When is homeomorphic to ?
Abstract.
We identify a class of linearly ordered topological spaces that may satisfy the property that is homeomorphic to or can be embedded into a linearly ordered space with the stated property. We justify the conjectures by partial results.
Key words and phrases:
linearly ordered topological space, lexicographical product, homeomorphism, ordinal1991 Mathematics Subject Classification:
06B30, 54F05, 06A05, 54A101. Questions
In this paper we provide a discussion that justifies our interest in the question of the title. We also identify more specific questions that may lead to affirmative resolutions. We back up our curiosity by some partial results and examples. The main result of this work is Theorem 2.7. To proceed further let us agree on some terminology. A linear order will also be called an order. An order on is compatible with the topology of , if the topology induced by is equal to the topology of . A linearly ordered topological space (abbreviated as LOTS) is a pair of a topological space and a topologycompatible order on . A topological space is orderable if its topology can be induced by some order on . When we consider the lexicographical product of two LOTS and , we first take the lexicographical products of the ordered sets and and then induce the topology as determined by the lexicographical order on . For the purpose of readability we will assume an informal style when describing some folkloretype structures or arguments.
The operations of Cartesian product and lexicographical product produce (more often than not) completely different structures. The former results in a visually ”more voluminous” structure, while the latter keeps ”visual linearity” but introduces ”stretches”. In rare cases, however, both operations produce the same results from a topological point of view. For example, is homeomorphic to . Also, is homeomorphic to , where . Note that is homeomorphic to the space of natural numbers. However, is not homeomorphic to . Indeed, the former is discrete while the latter has nonisolated points such as , etc. Following this discussion, it is not hard to see that given any discrete space , it is possible to find a topologycompatible order on such that is discrete and is homeomorphic to . Our discussion prompts the following general problem.
Problem 1.1.
What conditions on guarantee that there exists a topologycompatible order on such that is homeomorphic to ?
Note that homogeneity is not a necessary condition as follows from the following folklore fact.
Example 1.2.
(Folklore) is homeomorphic to .
Proof.
First observe that is homeomorphic to . We will, therefore, provide a homeomorphism between and . We define our homeomorphism in three stages as follows:
 Stage 1:

For every , fix a bijection between and . Such a homeomorphism exists since both subspaces are homeomorphic to .
 Stage 2:

For every , fix a bijection between and .
 Stage 3:

Define the promised homomorphism from to as follows:
Visually, maps the th vertical at or above the diagonal in onto the th copy of in . Also maps the th horizontal under the diagonal in onto the st copy of in . Finally, maps the upper right corner point of the Cartesian product to of the lexicographical product, which is the only point that is the limit of a sequence of nonisolated points.
Clearly, is a bijection. Let us show that and are continuous. Since the domains and images of and are clopen in the respective superspaces, it remains to show that is continuous at and is continuous at . For this let . Then , which is an open neighborhood of in . We have is a basis at in and is a basis at in . Since is bijective, is continuous at . We proved that is homeomorphic to , and therefore, to . ∎
Even though is not homogeneous, it is homogeneous at all nonisolated points (since there is only such point). But even this property is not necessary for the two types of products to be homeomorphic. A similar argument can be used to verify the presence of the studied phenomenon in the following example.
Example 1.3.
is homeomorphic to , where .
The limit points in this example have different natures. The leftmost point cannot be carried by a homeomorphism to any internal limit point. We omit the proof of the statement of Example 1.3 since we will prove a more general one later (Lemma 2.6). Following Example 1.2 and the fact that any discrete space has the property under discussion, one may wonder if any linearly ordered space with a single nonisolated point has the property. The following example shows that the answer is negative and opens another direction for our study.
Example 1.4.
Let , where is an sized discrete space. Then the following hold:

is orderable.

is not homogeneous to for any topologycompatible order on .
Proof.
To see why is orderable, first observe that we can think of as the subspace of that contains only all isolated ordinals of and the ordinal . To order , simply reverse the order of every sequence in form , for each limit ordinal greater than .
To prove part (2), fix an arbitrary topologycompatible order on . The space has at least one of extreme points or neither. Let us consider all possibilities.
 Case ( has neither minimum nor maximum):

Then is clopen in for each . Therefore, is the free sum of many topological copies of . Hence, is not homeomorphic to .
 Case ( has minimum but not maximum):

Assume first that has a strictly increasing sequence converging to . Then any neighborhood of contains . Therefore, any neighborhood of has size , while no point in has such populous base neighborhoods.
We now assume that has no strictly increasing sequences converging to . This and the absence of a maximum imply that does not have a topological copy of . However, does, which is . In other words, the second derived set of the lexicographical product is empty but .
 Case ( has maximum but not minimum):

