Inverse-producing extensions of Topological Algebras/circular set

Definition: Circular Set

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

Lemma: neighborhood base of the zero vector with circular sets

In a topological $𝕂$ vector space $(V,𝒯)$ there is a neighborhood base of zero vector consisting of balanced (circular) sets.

Proof

Be $U\in {𝔘}_{𝒯}(0_V)$ as desired. A $\epsilon >0$ and a zero environment $U_{\epsilon }\in {𝔘}_{𝒯}(0_V)$ with

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

with $B_{\epsilon }^{|\cdot |}(0):=\{\lambda \in 𝕂:|\lambda |<\epsilon \}$. The quantity $\hat{U}:=B_{\epsilon }^{|\cdot |}(0)\cdot U_{\epsilon }$ is circular.

Proof by contradiction

We now show that $\hat{U}:=B_{\epsilon }^{|\cdot |}(0)\cdot U_{\epsilon }$ is also a zero environment in ${𝔘}_{𝒯}(0_V)$. The assumption is that $\hat{U}$ is not a zero environment. $\epsilon <1$ is without restriction.

Proof 1: existence of a network

If $\hat{U}=B_{\epsilon }^{|\cdot |}(0)\cdot U_{\epsilon }$ is not a zero environment, there is a network $(x_i{)}_{i\in I}$ which is converged against the zero vector $0_V$

Proof 2: Convergence against zero vector

If a network $(x_i{)}_{i\in I}$ is converged against the zero vector $0_V\in V$, there is also an index barrier $U_{\epsilon }\in {𝔘}_{𝒯}(0_V)$ "$\ge$" means the partial order on the index quantity $I$.

Proof 3: scalar multiplication for convergent networks

If a network $(x_i{)}_{i\in I}$ is converged against the zero vector $0_V\in V$, it also converges $(\lambda \cdot x_i{)}_{i\in I}$ 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 $(y_i{)}_{i\in I}$ with $y_i:=\frac{1}{{\epsilon }^2}\cdot x_i\in V$ for all $i\in I$, which after proof step 3 also against the zero vector $0_V$ Then there is again an index cabinet $i_1\in I$, for which all $y_i\in U_{\epsilon }$ are valid if $i\ge i_1$ applies. Here too, "$\ge$" refers to the partial order on the index quantity $I$.

Proof 5: contradiction

Select $i_2\in I$ in the index quantity such that $i_2\ge i_0$ and $i_2\ge i_1$. For all $i\ge i_2$ the following applies with proof step 1, 4 and ${\epsilon }^2<\epsilon <1$:

• $x_i\notin \hat{U}$
• $x_i={\epsilon }^2\cdot \frac{1}{{\epsilon }^2}\cdot x_i={\epsilon }^2\cdot y_i\in B_{\epsilon }^{|\cdot |}(0)\cdot U_{\epsilon }=\hat{U}$.

Proof 4: circular zero environment

$\hat{U}\in {𝔘}_{𝒯}(0_V)$ is also a circular zero environment and any environment $U\in {𝔘}_{𝒯}(0_V)$ contains a circular zero environment $\hat{U}\in {𝔘}_{𝒯}(0_V)$ with $\hat{U}\subseteq U$. The quantity $\{\hat{U}:U\in {𝔘}_{𝒯}(0_V)\}$ is a zero environmental base of circular quantities. $◻$

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 $\{U_{\alpha }:\alpha \in 𝒜\}$ of circular quantities $U_{\alpha }$. The circular configuration provides the absolute homogeneity of the Gaugefunctional.

Lemma: Cut circular zero environments

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

Proof

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

Intersection of open sets

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

Intersection circular

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

Learning Tasks

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

Page Information

You can display this page as Wiki2Reveal slides

Wiki2Reveal

The Wiki2Reveal slides were created for the Inverse-producing extensions of Topological Algebras' and the Link for the Wiki2Reveal Slides was created with the link generator.

• This page is designed as a PanDocElectron-SLIDE document type.
• Source: Wikiversity https://en.wikiversity.org/wiki/Inverse-producing%20extensions%20of%20Topological%20Algebras/circular%20set
• see Wiki2Reveal for the functionality of Wiki2Reveal.