In 1985, Ramachandran introduced the concept of semi-well ordered sets during his investigation of anti-homogeneous topological spaces. A partially ordered set is said to be semi-well ordered [11, 12] if every non-empty subset of it has a least element or a greatest element. Sharkovsky's ordering [3] on the set of positive integers introduced by Oleksandr Sharkovsky in his result on the discrete dynamical system is an example of a semi-well ordered set. The concept of semi-well ordered sets is an extension of the concept of well ordered sets. It is easy to see that a semi-well ordered set is linearly ordered. To generalise the idea of well-ordered sets, many attempts have been made in the field of set theory, including those by [5, 8, 7, 4, 17, 9, 10, 13]. One such generalization is the idea of partially well ordered sets [17]. However, the collection of partially well ordered sets and the collection of semi-well ordered sets are two separate extensions of the collection of well ordered sets, whose only intersection is the collection of well ordered sets. A detailed study on semi-well ordered sets and semi-ordinals can be found in [16].

This paper is an attempt to explore the properties of order topology on semi-well ordered sets. Since a semi-well ordered set is linearly ordered, it is Hausdorff, completely normal [2] and Sober [6]. In [14] Sreeja proved that the order topology on a semi-well ordered set is scattered and anti-rigid. Section 2 is a collection of definitions, notations, and basic results which we will use further on. In Section 3, a detailed study on semi-well ordered sets with order topology is made. We prove that a semi-well ordered set together with its order topology is totally disconnected, zero dimensional, and strongly locally compact. Also, we identify the isolated points of a semi-well ordered set. A characterization of compact semi-well ordered sets and infinite homogeneous semi-well ordered sets is also provided. Several other properties of this topological space are also discussed.

In this section some basic definitions, notations, and results which we will use further on are collected.

A partially ordered set is said to be well ordered if every non-empty subset of it has a least element and it is said to be co-well ordered [11] if every non-empty subset of it has a greatest element. A semi-well ordered set [11] is a partially ordered set in which every non-empty subset of it has a least element or a greatest element. It is obvious that every well ordered and co-well ordered set is semi-well ordered.

Let (X, ≤q1) and (Y, ≤q2) be two disjoint linearly ordered sets, then X+Y denote the partially ordered set X∪Y, where the order on X∪Y is ≤q1∪≤q2∪{(x, y): x∈X and y∈Y}. In [11] Ramachandran proved that: ``Every semi-well ordered set (X, ≤) can be represented as A+B, where A and B are well ordered and co-well ordered subsets of X respectively". In this representation A is the set consisting of all elements of X with no immediate predecessor and those elements of X which is an nth successor of an element with no immediate predecessor and B is the complement of A in X. The interval topology [18] on a partially ordered set (X, ≤) is a topology in which {x∈X: x≤a} and {x∈X: x≥a} forms a sub-base for the closed sets. Interval topology is equivalent to the order topology on chains [1].

A topological space is said to be zero dimensional [15] if it has a basis consisting of clopen subsets. If every point in a topological space is contained in an open set whose closure is compact, the space is said to be strongly locally compact [15]. An element of a subset A of a topological space is said to be isolated [15] if it is contained in an open set containing no other points of A. A topological space (X, T) is said to be homogeneous if for any x,y∈X there exists a homeomorphism f on X such that f(x)=y.

In the sequel, we analyse the order topology on a semi-well ordered set and abbreviate semi-well ordered as swo.

Proposition 3.1. A swo set is connected if and only if it is empty or singleton.

Proof. Let (X, ≤) be a swo set such that X=A+B. Suppose that X is connected. Assume the contrary that X has more than one element. Then B should be non-empty. Let b be the greatest element of B. Then (b-1, b]={b} is an open set in X, also it is closed in X. Hence {b} is a clopen subset of X, a contradiction. The converse part is trivial.

Remark 3.1. A non-empty semi-well ordered set is totally disconnected.

Lemma 3.1. Let X=A+B be a swo set with A non-empty. Then,

• (a)

  for each x∈A where x is neither the least nor the greatest element of A and open set I containing x, there exists an open set in X of the form (y, x] where y∈A such that x∈(y, x]⊂I.

• (b)

  for each y∈B where y is not the greatest element of B and open set I containing y, there exists an open set in X of the form [y, z) where z∈B such that y∈[y, z)⊂I.

Proof.

• (a)

  Let x0 be the least element of X and x∈A. Then,

  • (i)

    x∈[x0, y) implies x∈(x0, x]⊂[x0, y).

  • (ii)

    x∈(y1, y2) implies x∈(y1, x]⊂(y1, y2)

  • (iii)

    If X has a greatest element say y0, then x∈(y, y0] implies
    x∈(y, x]⊂(y, y0].

