is a semi-order set as shown in Fig. 1.5. Let 
. From Example 1.14, we have that 
is a topology on [X] corresponding to R, i.e., 
and 
have the same neighborhood system on [X]. Based on the mergence method, we have a revised partition 
. The example shows that undoubtedly by the mergence method, an incompatible equivalence relation can be transformed into a compatible and the finest one. But in some cases, the revised grain-size is too coarse. In the example, 
only has one equivalence class, this is an extreme case. This is not our expectation.
is compatible. Of course, such exists, since X itself has already satisfied the compatible requirement, but it is too fine. So the point is to find the coarsest one among all refined spaces, i.e., the refinement method has minimality. This is the decomposition method that we will discuss below. is a finite semi-order set, its topology is a right-order topology of T. 
is a projection, where [X] is a quotient space. is a (pseudo) semi-order set. For 
, let
{
there exist 
}.
is called a (pseudo) semi-order closure of A on X. is a semi-order set. For 
and 
, we define 
{
, if 
and 
, then 
}. 
is called a boundary point of A with respect to B, simply denoted as 
.
is a quotient space of , R is its corresponding equivalence relation. From Section 1.5, we have 
right-order topology 
quotient topology 
. 
is called a pseudo semi-order on . When R is compatible, is a semi-order. If 
on under the topology , where 
, then it is denoted as 
. and its corresponding quotient space is [X]. For , if 
, then a and b are pseudo-equivalent and denoted by 
.
letting 
, 
is called a pseudo-equivalence class corresponding to a.
, or simply C(A) if it does not cause any confusion.
. From 
and the definition of C(A), we can see that 
such that 
. Since 
and 
are pseudo-equivalent 
. We have 
, then, 
and are pseudo-equivalent, i.e., 
. This is in contradiction to 
. is a semi-order set. is a quotient set of X. For , there exists a minimal open set 
containing a. 
is a topology on . Set can be constructed as follows.
, where 
on 
is a minimal open set containing x. 
is a right-order topology of T.
, is a projection.
. is finite, there exists a minimal integer n such that 
.
. Set is just a minimal open set on 
containing a.
is an open set.
, have 
. Thus, 
.
is an open set, is open.
is an arbitrary open set containing a, then 
.
.
.
is an open set 
.
. Therefore 
is a minimal open set containing a.. Let , there exist 
such that 
. be a minimal open set containing a, since , have 
.
.
, we have 
, i.e., 
.
, have 
.
such that 
.
, have 
, 
and 
, 
= 1, 2,…,
.
., then 
, where 
is a pseudo semi-order on quotient space 
.
, 
, have 
.
be a natural projection, from the order-preserving ability of p, have 
.
. From Lemma 1.6, 
, such that 
. From the definition of ‘
’, we have 
. is a semi-order space and [X] is a quotient space. Let 
.
, then stop and A is what we want. Otherwise, go to (2).
, let 
. If 
, letting 
and 
, then 
and 
, go to (1). Otherwise, go to (3).
. Let 
.
, we have the decomposition 
. Letting 
, then and , go to (1).
, letting 
and , then and 
, go to (3).
. Its corresponding equivalence relation is 
. is what we want. is called a normed decomposition space of [X]. We will prove below that is compatible with X and has minimality. is a semi-order space. For , if 
, A is called semi-order closed on , or semi-order closed for short.
are as follows. is a semi-order space. For , is semi-order closed.
, there exist 
and 
, i.e., 
. We have 
. Thus, is semi-order closed. is a semi-order space. Assuming , then 
is semi-order closed.
, have 
. 
and 
, have 
. From 
, have 
. And from the definition of boundary points, have 
, i.e., is semi-order closed.
.
.
, there exists 
. Since A is semi-order closed, and 
. From the definition of , have 
. is a semi-order space. For 
, a is a semi-order closed. Letting its quotient space be 
, then 
is also a semi-order space, where 
is a quotient structure (topology) corresponding to 
.
. If 
, have 
on . If 
, from Lemma 1.5, have 
. From a is semi-order closed, have 
. This is a contradiction.
