in I}:t_{i}\in T{\mbox{ for all }}i\in I\}}$$

Let ${\displaystyle V}$ be a vector space, then ${\displaystyle c_{oo}(V)}$ denotes the set of all finite sequences with elements in ${\displaystyle V}$:

$${\displaystyle c_{oo}(V):=\left\{(v_{n})_{n\in \mathbb {N} _{0}}\in V^{\mathbb {N} _{0}}:\exists _{\displaystyle N\in \mathbb {N} _{0}}\forall _{\displaystyle n\geq N}:\,v_{n}=0\right\}.}$$

An algebra ${\displaystyle A}$ over the field ${\displaystyle \mathbb {K} }$ is a vector space over ${\displaystyle \mathbb {K} }$ in which a multiplication is an inner join

$${\displaystyle \cdot :A\times A\longrightarrow A\quad (v,w)\longmapsto v\cdot w}$$

is defined where for all ${\displaystyle x,y,z\in A}$ and ${\displaystyle \lambda \in \mathbb {K} }$ the following properties are satisfied:

${\displaystyle {\begin{array}{rcl}x\cdot (y\cdot z)&=&(x\cdot y)\cdot z\\.x\cdot (y+z)&=&x\cdot y+x\cdot z\\(x+y)\cdot z&=&x\cdot z+y\cdot z\\\lambda \cdot (x\cdot y)&=&(\lambda \cdot x)\cdot y=x\cdot (\lambda \cdot y)\end{array}}}$

A topological algebra ${\displaystyle (A,{\mathcal {T}}_{A})}$ over the field ${\displaystyle \mathbb {K} }$ is a topological vector space ${\displaystyle (A,{\mathcal {T}}_{A})}$ over ${\displaystyle \mathbb {K} }$, where also multiplication is

$${\displaystyle \cdot :A\times A\longrightarrow A\quad (v,w)\longmapsto v\cdot w}$$

is a continuous inner knotting.

Continuity of multiplication means here:

$${\displaystyle \forall _{\displaystyle U\in {\mathfrak {U}}\,(0)}\exists _{\displaystyle V\in {\mathfrak {U}}\,(0)}:V\cdot V=V^{2}\subset U}$$

The topology is called multiplicative if holds:

$${\displaystyle \forall _{\displaystyle U\in {\mathfrak {U}}\,(0)}\exists _{\displaystyle V\in {\mathfrak {U}}\,(0)}:V^{2}\subset V\subset U}$$

In describing topology, the Topologization Lemma for Algebras shows that the topology can also be described by a system of Gaugefunctionals

The algebra ${\displaystyle A}$ is called unital if it has a neutral element ${\displaystyle e}$ of multiplication. In particular, one defines ${\displaystyle x^{o}:=e}$ for all ${\displaystyle x\in A}$. The set of all invertible (regular) elements is denoted by ${\displaystyle {\mathcal {G}}(A)}$. Non-invertible elements are called singular.

Consider the set ${\displaystyle V}$ of square ${\displaystyle 2\times 2}$ matrices with matrix multiplication and the maxmum norm of the components of the matrix. Try to prove individual properties of an algebra (${\displaystyle V}$ is a non-commutative unitary algebra). For the proof that ${\displaystyle V}$ with matrix multiplication is also a topological algebra, see Topologization Lemma for Algebras.

Let ${\displaystyle (A,{\mathcal {T}})}$ be a topological algebra over the field ${\displaystyle \mathbb {K} }$, ${\displaystyle \Lambda \subset \mathbb {K} }$ and ${\displaystyle M_{1},M_{2}}$ be subsets of ${\displaystyle A}$, then define

${\displaystyle {\begin{array}{rcl}M_{1}\times M_{2}&:=&\{(m_{1},m_{2})\in A\times A\,:\,m_{1}\in M_{1}\wedge m_{2}\in M_{2}\}\\M_{1}+M_{2}&:=&\{m_{1}+m_{2}\,:\,m_{1}\in M_{1}\wedge m_{2}\in M_{2}\}\\M_{1}\cdot M_{2}&:=&\{m_{1}\cdot m_{2}\,:\,m_{1}\in M_{1}\wedge m_{2}\in M_{2}\}\\\Lambda \cdot M_{1}&.=&\{\lambda \cdot m_{1}\,:\,m_{1}\in M_{1}\wedge \lambda \in \Lambda \}.\\\end{array}}}$

Draw the following set ${\displaystyle M_{k}}$ of vectors as sets of points in the Cartesian coordinate system ${\displaystyle \mathbb {R} ^{2}}$ with ${\displaystyle M_{1}:=\left\{{\begin{pmatrix}1\ 2\end{pmatrix}},{\begin{pmatrix}1\ 0\end{pmatrix}}\right\}}$ and ${\displaystyle M_{2}:=\left\{{\begin{pmatrix}3\ 2\end{pmatrix}},{\begin{pmatrix}0\ 1\end{pmatrix}}\right\}}$ and the following intervals ${\displaystyle [a,b]\in \mathbb {R} }$:

• ${\displaystyle [1,4]\times [2,3]}$.
• ${\displaystyle M_{1}+M_{2}}$.
• ${\displaystyle [1,2]\cdot M_{1}}$.

• Topological vector space

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