• (b)

  Let y0 be the greatest element of X and y∈B. Then,

  • (i)

    y∈(a, y0] implies y∈[y, y0)⊂(a, y0].

  • (ii)

    y∈(a, b) implies y∈[y, b)⊂(a, b).

  • (iii)

    y∈[x0, b) implies x∈[y,b)⊂[x0,b).

We know that for a well ordered set (X, ≤) the isolated points are exactly the least element, and those elements of X that has an immediate predecessor. Now we have the following theorem:

Theorem 3.1. Let (X=A+B, ≤) be a swo set, where A has no greatest element. Then the isolated points of X are exactly the least element of A(if it exists), the greatest element of B(if it exists), the elements of A with an immediate predecessor, and the elements of B with an immediate successor.

Proof. To prove this we have the following cases:

• (i)

  A is non-empty and B is empty.

  Then X is a well ordered set and the theorem follows.

• (ii)

  A is empty and B is non-empty.

  Then X=C is a co-well ordered set. Let x0 be the greatest element of X. Then {x0} is open in X, x0 is an isolated point. Let x be an element of X with an immediate successor say a then {x}=(x-1, a) is an open set in X, where x-1 is the immediate predecessor of x. So x is an isolated point of X. Again let y be an element with no immediate successor, then by Lemma 3.1, any open set containing y contains a subset of the form [y, b). So y is not an isolated point.

• (iii)

  Both A and B are non-empty.

  Let x0 and y0 be the least and greatest elements of X respectively, then {x0} and {y0} are both open in X, so both x0 and y0 are isolated points of X. Let x≠x0 be an element in A with no immediate predecessor, then by Lemma 3.1 it is not an isolated point, and if x has an immediate predecessor, then x is an isolated point. Also, if y≠y0 is an element in B with an immediate successor, then y is an isolated point of X and if y has no immediate successor, then by Lemma 3.1, it is not an isolated point of X.

Remark 3.2. In Theorem 3.1, if A has the greatest element a and B is non-empty, then a is not an isolated point of X.

A well ordered set with order topology is compact if and only if it has the greatest element. But a semi-well-ordered set need not be compact, even if it has the largest element. Compact semi-well ordered sets are characterised by the following theorem.

Theorem 3.2. Let (X, ≤) be a non-empty semi-well ordered set where X=A+B, then X is compact if and only if A has a greatest element.

Proof. Suppose that X is compact. Again suppose A is non-empty and has no greatest element. Clearly then B is non-empty. Let {x0} and {y0} be the least and greatest element of X respectively. Let U={[x0, x): x∈A} and V={(y, y0]: y∈B}. Clearly U∪V is an open cover of X. In particular U and V are open covers of A and B respectively with open sets in X . Now X is compact so U∪V has a finite sub-cover. Since U and V are nested intervals there exist single sets [x0, x') and (y', y0] such that A= [x0, x') and B= (y', y0]. So X=[x0, x')∪(y', y0], which implies x' and y' does not belong to X, a contradiction. So A has a greatest element.

Conversely, suppose that A has a greatest element, say a. If B is empty, then X is a well ordered set with a greatest element. So it is compact. Suppose B is non-empty. Let x0 and y0 be the least and greatest elements of X respectively. Let U be a open cover of X by basic open sets. Then we have the following cases:

• (i)

  The open set in U containing a is the basic open set (x, y), where x∈A and y∈B.

  Assume without loss of generality that all elements of U other than (x, y) is the basic open sets of the form {x0}, {y0}, (x, x']⊂A or [y, y')⊂B. Clearly then {{x0}, {y0}, (x, y)}⊂U. If x≠x0, then let b-1=x. If this is the case, then there exists a-1∈A such that (a-1, b-1]∈U. If a-1=x0, then {{x0}, (a-1, b-1], (x, y)} covers A. If a-1≠x0 put a-1=b-2 and then there exists (a-2, b-2]∈U such that a-1∈(a-2, b-2]. If a-2=x0 then we are done. If not, the process continues. Now a-n=x0 for some n because otherwise {a-1, a-2, a-3, ⋯} will be a strictly decreasing sequence in A, which is not possible.

  Similarly, if y≠y0 put y=b1, then there exists a1∈B with [b1, a1)∈U. If a1=y0, {{x0}, (x-1, y-1], ⋯, (x-n, y-n], (x, y), [y, y0), {y0}} is a finite sub-cover of U. If a1≠y0 put a1=b2 and then there exists [b2, a2)∈U such that a1∈[b2, a2) and a1<a2. If a2=y0 we are done. Otherwise, this process continues. Now am=y0 for some m because otherwise, {a1, a2, ⋯} will be a strictly increasing sequence in C which is not possible. So {{x0}, (x-1, y-1], ⋯, (x-n, y-n], (x, y), [b1, a1), ⋯, [bm, am), {y0}} is a finite sub-cover of U.

