# Inverse-producing extensions of Topological Algebras/circular set
## Definition: Circular Set

Let V{\textstyle V} a vector space over K{\textstyle \mathbb {K} }, then N⊂V{\textstyle N\subset V} is circular, if and only if for all |λ|≤1{\textstyle |\lambda |\leq 1} and for all x∈N{\textstyle x\in N} also λ⋅x∈N{\textstyle \lambda \cdot x\in N} is valid.

## Lemma: neighborhood base of the zero vector with circular sets

In a topological K{\textstyle \mathbb {K} } vector space (V, T){\textstyle (V, {\mathcal {T}})} there is a neighborhood base of zero vector consisting of balanced (circular) sets.

## Proof

Be U∈U$_T$(0$_V$){\textstyle U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} as desired. A ε>0{\textstyle \varepsilon >0} and a zero environment U$_ε$∈U$_T$(0$_V$){\textstyle U_{\varepsilon }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} with
    $B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }\subseteq U$
with B$_ε$$^{|⋅|}$(0):={λ∈K:|λ|<ε}{\textstyle B_{\varepsilon }^{|\cdot |}(0):=\{\lambda \in \mathbb {K} \, :\, |\lambda |<\varepsilon \}}. The quantity U$^^$:=B$_ε$$^{|⋅|}$(0)⋅U$_ε${\textstyle {\widehat {U}}:=B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }} is circular.

### Proof by contradiction

We now show that U$^^$:=B$_ε$$^{|⋅|}$(0)⋅U$_ε${\textstyle {\widehat {U}}:=B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }} is also a zero environment in U$_T$(0$_V$){\textstyle {\mathfrak {U}}_{\mathcal {T}}(0_{V})}. The assumption is that U$^^${\textstyle {\widehat {U}}} is not a zero environment. ε<1{\textstyle \varepsilon <1} is without restriction.

### Proof 1: existence of a network

If U$^^$=B$_ε$$^{|⋅|}$(0)⋅U$_ε${\textstyle {\widehat {U}}=B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }} is not a zero environment, there is a network (x$_i$)$_{i∈I}${\textstyle (x_{i})_{i\in I}} which is converged against the zero vector 0$_V${\textstyle 0_{V}}

### Proof 2: Convergence against zero vector

If a network (x$_i$)$_{i∈I}${\textstyle (x_{i})_{i\in I}} is converged against the zero vector 0$_V$∈V{\textstyle 0_{V}\in V}, there is also an index barrier U$_ε$∈U$_T$(0$_V$){\textstyle U_{\varepsilon }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} " ≥{\textstyle \geq }" means the partial order on the index quantity I{\textstyle I}.

### Proof 3: scalar multiplication for convergent networks

If a network (x$_i$)$_{i∈I}${\textstyle (x_{i})_{i\in I}} is converged against the zero vector 0$_V$∈V{\textstyle 0_{V}\in V}, it also converges (λ⋅x$_i$)$_{i∈I}${\textstyle (\lambda \cdot x_{i})_{i\in I}} because of the stiffness of the multiplication with scalers in a topological vector space against the zero vector.

### Proof 4: scalar multiplication for convergent networks

Define now a network (y$_i$)$_{i∈I}${\textstyle (y_{i})_{i\in I}} with y$_i$:=1/ε$^2$ ⋅x$_i$∈V{\textstyle y_{i}:={\frac {1}{\varepsilon ^{2}}}\cdot x_{i}\in V} for all i∈I{\textstyle i\in I}, which after proof step 3 also against the zero vector 0$_V${\textstyle 0_{V}} Then there is again an index cabinet i$_1$∈I{\textstyle i_{1}\in I}, for which all y$_i$∈U$_ε${\textstyle y_{i}\in U_{\varepsilon }} are valid if i≥i$_1${\textstyle i\geq i_{1}} applies. Here too, " ≥{\textstyle \geq }" refers to the partial order on the index quantity I{\textstyle I}.

### Proof 5: contradiction

Select i$_2$∈I{\textstyle i_{2}\in I} in the index quantity such that i$_2$≥i$_0${\textstyle i_{2}\geq i_{0}} and i$_2$≥i$_1${\textstyle i_{2}\geq i_{1}}. For all i≥i$_2${\textstyle i\geq i_{2}} the following applies with proof step 1, 4 and ε$^2$<ε<1{\textstyle \varepsilon ^{2}<\varepsilon <1}:
* x$_i$∉U$^^${\textstyle x_{i}\notin {\widehat {U}}}
* x$_i$=ε$^2$⋅ 1/ε$^2$ ⋅x$_i$=ε$^2$⋅y$_i$∈B$_ε$$^{|⋅|}$(0)⋅U$_ε$=U$^^${\textstyle x_{i}=\varepsilon ^{2}\cdot {\frac {1}{\varepsilon ^{2}}}\cdot x_{i}=\varepsilon ^{2}\cdot y_{i}\in B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }={\widehat {U}}}.

