Proof
For all such that .
For all , we have an arbitrary element of .
The Archimedean property [
21] states that for any field
F, with
:
and this allows us to ensure that
From now on, understanding the axiomatic foundation of the base, we can simplify the notation of the set
, such that
therefore:
Recognizing this, we can further simplify the notation without compromising the rigor of the proof:
Let be the integer part of a number , such that and . It always holds that , for .
Taking we have
Since
we denote
. Consequently, we have:
Now, resuming our notation:
Multiplying by
we obtain:
Since the equivalence classes and will be real. And since this implies that between any pair of elements of the set we can always find an element of the set
Using our logical notation, consider the direction of one of the components of a treon between an arbitrary pair of treons and , denoted as and , respectively, there always exists a , denoted as .
Extending this analysis to the topological analysis, we have: Let an arbitrary element
and let
such that
:
Therefore, as for each pair
with
, there exists a
such that
, we have:
This implies that:
where
denotes an open interval.
Hence, is an accumulation point of
Since we had chosen an arbitrary element this implies that all reals are accumulation points of Then: and therefore, as we can conclude that