Inverse-producing extensions of Topological Algebras/circular set

Definition: Circular Set
Let $ V $ a vector space over $ \mathbb{K} $, then $ N\subset V $ is circular, if and only if for all $ |\lambda|\leq 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 $ \mathbb{K} $ vector space $ (V,\mathcal{T}) $ there is a neighborhood base of zero vector consisting of balanced (circular) sets.

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



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

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

Proof 1: existence of a network
If $ \widehat{U} = B_{\varepsilon}^{|\cdot |}(0) \cdot  U_{\varepsilon} $ 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_{\varepsilon} \in \mathfrak{U}_{\mathcal{T}}(0_V) $ "$   \geq $ " 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}{\varepsilon^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_{\varepsilon} $ are valid if $ i \geq i_1 $ applies. Here too, "$  \geq $ " 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 \geq i_0 $ and $ i_2 \geq i_1 $ . For all $ i \geq i_2 $ the following applies with proof step 1, 4 and $ \varepsilon^2 < \varepsilon < 1 $ :  x_i \notin \widehat{U} $   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
$ \widehat{U} \in \mathfrak{U}_{\mathcal{T}}(0_V) $ is also a circular zero environment and any environment $ U\in \mathfrak{U}_{\mathcal{T}}(0_V) $ contains a circular zero environment $ \widehat{U} \in \mathfrak{U}_{\mathcal{T}}(0_V) $ with $ \widehat{U} \subseteq U $. The quantity $ \{ \widehat{U} \, : \, U \in \mathfrak{U}_{\mathcal{T}}(0_V) \} $ is a zero environmental base of circular quantities. $ \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_\alpha \, : \, \alpha \in \mathcal{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 {\mathfrak{U}}_{\mathcal{T}} ( 0_V ) $ circular zero environments in a topological vector space $ (V,\mathcal{T}) $, then also $ U_\alpha \cap U_\beta \in {\mathfrak{U}}_{\mathcal{T}} ( 0_V ) $ is a balanced neighborhood of zero.

Proof
from $ U_\alpha, U_\beta \subset V $ follows circularly, for all $ \lambda \in \mathbb{K} $ with $ |\lambda|\leq 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,\mathcal{T}) $, the intersection of two open quantities is again open, i.e. $ U_\alpha \cap U_\beta \in \mathcal{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 {\mathfrak{U}}_{\mathcal{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 \mathbb{K} $ with $ |\lambda|\leq 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
\widehat{U} \in \mathfrak{U}_{\mathcal{T}}(0_V) $, show that the set $ \widehat{U} $ is circular. 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 \mathfrak{U}_{\mathcal{T}}(0_V) $ is again a circular neighborhoods of the zero vector. U_\alpha \cup U_\beta $ of two circular neighborhoods of the zero vector  $ U_\alpha, U_\beta \in \mathfrak{U}_{\mathcal{T}}(0_V) $ are circular again
 * For the definition of $
 * Check if the sum $
 * Check if the union $

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