(AS(k)) holds for all such that . Let
and . First
we assume that . Then we obtain
. Thence by Property
(WT), we deduce . So, by inductive
assumption, we have . The equality
together with the correlation, obtained before
leads to the correlation , which, by
Property (WT) leads to the contradiction .
Thus, the assumption is false. In
the case where or or
Property (AS(n)) can be reduced to Property
(AS(n-1)), which is valid, according to inductive
assumption. Therefore, the inductive transition is
well-founded. That is why, Property (AS(n)) holds for
each .
Let be a transitive closure (transitive hull) of
the relation in the sense of [18] (see page 337) or
[19], (see page 69), that is binary relation on
satisfying the following condition:
(PO) For the correlation is valid
if and only if there exist and
such, that , and .
The following inclusion holds:
(8)
Indeed, if assume that and , then
we can put , , . And, according
to (PO), we obtain .
It follows from the reflexivity of the relation
and the inclusion (8) that the relation is also
reflexive. According to Property (AS(n)) that the
relation is asymmetric (that is, if and
then ). Being a transitive closure of the relation
, the relation is transitive (according to [18] (see
Theorem and Definition 28.18) or [19], (see Theorem
5.7)). So if and then . Note that
the transitivity of the relation is not difficult to
check also by the direct verification method. Thus, the
relation is a partial order on . Therefore, by
Assertion 4, there exists a linear order relation
on such that . Then, using (8), we get the
inclusion:
(9)
Denote: . Also we define the mapping
by formula (5). That is for an arbitrary
we put . We
are going to prove that the mapping is an one-point
time on the oriented set .
1. According to formula (5), for any
we obtain:
where
Hence, the first condition of Definition 3 is satisfied.
2. Let , and
. Denote, , . Then for
elements and it is performed Condition
(Preo2). Therefore . Thence, by inclusion (9),
we deduce the inequality . And, since ,
we have . Moreover, by formula (5), we have
, . Hence, the second
condition of Definition 3 also is satisfied.
Thus, in accordance with Definition 3, the mapping
is a time on the oriented set .
3. Let’s prove that the time is one-point.
3.1. According to formula (5) the set consists of
one element. Hence, by Definition 4, the time is
quasi one-point.
3.2 Suppose that , and
. Then it follows from the quasi-one-pointness of
time that , . In
accordance with the formula (5), the last two
equalities are possible only if there exist numbers
such that ,
. Denote, , . Then,
according to the conditions (Preo1), (Preo2), taking
into account the reflexivity of the relation , we
obtain and . Hence, taking into
account the inclusion (9), we have and
(where , according to the above). That is
why:
(10)
Also, by formula (5), we obtain ,
. Now we are going to prove that
. Assume the the contrary, . Then, since the
oriented set is a quasi-chain, we deduce
.
Consequently, according to condition (Preo3), we
obtain and therefore . The last
inequality together with (10) ensures , which
is impossible, because , .
The obtained contradiction proves that .
Thus the both conditions of Definition 4 are
satisfied. Therefore the time is one-point.
4. Now we are going to prove that the time is
WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2023.22.109