Similar to Case 2.
 Case ( has both maximum and minimum):

Similar to the first part of Case 2.
Since we have exhausted all cases, the roof is complete. ∎
It is known (see, for example [2]) that given a subspace of an ordinal, the square of is homeomorphic to a subspace of a linearly ordered space if and only if has no stationary subsets and is character homogeneous at all nonisolated points. This statement and the preceding discussion lead to the following question.
Question 1.5.
Let be a subset of an ordinal, character homogeneous at nonisolated points, and have no stationary subsets. Can be embedded in a linearly ordered space for which and are homeomorphic?
Note that even though spaces in examples 1.2 and 1.3 are not homogeneous, each point has a basis of mutually homeomorphic neighborhoods. This observation prompts the following question.
Question 1.6.
Let be a subspace of an ordinal and every point of has a basis of mutually homeomorphic neighborhoods. Can be embedded in a linearly ordered space for which and are homeomorphic?
In the next section we will justify the discussed questions by proving a statement that generalizes Example 1.2. Namely, we will show that if is a subspace of an ordinal and is homogeneous on its derived set , then is embeddable in a linearly ordered space that has homeomorphic Cartesian and lexicographical products (Theorem 2.7). To prove this we will first identify a special class of spaces for which the two types of products are homeomorphic (Lemma 2.6). The structure of these spaces is similar to that of the space in Example 1.3. We, therefore, generalize Example 1.3 too.
In notations and terminology we will follow [3]. If is a linearlyordered set, by we denote the closed interval in . If it is clear that the interval is considered in but not in some larger ordered set, we simply write . The same concerns other types of intervals. By we will denote the set of all nonisolated points of , that is, the derived set of . We also say that is homogeneous on its subset if for every there exists a homeomorphism such that and .
2. Partial Results
In what follows, by we denote the class of all subspaces of ordinals that are homeomorphic on their derived sets.
To prove our main statement (Theorem 2.7), we start with two technical lemmas about the key properties of the members of that will be used in further arguments.
Lemma 2.1.
Let . Then, for any there exists such that is the single nonisolated point of .
Proof.
By homogeneity of on it suffices to show that the conclusion holds for some element of . We may assume that is not empty. Then is defined. Then is as desired for . ∎
Lemma 2.2.
Let . Then, can be written as so that the following hold:

is clopen and discrete,

and are homeomorphic for any ,

is the only nonisolated point of for each .
Proof.
For any , let be as in Lemma 2.1. We can find between and such that has the same cardinality as any smaller neighborhood of . Then is a clopen discrete subset of and is a desired representation. ∎
To prove our target statement, first for each infinite cardinal , we identify a linearly ordered topological space for which is homeomorphic to . Next, we will direct our efforts on the task of embedding the members of into such spaces.
Construction of for an infinite cardinal .
Definition of . Denote by the ordinal . Define as the subspace of that consists of all points that fall into one of the following three categories:


is orderisomorphic to for some .

is isolated.
Remark. To help visualize , put . Then can be thought of as a long sequence of many clopen copies of converging to .
Definition of . If , then and we let be equal to the existing ordering . For , we will define using a folklore ordering procedure. We first define the order formally and then follow up with a simple demonstration. For each , put . By the definition of and the fact that , we conclude that is a closed subset of . Define on as follows: . Define as follows:

if and .

if is not a subset of for any and .
Construction of is complete.
To convince a reader that the above definition is legal without going into painful details, let us demonstrate a folklore construction of a topologycompatible order for the space . The space is not a linearly ordered space but there are many simple topologycompatible orders on . The one that mimics the above construction is defined as follows. First, reverse the order on . The resulting set becomes order isomorphic to and is homeomorphic to . This short construction is formalized in the above definition in which we top every ”missing limit point” by the reversed sequence ”converging to the next missing limit point”.
Note that in our definition of for we do not change the order position of limit points of , which means that the new order coincides with the natural order when one of the compared elements is in . In a sense, the new order on is almost indistinguishable from the standard order if ”observed from far away”. Also note that if is a subspace of an ordinal that is homogeneous on the derived set, then by Lemma 2.2, can be embedded into for some . Let us record these observations for future reference.
Lemma 2.3.
The following hold:

Every embeds in for some cardinal .

If , , and , then .

