Inverse-producing extensions of Topological Algebras/circular set

Definition: Circular Set

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

Lemma: neighborhood base of the zero vector with circular sets

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

Proof

Be ${\textstyle U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}$ as desired. A ${\textstyle \varepsilon >0}$ and a zero environment ${\textstyle U_{\varepsilon }\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}$ with

$$B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }\subseteq U$$

with ${\textstyle B_{\varepsilon }^{|\cdot |}(0):=\{\lambda \in \mathbb {K} \,:\,|\lambda |<\varepsilon \}}$. The quantity ${\textstyle {\widehat {U}}:=B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }}$ is circular.

Proof by contradiction

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

Proof 1: existence of a network

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

Proof 2: Convergence against zero vector

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

Proof 3: scalar multiplication for convergent networks

If a network ${\textstyle (x_{i})_{i\in I}}$ is converged against the zero vector ${\textstyle 0_{V}\in V}$, it also converges ${\textstyle (\lambda \cdot x_{i})_{i\in I}}$ 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 ${\textstyle (y_{i})_{i\in I}}$ with ${\textstyle y_{i}:={\frac {1}{\varepsilon ^{2}}}\cdot x_{i}\in V}$ for all ${\textstyle i\in I}$, which after proof step 3 also against the zero vector ${\textstyle 0_{V}}$ Then there is again an index cabinet ${\textstyle i_{1}\in I}$, for which all ${\textstyle y_{i}\in U_{\varepsilon }}$ are valid if ${\textstyle i\geq i_{1}}$ applies. Here too, "${\textstyle \geq }$" refers to the partial order on the index quantity ${\textstyle I}$.

Proof 5: contradiction

Select ${\textstyle i_{2}\in I}$ in the index quantity such that ${\textstyle i_{2}\geq i_{0}}$ and ${\textstyle i_{2}\geq i_{1}}$. For all ${\textstyle i\geq i_{2}}$ the following applies with proof step 1, 4 and ${\textstyle \varepsilon ^{2}<\varepsilon <1}$:

• ${\textstyle x_{i}\notin {\widehat {U}}}$
• ${\textstyle x_{i}=\varepsilon ^{2}\cdot {\frac {1}{\varepsilon ^{2}}}\cdot x_{i}=\varepsilon ^{2}\cdot y_{i}\in B_{\varepsilon }^{|\cdot |}(0)\cdot U_{\varepsilon }={\widehat {U}}}$.

Proof 4: circular zero environment

${\textstyle {\widehat {U}}\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}$ is also a circular zero environment and any environment ${\textstyle U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}$ contains a circular zero environment ${\textstyle {\widehat {U}}\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})}$ with ${\textstyle {\widehat {U}}\subseteq U}$. The quantity ${\textstyle \{{\widehat {U}}\,:\,U\in {\mathfrak {U}}_{\mathcal {T}}(0_{V})\}}$ 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 ${\textstyle \{U_{\alpha }\,:\,\alpha \in {\mathcal {A}}\}}$ of circular quantities ${\textstyle U_{\alpha }}$. The circular configuration provides the absolute homogeneity of the Gaugefunctional.

Lemma: Cut circular zero environments

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

Proof

from ${\textstyle U_{\alpha },U_{\beta }\subset V}$ follows circularly, for all ${\textstyle \lambda \in \mathbb {K} }$ with ${\textstyle |\lambda |\leq 1}$, ${\textstyle x_{\alpha }\in U_{\alpha }}$ and ${\textstyle x_{\beta }\in U_{\beta }}$

Intersection of open sets

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

Intersection circular

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

Learning Tasks

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

See also

• Net (Mathematics)
• Inverse-producing extensions of Topological Algebras/Absorbing sets and Minkowski functionals
• Inverse-producing extensions of Topological Algebras/Sequences for continousand balanced sets
• absolute pseudo-convex

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