Definition: Circular Set
Let
a vector space over
, then
is circular,
if and only if for all
and for all
also
is valid.
Lemma: neighborhood base of the zero vector with circular sets
In a topological
vector space
there is a neighborhood base of zero vector consisting of balanced (circular) sets.
Proof
Be
as desired. A
and a zero environment
with

with
. The quantity
is circular.
Proof by contradiction
We now show that
is also a zero environment in
. The assumption is that
is not a zero environment.
is without restriction.
Proof 1: existence of a network
If
is not a zero environment, there is a network
which is converged against the zero vector
Proof 2: Convergence against zero vector
If a network
is converged against the zero vector
, there is also an index barrier
"
" means the partial order on the index quantity
.
Proof 3: scalar multiplication for convergent networks
If a network
is converged against the zero vector
, it also converges
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
with
for all
, which after proof step 3 also against the zero vector
Then there is again an index cabinet
, for which all
are valid if
applies. Here too, "
" refers to the partial order on the index quantity
.
Proof 5: contradiction
Select
in the index quantity such that
and
. For all
the following applies with proof step 1, 4 and
:

.
Proof 4: circular zero environment
is also a circular zero environment and any environment
contains a circular zero environment
with
. The quantity
is a zero environmental base of circular quantities.
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
of circular quantities
. The circular configuration provides the absolute homogeneity of the Gaugefunctional.
Lemma: Cut circular zero environments
Be
circular zero environments in a topological vector space
, then also
is a balanced neighborhood of zero.
Proof
from
follows circularly, for all
with
,
and
Intersection of open sets
In a topological space (in particular also in a topological vector space)
, the intersection of two open quantities is again open, i.e.
(see Norms, metrics, topology).
are neighborhood of the zero vector, then
is valid. Thus,
is an open set containing the zero vector and this yields
.
Intersection circular
We now show that
is circular. Be selected as
and
with
. This applies to
and
. The circularity of
and
then supplies
and
and thus also
.
Learning Tasks
- For the definition of
, show that the set
is circular.
- Check if the sum
of two circular neighborhoods of the zero vector
is again a circular neighborhoods of the zero vector.
- Check if the union
of two circular neighborhoods of the zero vector
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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