Order as a Way to Organize a Set
An order lets us compare elements and describe the elements lying between them. These descriptions are useful even before any topology is assigned: an interval is first a subset specified by order inequalities. In particular, the word “open” in open interval refers to excluding its endpoints; it does not by itself assert that the set is open in a topology.
We begin with an ordered set \((X,\leq)\). The relation \(\leq\) is a relation on the set \(X\); no topology on \(X\) is needed for the definitions in this tutorial. When we use \(\mathbb{R}\), it has its usual order and its standard topology, but the interval calculations below use only its order.
Antisymmetry means that two different elements cannot each be at most the other. Comparability says that every pair can be placed in order. Both are required: for example, inclusion on the power set of a two-element set is reflexive, antisymmetric, and transitive, but not total, since neither of the two distinct singleton subsets contains the other.
The strict relation inherits useful properties from \(\leq\). If \(x<y\), then \(y\not<x\); if \(x<y\) and \(y<z\), then \(x<z\). Also, for any \(x,y\), exactly one of \(x<y\), \(x=y\), or \(y<x\) holds. These facts follow from the order axioms: comparability gives \(x\leq y\) or \(y\leq x\); if both hold, antisymmetry gives equality; and transitivity gives \(x\leq z\) when \(x\leq y\leq z\). If also \(x\ne z\), the definition gives \(x<z\).
Intervals and Rays
Fix \(a,b\in X\). The interval notation records which inequalities a point must satisfy. Square brackets include the endpoint named by that bracket; parentheses exclude it.
These definitions work in any total order; they do not require that there be elements between distinct endpoints. If \(a>b\), each of the four bounded intervals from \(a\) to \(b\) is empty: membership would require an element \(x\) with \(a\leq x\leq b\), or with the corresponding strict inequalities, which contradicts transitivity and \(a>b\). If \(a=b\), then \([a,a]=\{a\}\), while \((a,a)\), \([a,a)\), and \((a,a]\) are empty. The ray definitions also apply when \(a\) is a least or greatest element: for example, if \(a\) is least, then \((-\infty,a)=\varnothing\).
Proof. First consider a bounded interval. For \([a,b]\), membership of \(u\) and \(v\) gives \(a\leq u\) and \(v\leq b\). Since \(u\leq w\leq v\), transitivity gives \(a\leq w\leq b\), so \(w\in[a,b]\). For \((a,b)\), we have \(a<u\leq w\leq v<b\), hence \(a<w<b\). For \([a,b)\), we have \(a\leq u\leq w\leq v<b\), hence \(a\leq w<b\). For \((a,b]\), we have \(a<u\leq w\leq v\leq b\), hence \(a<w\leq b\). For a left ray, if \(u,v<a\) and \(u\leq w\leq v\), then \(w\leq v<a\), so \(w<a\); if \(u,v\leq a\), then \(w\leq v\leq a\). For a right ray, if \(a<u,v\) then \(a<u\leq w\), so \(a<w\); if \(a\leq u,v\), then \(a\leq u\leq w\). These are the required conditions for all eight sets. \(\square\)
Finite Chains
A finite totally ordered set can be written as a list from least to greatest. A finite chain is a finite set equipped with a total order. The list is more than a convenient notation: totality ensures that every pair can be compared, and transitivity makes the comparisons consistent.
Proof. We prove existence by induction on the number \(n\) of elements of \(A\). For \(n=1\), the single element gives the required list. Suppose the statement holds for sets of \(n\) elements, and let \(A\) have \(n+1\) elements. Choose \(a\in A\). By the induction hypothesis, the set \(A\setminus\{a\}\) has a least element \(m\) in its increasing list. Compare \(a\) and \(m\). If \(a<m\), then \(a\) is less than every other element of \(A\), because \(m\) is least in \(A\setminus\{a\}\); place \(a\) first. If \(m<a\), then \(m\) is less than \(a\) and every other element of \(A\), so place \(m\) first. The case \(a=m\) cannot occur because \(a\notin A\setminus\{a\}\). In either case, the remaining elements have an increasing list by the induction hypothesis, giving an increasing list for \(A\). For uniqueness, suppose two increasing lists of \(A\) differ. At their first differing position, the two entries are distinct; totality makes one smaller than the other. The list with the larger entry at that position cannot be increasing through all the elements of \(A\), since the smaller entry appears later in that list. This contradicts that the two lists contain the same elements in increasing order. Thus the list is unique. \(\square\)
Worked Example: Intervals in a Four-Element Chain
Let \(X=\{p,q,r,s\}\) with total order \(p<q<r<s\). Find \((q,s)\), \([q,s)\), \((q,s]\), \([q,s]\), and the rays \((-\infty,r]\) and \((r,\infty)\).
For \((q,s)\), the defining condition is \(q<x<s\). Among the four elements, \(p\) fails \(q<p\), \(q\) fails \(q<q\), \(r\) satisfies both inequalities, and \(s\) fails \(s<s\). Thus \((q,s)=\{r\}\).
