Proof. The sketch of proof is: Firstly, by Theorem 22 and the property of -locally finite base, we will generate a countable family of Deng pseudo-metric spaces . Secondly, by Theorem 16, we will construct a Deng pseudo-metric on and prove that the space is exactly the product space of the family . Finally, we will deduce that can be embedded into .
Now let us complete the proof step by step. First, for each let be locally finite in and let the union be a base of .
Arbitrarily select a pair of positive integers
. Let
and let it be fixed for the moment. We consider the following open set
For convenience, we denote as . Because is locally finite, by Theorem 4 we have . Next, the proof shall be divided into the following five steps.
Step 1. By Theorem 22 there exists a family
corresponding to
and
. Therefore, we can define a mapping
by
For any
let
and for each
let
Then there exists a family
. Therefore, we can define a mapping
by
Let Then by Theorem 22, we have .
Step 2. Let
Then
is a Deng pseudo-metric on
. This is because both
and
satisfy the properties (D1)-(D3), and so does
. Besides, by Theorem 22, we have the following two equalities:
Therefore, satisfies (D4).
Step 3. Since
is locally finite, for any
,
there are two open sets: an open neighborhood
of
and an open neighborhood
of
such that they are quasi-coincident with only a finite family
and a finite set
, respectively. Let
Therefore, either or is true for each i () and corresponding . This implies that there exists at most a finite family such that or . It follows that or (). On the other hand, when and (), it is easy to show that and , and then and . Hence . In other words, when , it may be correct that only if i belongs to the finite index set .
Similarly, for any , , there exists at most a finite family such that or . It follows that or holds (). When and , we have and , and then and . Therefore, . In other words, when , it may be correct that only if i belongs to the finite index set .
Let
. Then for any
, there exists at most a finite index set
J such that when
, it may be correct that
. Therefore, for the two positive integers
, we can define a mapping
by
Next, we will prove that each is also a Deng pseudo-metric on . The proof is as follows:
(D1) Because each
satisfies (D1), when
, we have
(D2) Let
. Because
satisfies (D2) for each
i, we have
(D3) Because and satisfy (D3), i.e., we have the following two equalities:
Note that the above two formulas are finite sums. Therefore,
(D4) Because
satisfies (D4) for each
, we have the following equalities:
Therefore, is a countable family of Deng pseudo-metrics. Meanwhile, we denote the topology generated by as .
Step 4. We will prove that is a base of . For this purpose, we only need to prove the following (a) and (b).
- (a)
.
By Theorem 1, it is sufficient to find an open set such that for any open set .
Since
is locally finite, there is an open neighborhood
of
which is only quasi-coincident with finitely many members:
. Therefore,
Since
, we may select an open set
with
such that
. Therefore, when
, we have
(
). Thus
When with , it must hold , and then .
Similarly, there is an open neighborhood
of
which is only quasi-coincident with finitely many members:
. This implies that
for each
, so that
Let
. Then we can select an open set
with
(
) such that
. Take a number
of
satisfying
such that
(
). Hence
(
). Because
, it is true that
. Therefore, by the property of
we have
, i.e.,
. Furthermore, because
, it is true that
. Hence
. Let
. Then
. If
, then
. Hence
Since
and
are not quasi-coincident for each
(
), when
, we have
and then
. And consequently
. If
, then
In either case . Therefore, (a) is proved.
Incidentally, to make the above proof more perfect, we add the following two points. If for all , then . Meanwhile, let us consider two more special cases below.
(i) If there exists a nonempty set
such that each element
satisfies
, then
(ii) If for all , then . Let . Then for any we have , and thus , i.e., .
- (b)
Each member in is the union of some members in .
Let
. Because
is regular, there exists an open set
v belonging to
such that
. Therefore, it is easy to show that there is a natural member
n such that
. For convenience, we denote
v as
. Similarly, for
, there are another natural member
m and an open set
belonging to
such that
Let . Clearly, . Therefore, by Theorem 22 there exists a corresponding family relative to and B such that for all . And consequently . If , then . Thus , and then . Therefore, we assert that . In other words, as long as , it must hold that . This implies that for each there exists belonging to such that . Thus . That is to say, if , then there is such that . Therefore, (b) is proved.
Step 5. Based on the discussions above, we renumber the countable set
as
. Let
. By Theorem 4.2, we define a mapping
by
where
is the
n-th projection, and affirm that
p is a Deng pseudo-metric on
and
is the product space of
, where
is generated by
as a subbase. Now let us prove that
can be embedded into
.
Let and denote as the fuzzy point whose support and value are and , respectively. All these fuzzy points are denoted by . Let . A mapping is defined by Obviously, e is a bijection and its inverse mapping embeds into . Let . Consequently, we regenerate a new mapping It is easy to prove that is a Deng pseudo-metric on , and is a subspace of . Because is a subbase of , is certainly a subbase of . Moreover, because of Step 4, is exactly a base of . Hence and are homeomorphic. In fact, let for any . Then can be embedded into and the mapping is a Deng pseudo-metric on which metricizes the -topological space . Consequently, . In summary, the proof has been completed. □