, from Lemma 1.5, have 
. Thus, 
and 
. This is a contradiction. is a semi-order space. is a semi-order set. is a quotient space. is a normed decomposition of . Then 
is a semi-order space, where 
is a quotient structure corresponding to .
is the decomposed partition of .
. From Property 1.3, each 
is semi-order closed on 
. From Proposition 1.10, we have that 
is a semi-order space. Letting 
, have 
. Namely, quotient space is a semi-order one. is a quotient space of semi-order set . Let . Assume that 
is its normed decomposition space. Then when 
, we have 
.
, 
and 
such that 
. From the definition of , there must exist 
, but . Otherwise, from the definition of , 
. This is a contradiction. And from 
, we have and . Then, 
. Otherwise, 
. This is in contradiction with 
. From the definition of quotient structures, we have 
.
, we have .. is a quotient space. Let be an equivalence relation corresponding to the normed decomposition space of . We have
and are compatible. has minimality, i.e., for 
, if 
and 
are compatible, then 
. is compatible. Now, we prove that is the minimum.
and are compatible. If 
, obviously, for 
, 
and 
belong to the same equivalence class on 
but are not equivalent on . Since and are equivalent on and 
, then and are also equivalent on R.
denoted 
, where and 
, , are obtained from the normed decomposition of a. From Lemma 1.7, there exists 
. Now we regard as a quotient space of X. Let 
., we have p(x)<p(y)<p(z). Thus, 
and 
belong to the same equivalence class of denoted by b. Then, 
.
, where 
is the complement set of a, this contradicts with .
, this is in contradiction with the assumption that and are not equivalent in ..
, since 
, we have . Let 
.
, since 
, we have . Let 
.
, since 
, we have 
. Then is decomposed into 
and 
.
. 
is a semi-order space. Its corresponding equivalence relation is the decomposed one of R and compatible.
. 
is its corresponding quotient space. 
is a (pseudo) semi-order closure of a on 
. Then, R and T are compatible 
.
is a quotient space of Y, then 
, 
.
: Assume . Decompose [X] according to the following order 
. Obviously, if it is 
′s turn for decomposition, assuming that 
is the decomposed space and is a quotient set of , then 
. And 
, then 
, i.e., needs not be decomposed. R is compatible.: If R is compatible, then 
, we leave the decomposition of to the end. From the compatibility of R, 
needs not be decomposed. Therefore, 
. Since needs not be decomposed as well, then 
. is a semi-order space. From the previous discussion, we have the following results.
and satisfies
, (b) is compatible, and (c) is the maximum (finest). and satisfies, (b) is compatible, and (c) is the minimum. is induced from T. From , a quotient topology on is induced. Finally, from , a quotient (pseudo) semi-order on [X] is induced. Now, the point is whether there exists another method for inducing quotient semi-order. If the method exists, the induced quotient semi-order is as the former, i.e., the uniqueness of quotient semi-order. Moreover, when R and T are incompatible, we show that the method above cannot induce quotient semi-order on [X]. Then, whether there is another method that can induce quotient semi-order on [X] such that its projection is order-preserving. That is, the existence of quotient semi-order.. R and T are compatible. Let [T] be the quotient semi-order obtained from the above method. 
is any quotient semi-order that has the order-preserving projection , then 
., then 
. From Lemma 1.5, we have
. From the assumption that p is order-preserving, have 
and 
.
, i.e., there exists on 
. is order-preserving.
, 
does not hold. From 
, 
satisfy 
and 
.
, we have 
and 
. Thus, 
.
is semi-order, 
. This is in contradiction with that 
does not hold., when the quotient structure [T] of its quotient space 
is represented by a directed graph, if there is no directed loop in the graph, then the corresponding structure is semi-order; otherwise, it is not semi-order. There are two methods to ‘remove’ the loops on the graph. One is to merge all notes on a loop into a new note and carry on until the graph becomes an acyclic one. Then the quotient structure obtained is a semi-order structure. This is the mergence method. The other one is to choose an arbitrary note on a loop and cut the loop open from the note, carry on until the graph becomes an acyclic one. This is just the idea of decomposition. Due to the different cutting sequences, the acyclic graph obtained may be different. So the semi-order structure obtained by the decomposition method is not unique. We can prove the structure is local minimum but not a global one.