If , , and , then .
We will often use the facts in this summary lemma without explicit referencing. Our next goal is to show that lexicographical and Cartesian product operations produce topologically equivalent results when applied to an . We start by considering the two operations on smaller pieces of ’s. In the following three statements the arguments will be very similar to each other. For clarity, we will also use similar wording.
Lemma 2.4.
Let be an infinite cardinal. Then is homeomorphic to .
Proof.
To prove the statement we will visualize and as described in the remark after the definition of . Namely, and is a long sequence of many clopen copies of converging to . We can write then , where every neighborhood of contains all ’s starting from some moment. Having these visuals in mind we will construct a desired homeomorphism in three stages as follows:
 Stage 1:

Partition the set of isolated ordinals of into pairs so that and indexing agrees with the natural wellordering of the partitioned set.
 Stage 2:

Since is an infinite cardinal, is homeomorphic to for any . Therefore, for each isolated we can fix homeomorphisms and . That is, maps the ’s vertical of at or above the diagonal onto ’s copy of in and maps the ’s horizontal strictly below the diagonal onto ’s copy of .
 Stage 3:

Define a homomorphism from to as follows:
The argument similar to that in Example 1.2 shows that is a homeomorphism. ∎
Lemma 2.5.
Let be an infinite cardinal. Then is homeomorphic to .
Proof.
Denote the spaces in the statement by and , respectively. Since is homeomorphic to , it suffices to construct an isomorphism from to , which we will do next.
When treating and as topological spaces with regard to order, we will visualize them as described in Lemma 2.4. For convenience, let us copy our notation from Lemma 2.4 next:
where is the maximum element of in either of the two orders. We are now ready to construct a desired homeomorphism in three stages as follows:
 Stage 1:

Partition the set of isolated ordinals of into pairs so that and indexing agrees with the natural wellordering of the partitioned set.
 Stage 2:

By Lemma 2.4, for each isolated ordinal there exists a homeomorphism of onto . Since is an infinite cardinal, is homeomorphic to . Hence, we can find a homeomorphism from onto .
 Stage 3:

Define a homomorphism from to as follows:
In words, maps most of the ’s horizontal strip corresponding to onto the ’s copy of , most of the ’s vertical onto ’s copy of , and the corner point of the Cartesian product to the maximum of . The argument similar to that in Example 1.2 shows that is a homeomorphism. ∎
We are now ready to prove a generalization of the statement of Example 1.3.
Lemma 2.6.
For every infinite cardinal , the space is homeomorphic to .
Proof.
Denote by and the two spaces in the statement. As in Lemma 2.5, it suffices to construct a homeomorphism from to As in the previous two lemmas, we visualize and as follows:
where is the maximum element of in either of the two orders. We will closely follow our constructions in the previous two lemmas and construct the promised homeomorphism in three stages as follows:
 Stage 1:

Partition the set of isolated ordinals of into pairs so that and indexing agrees with the natural wellordering of the partitioned set.
 Stage 2:

By Lemma 2.5, for each isolated , we can fix two homeomorphisms:
 Stage 3:

Define a homomorphism from to as follows:
In words, maps most of the ’s horizontal strip onto the ’s copy of , most of the ’s vertical strip onto the ’s copy of , and the corner point of the Cartesian product to the maximum of . An argument similar to one of Example 1.2 shows that is a homeomorphism. ∎
Theorem 2.7.
Let be a subspace of an ordinal that is homogeneous on the derived set. Then can be embedded into a LOT such that is homeomorphic to .
In search for candidates with the discussed phenomenon, it is clear that we should immediately eliminate any ordered spaces with stationary subsets. Indeed, the square of such a space is not orderable as follows from a standard generalization of Katetov’s example [5]. Therefore, by the characterization of hereditary paracompactness for GOspaces due to Engelking and Lutzer ([1] or [4]), we should consider only hereditary paracompact ordered spaces. It is clear that if has no stationary subset, then does not have such either. Thus, we need to concentrate on spaces with orderable hereditary paracompact squares. While the EngelkingLutzer characterization is incredibly handy for testing an ordered space for hereditary paracompactness, the author is not aware of any criterion for the square of a LOTS to be hereditary paracompact. Is there such a criterion? If not, let us find one!
References
 [1] H. Bennet and D. Lutzer, Linearly Ordered and Generalized Ordered Spaces, Encyclopedia of General Topology, Elsevier, 2004.
 [2] R. Buzyakova, Ordering a Square, Topology and its Appl., vol 191, 2015, 7681.
 [3] R. Engelking, General Topology, PWN, Warszawa, 1977.
 [4] D. Lutzer, Ordered Topological Spaces, Surveys in General Topology, G. M. Reed., Academic Press, New York (1980), 247296.
 [5] M. Katetov, Complete Normality of Cartesian Products, Fund. Math., 36 (1948), 271274.