• (ii)

  The open set in U containing a is the basic open set (x, y0], where x∈A.

• (iii)

  The open set in U containing a is the basic open set [x0, y), where y∈B.

  The cases (ii) and(iii) can be proved easily. So we get a finite sub-cover of U. Since U is arbitrary X is compact.

Proposition 3.2. If (X, ≤) is a swo set, where X=A+B, then B is always open in X.

Proof. Let x0∈B be the greatest element. Then {x0} is an open set in X contained in B. Now for all x∈B and x≠x0, [x, y) where y∈B is an open set in X containing x contained in B. So B is open in X.

Remark 3.3. If X=A+B is a swo set, then A need not be open in X. For example, consider X={1, 2, 3, ⋯, 0, ⋯, -3, -2, -1} then A={1, 2, 3, ⋯, 0} and is not open in X since int(A)={1, 2, 3, ⋯}.

Proposition 3.3. Every swo set is zero dimensional.

Proof. Let (X, ≤) be a swo set where X=A+B. Now we have the following three cases:

• (i)

  A is non-empty and B is empty.

  Let a0 be the least element of A. Then {{a0}}∪{(a, x]: a, x ∈A, x≠a0} is a clopen basis for X.

• (ii)

  A is empty and B is non-empty.

  Let b0 be the greatest element of B. Then {{b0}}∪{[y, b): y, b ∈B, y≠b0} is a clopen basis for X.

• (iii)

  A is a non-empty set with no greatest element and B is non-empty.

  Let a0 and b0 be the least and greatest elements of X respectively. Then {{a0}}∪{(x1, x2]: x1, x2 ∈A, x2≠a0}∪{{b0}}∪{[y1, y2): y1, y2 ∈B, y1≠b0} is a clopen basis for X.

• (iii)

  A is a non-empty set with a greatest element say a and B is non-empty.

  Here, if a≠a0 then {{a0}}∪{(x1, x2]: x1, x2 ∈A, x2≠a0, x2≠a}∪{{b0}}∪{[y1, y2): y1, y2 ∈B, y1≠b0}∪{(a1, b1): a1∈A, b1∈B, a1<a} is a clopen basis for X. If a=a0, then {[a, b1): b1∈B}∪{{b0}}∪{[y1, y2): y1, y2 ∈B, y1≠b0} is a clopen basis for X.

Hence in all the cases X has a basis with clopen subsets of X, so X is zero dimensional.

Proposition 3.4. Every swo set is strongly locally compact.

Proof. Let X=A+B be a swo set, where A is a well ordered set and B is a co-well ordered set. If B is empty, then X is well ordered and hence it will be strongly locally compact.

Now suppose that B is non-empty. If A has a greatest element, then X is compact and hence strongly locally compact. Suppose that A has no greatest element. Let a0 be the least element of A(hence of X) and b0 be the greatest element of B( and hence of X). For the element a0 of X, {a0} is an open set in X whose closure {a0} itself is compact. Again let a∈A and a≠a0, then (x, a] where x∈A, is a clopen neighbourhood of a which is compact. Now for b0, {b0} is a clopen set which is compact. Again, for b∈B and b≠b0 [b, y), y∈B is a clopen neighbourhood of b which is compact. So X is strongly locally compact.

Remark 3.4. Since every strongly locally compact space is locally compact, every swo set is locally compact.

Theorem 3.3. Let (X, ≤) be a swo set, where X=A+B. Then X is complete if and only if A has a greatest element.

Proof. Suppose that A has a greatest element. If B is empty, then A is well ordered and so X is complete. Suppose that B is non-empty. Let Y be a non-empty subset of X which is bounded above. Then Y=W+C, where W⊂A and C⊂B. If C is empty, then W⊂A has a least upper bound. If C is non-empty, then C has a greatest element which is the least upper bound of W+C. So X is complete.

Conversely, suppose that X is complete. If B is empty, then since A is well ordered it has a greatest element. Suppose that B is non-empty. If A=∅, then B=[0, α)​*, for some limit ordinal α, where [0, α)​* is the dual of [0, α). Now α is an upper bound for B. So B = [0, β]​*, but this is not possible by the definition of B. So A should be non-empty. So, B has lower bound by elements of A. If this lower bounds has a supremum, then it should belong to A and so it is the greatest element of A.

We have the following theorem for an arbitrary partially ordered set.