### Proof 4: circular zero environment

U$^^$∈U$_T$(0$_V$){\textstyle {\widehat {U}}\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} is also a circular zero environment and any environment U∈U$_T$(0$_V$){\textstyle U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} contains a circular zero environment U$^^$∈U$_T$(0$_V$){\textstyle {\widehat {U}}\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} with U$^^$⊆U{\textstyle {\widehat {U}}\subseteq U}. The quantity {U$^^$:U∈U$_T$(0$_V$)}{\textstyle \{{\widehat {U}}\, :\, U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})\}} is a zero environmental base of circular quantities. ◻{\textstyle \Box }

### Remark: circular zero environment base

With this statement, a zero-environmental basis of circular quantities exists in each topological vector space.

## Cut circular zero environments

In topological vector spaces (and thus also topological algebras), it is shown that there is a zero environmental base {U$_α$:α∈A}{\textstyle \{U_{\alpha }\, :\, \alpha \in {\mathcal {A}}\}} of circular quantities U$_α${\textstyle U_{\alpha }}. The circular configuration provides the absolute homogeneity of the Gaugefunctional.

## Lemma: Cut circular zero environments

Be U$_α$, U$_β$∈U$_T$(0$_V$){\textstyle U_{\alpha }, U_{\beta }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} circular zero environments in a topological vector space (V, T){\textstyle (V, {\mathcal {T}})}, then also U$_α$∩U$_β$∈U$_T$(0$_V$){\textstyle U_{\alpha }\cap U_{\beta }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} is a balanced neighborhood of zero.

## Proof

from U$_α$, U$_β$⊂V{\textstyle U_{\alpha }, U_{\beta }\subset V} follows circularly, for all λ∈K{\textstyle \lambda \in \mathbb {K} } with |λ|≤1{\textstyle |\lambda |\leq 1}, x$_α$∈U$_α${\textstyle x_{\alpha }\in U_{\alpha }} and x$_β$∈U$_β${\textstyle x_{\beta }\in U_{\beta }}

### Intersection of open sets

In a topological space (in particular also in a topological vector space) (V, T){\textstyle (V, {\mathcal {T}})}, the intersection of two open quantities is again open, i.e. U$_α$∩U$_β$∈T{\textstyle U_{\alpha }\cap U_{\beta }\in {\mathcal {T}}} (see Norms, metrics, topology). U$_α$, U$_β${\textstyle U_{\alpha }, U_{\beta }} are neighborhood of the zero vector, then 0$_V$∈U$_α$, 0$_V$∈U$_β${\textstyle 0_{V}\in U_{\alpha }, 0_{V}\in U_{\beta }} is valid. Thus, U$_α$∩U$_β${\textstyle U_{\alpha }\cap U_{\beta }} is an open set containing the zero vector and this yields U$_α$∩U$_β$∈U$_T$(0$_V$){\textstyle U_{\alpha }\cap U_{\beta }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}.

### Intersection circular

We now show that U$_α$∩U$_β${\textstyle U_{\alpha }\cap U_{\beta }} is circular. Be selected as x∈U$_α$∩U$_β${\textstyle x\in U_{\alpha }\cap U_{\beta }} and λ∈K{\textstyle \lambda \in \mathbb {K} } with |λ|≤1{\textstyle |\lambda |\leq 1}. This applies to x∈U$_α${\textstyle x\in U_{\alpha }} and x∈U$_β${\textstyle x\in U_{\beta }}. The circularity of U$_α${\textstyle U_{\alpha }} and U$_β${\textstyle U_{\beta }} then supplies λ⋅x∈U$_α${\textstyle \lambda \cdot x\in U_{\alpha }} and λ⋅x∈U$_β${\textstyle \lambda \cdot x\in U_{\beta }} and thus also λ⋅x∈U$_α$∩U$_β${\textstyle \lambda \cdot x\in U_{\alpha }\cap U_{\beta }}.

### Learning Tasks

* For the definition of U$^^$∈U$_T$(0$_V$){\textstyle {\widehat {U}}\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}, show that the set U$^^${\textstyle {\widehat {U}}} is circular.
* Check if the sum U$_α$+U$_β$:={u$_α$+u$_β$:u$_α$∈U$_α$∧u$_β$∈U$_β$}{\textstyle U_{\alpha }+U_{\beta }:=\{u_{\alpha }+u_{\beta }\, :\, u_{\alpha }\in U_{\alpha }\wedge u_{\beta }\in U_{\beta }\}} of two circular neighborhoods of the zero vector U$_α$, U$_β$∈U$_T$(0$_V$){\textstyle U_{\alpha }, U_{\beta }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} is again a circular neighborhoods of the zero vector.
* Check if the union U$_α$∪U$_β${\textstyle U_{\alpha }\cup U_{\beta }} of two circular neighborhoods of the zero vector U$_α$, U$_β$∈U$_T$(0$_V$){\textstyle U_{\alpha }, U_{\beta }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})} are circular again