For \([q,s)\), the condition is \(q\leq x<s\). The elements \(q\) and \(r\) satisfy it; \(p\) is too small and \(s\) is excluded. Hence \([q,s)=\{q,r\}\). For \((q,s]\), the condition is \(q<x\leq s\), so \(r\) and \(s\) satisfy it and \((q,s]=\{r,s\}\). For \([q,s]\), both endpoints are allowed, giving \([q,s]=\{q,r,s\}\).
The ray \((-\infty,r]\) contains precisely the elements at most \(r\), so it is \(\{p,q,r\}\). The ray \((r,\infty)\) contains elements strictly greater than \(r\), so it is \(\{s\}\).
Check: In each bounded interval, every listed element satisfies its defining inequalities, and each excluded element fails at least one. In the two rays, the listed elements satisfy the corresponding one-sided inequality, while the only excluded element \(s\) fails the condition \(x\leq r\) for the left ray and the only excluded elements \(p,q,r\) fail \(r<x\) for the right ray.
Worked Example: Intervals in the Real Line
In \(\mathbb{R}\) with its usual order, compute \([-2,3)\cap(0,5]\), \([-2,3)\cup(0,5]\), and the ray \((3,\infty)\).
A real number belongs to the intersection exactly when it satisfies both \(-2\leq x<3\) and \(0<x\leq5\). Combining the lower bounds gives \(0<x\); combining the upper bounds gives \(x<3\). The endpoint \(0\) is excluded by the second interval, and \(3\) is excluded by the first. Therefore the intersection is \((0,3)\).
For the union, the first interval covers all real numbers from \(-2\), included, up to \(3\), excluded. The second covers numbers strictly above \(0\) through \(5\), included. Together they cover every number from \(-2\) through \(5\), including both outer endpoints. Thus the union is \([-2,5]\). The ray \((3,\infty)\) is the set of all real numbers strictly greater than \(3\).
Check: The number \(0\) belongs to the union but not the intersection; \(3\) belongs to the second interval and hence the union, but not to the intersection; \(-2\) and \(5\) belong to the union. These endpoint tests agree with \((0,3)\) and \([-2,5]\).
Worked Example: A Chain Ordered by Divisibility
Let \(X=\{1,2,4,8\}\), ordered by divisibility: \(x\leq y\) means that \(x\) divides \(y\). Compute \([2,8]\), \((2,8)\), \([1,4)\), and \((4,\infty)\) in this ordered set.
This relation is a total order on this particular \(X\): each element divides every element to its right in the list \(1,2,4,8\); divisibility is reflexive and transitive; and if two positive integers in this list divide each other, they are equal. Thus \(1<2<4<8\) in this order.
The condition for \([2,8]\) is \(2\leq x\leq8\), so the elements are \(2,4,8\). For \((2,8)\), both inequalities are strict, leaving only \(4\). For \([1,4)\), the condition \(1\leq x<4\) gives \(1,2\). Finally, \((4,\infty)\) asks for elements strictly greater than \(4\), which gives \(\{8\}\).
Check: In the divisibility order, \(2<4\) because \(2\) divides \(4\) and \(2\ne4\); \(4<8\) because \(4\) divides \(8\) and \(4\ne8\). The four computed sets therefore match the defining endpoint conditions in the stated chain.
A Common Endpoint Mistake
Common pitfall. Brackets do not mean that every endpoint written in an expression is automatically included. For example, in the finite chain \(p<q<r<s\), it is wrong to calculate \([q,s)\) as \(\{q,r,s\}\): the right parenthesis means the condition is \(x<s\), so \(s\) must be left out. The correct calculation is \([q,s)=\{q,r\}\). Conversely, \((q,s]\) excludes \(q\) but includes \(s\), giving \(\{r,s\}\). Always translate the notation into inequalities before listing elements.
The distinction between order intervals and topological open sets is equally important. The notation \((a,b)\) specifies elements strictly between \(a\) and \(b\). Whether that subset is open in a topology is a separate question. The order topology, which uses intervals and rays to define a topology, is a further construction; here we have only defined and computed the underlying ordered-set objects.
Check Your Understanding
In each question, begin by writing the relevant order inequalities, then test endpoint inclusion.
- In the chain \(a<b<c<d\), list \((a,c]\), \([b,d)\), and \((-\infty,b)\).
- For \(a,b\in X\), determine \([a,b]\) and \((a,b)\) when \(a=b\), and when \(a>b\).
- In \(\mathbb{R}\), compute \((-1,4]\cap[2,6)\), paying particular attention to the endpoints.
- On \(X=\{1,3,9,27\}\) ordered by divisibility, list \((3,27]\) and \([1,9)\).
- Explain why an open interval, as defined here, need not be described as an open set without first specifying a topology.