Theorem 3.4. [18] Let X be a partially ordered set, T be the interval topology on X and {Sn: n∈D} a net in X. If {Sn} converges to y in (X, T) and Sn≤y for all n∈D, then y=l.u.b(rangeSn).

Analogous to this theorem we have the following theorem:

Theorem 3.5. Let X be a partially ordered set, T be the interval topology on X and {Sn : n∈D} a net in X. If {Sn} converges to y in (X, T) and Sn≥y for all n∈D, then y=g.l.b(rangeSn).

Proof. Let A be the range set of {Sn: n∈D}. Suppose that y≠g.l.b(rangeSn). Let P={x∈X: x≤a for all a∈A}. Since y≠g.l.b(rangeSn), we can find z∈P such that y∉{x∈X: z≤x}(=Q). Now Q is closed, X∖Q is an open set containing y. So {Sn} is eventually in X∖Q. Which implies z∉P, a contradiction.

Theorem 3.6. Let (X, ≤) be a swo set where X=A+B, B≠φ. If {Sn} is a net in X which converges to y∈X, then

• (i)

  if y∈A and y is not the greatest element of A, then y=l.u.b{Sn: m≤n} for some m∈N.

• (ii)

  if y∈B, then y=g.l.b{Sn: m≤n} for some m∈N.

Proof.

• (i)

  Let K={x∈X: x≤y}. Since y is not the greatest element of A, K is an open set containing y. So Sn∈K for all n≥m for some m∈N. Hence, Sn≤y for all n≥m. So by Theorem 3.4, y=l.u.b{Sn: m≤n}.

• (ii)

  Let Q={x∈X: x≥y}. Clearly Q is open and contains y. So {Sn} is eventual in Q. So y=g.l.b{Sn: m≤n} for some m∈N.

For a subset Y of a semi-well ordered set X, the subspace topology on Y need not be equivalent to its order topology. For example, let X be a set with ordering 1<2<3<⋯<10<20<⋯<1'<2'<⋯<⋯<-3'<-2'<-1'<⋯<-3<-2<-1 and Y={20}+{ ⋯, -3, -2, -1} with the ordering inherited from X. Then {20}=(10,30)∩Y is an open set in the relative topology on Y but not open in the order topology on Y.

Since every swo set is linearly ordered, from Theorem 4 of [1], we have the following theorem.

Theorem 3.7. [1] Let (X, ≤) be a swo set and Y⊂X, then the order topology on Y is equivalent to the relative topology on Y as a subspace of X if and only if for any y∈Y and x∈X∖Y, there exists an element a∈Y between x and y, or if y<x, the elements greater than x in Y have the least element, or if x<y, the element less than x in Y has the greatest element.

Corollary 3.1. Let X=A+B be a swo set, where A has no greatest element. Let Y be a non-empty swo set. Then X+Y is a swo set and the subspace topology on X as a subspace of X+Y is equivalent to the order topology on X.

Proof. Let Y=C+D, where C is well ordered and D is co-well ordered. Then X+Y= (A+C)+(D+B). Now x∈X and y∈(X+Y)∖X implies that there exists an element between x and y (since A has no greatest element and B has no least element). Hence by Theorem 3.7, the corollary follows.

Proposition 3.5. Let (X, ≤) be an infinite swo set where X=A+B. Then, X is homogeneous if and only if the least element (if any) of X has an immediate successor, greatest element (if any) of X has an immediate predecessor and all other elements has both an immediate predecessor and an immediate successor.

Proof. The sufficient part trivially follows since then the order topology coincides with the discrete topology on X. Now for the necessary part suppose X has a least element a0 with no immediate successor. Then clearly B is non-empty and has an element say b with an immediate successor and an immediate predecessor. So there exists a homeomorphism f:X→X such that f(a0)=b. But {b} is open in X and f-1({b})={a0} is not open in X, a contradiction. So if X has a least element, then it should have an immediate successor.

Again suppose that X has a greatest element b0 with no immediate predecessor. Then clearly b0∈A and B is empty. Since A is infinite it has an element, say a with an immediate predecessor and successor. So there exists a homeomorphism g:X→X such that g(b0)=a. Now {a} is open in X but g-1({a})={b0} is not open in X, a contradiction.

Now if X contains an element say x with no immediate predecessor and no immediate successor, then x∈A and B is non-empty. So B contains an element say y with an immediate predecessor and successor. So there exists a homeomorphism h:X→X such that f(x)=y. But {y} is open in X and f-1({y})={x} is not open in X, a contradiction. Hence the theorem follows.

The authors are grateful to late Prof. P. T. Ramachandran for interesting discussions and useful comments. The first author is thankful for the financial support provided by the University Grants Commission, Government of India. The authors are also thankful to the reviewers and editors for their valuable comments, which helped to improve the presentation of the paper.