When Is a Point Isolated?
In an order topology, whether a point is isolated depends on the elements immediately around it. An interior point needs a gap on both sides: an immediate predecessor and an immediate successor. At a minimum or maximum, only one neighboring element is needed. These conditions refer to the order on the set, not to distances or to the order inherited from some larger set.
Let \(X\) be a linearly ordered set, equipped with its order topology. Its basic open sets are open intervals \((a,b)=\{z\in X:a<z<b\}\), initial rays \(\{z\in X:z<b\}\), final rays \(\{z\in X:a<z\}\), and \(X\) itself. When \(X\) has a least or greatest element, the corresponding endpoint neighborhoods are represented by these rays. A point \(x\in X\) is isolated if \(\{x\}\) is open.
An immediate predecessor or successor is an element of \(X\), not merely a bound outside \(X\). For instance, the real number halfway between two elements is useful only if it also belongs to the ordered set being considered.
The Isolation Criterion
- \(x\) is the minimum of \(X\) and has an immediate successor;
- \(x\) is the maximum of \(X\) and has an immediate predecessor;
- \(x\) has both an immediate predecessor and an immediate successor.
Proof. First suppose that \(X\) has more than one element and that \(x\) is isolated. Then \(\{x\}\) is open. By the basis property of the order topology, there is a basic open set \(B\) with \(x\in B\subseteq\{x\}\), so \(B=\{x\}\). We check the possible types of \(B\). If \(B=(a,b)\), then \(a<x<b\). Since the interval contains no other point, there is no element strictly between \(a\) and \(x\), and none strictly between \(x\) and \(b\). Thus \(a\) is an immediate predecessor and \(b\) is an immediate successor. If \(B=\{z\in X:z<b\}\), then \(x<b\), and the only element less than \(b\) is \(x\). Every element of \(X\) is therefore greater than or equal to \(x\), so \(x\) is the minimum; moreover, \(b\) is its immediate successor. If \(B=\{z\in X:a<z\}\), the corresponding reasoning shows that \(x\) is the maximum and \(a\) is its immediate predecessor. The remaining possibility \(B=X=\{x\}\) cannot occur when \(X\) has more than one element. This proves that isolation implies one of the three stated conditions.
Conversely, if \(x\) is the minimum and \(s\) is its immediate successor, then \(\{z\in X:z<s\}=\{x\}\). This is an open initial ray, so \(x\) is isolated. If \(x\) is the maximum and \(p\) is its immediate predecessor, then \(\{z\in X:p<z\}=\{x\}\), an open final ray. If \(p\) and \(s\) are an immediate predecessor and immediate successor of \(x\), then \((p,s)=\{x\}\), an open interval. In each case the equality follows from the definition of immediate neighbor. Finally, if \(X=\{x\}\), then \(X\) is open and equals \(\{x\}\), so \(x\) is isolated. \(\square\)
The three cases are stated separately because endpoints have only one side on which a neighbor is needed. In a two-element order, each point is both an endpoint and has a neighbor on its available side, so both points are isolated. The one-element case must also be included: it has no immediate predecessor or successor, but its only singleton is the whole open space.
Worked Examples in Different Orders
Worked Example: The Integers
Give \(\mathbb{Z}\) its usual order and the associated order topology. Determine whether \(n\in\mathbb{Z}\) is isolated.
For any \(n\in\mathbb{Z}\), \(n-1\) is an immediate predecessor: it is an integer less than \(n\), and there is no integer strictly between \(n-1\) and \(n\). Likewise, \(n+1\) is an immediate successor. The criterion therefore says that every \(n\) is isolated. Directly, the basic open interval \((n-1,n+1)\) contains exactly \(n\).
Check: if \(m\in(n-1,n+1)\), then \(n-1<m<n+1\), so the integer \(m-n\) satisfies \(-1<m-n<1\). The only integer strictly between \(-1\) and \(1\) is \(0\), hence \(m=n\).
Worked Example: A Finite Ordered Set
Let \(X=\{1,2,3,4\}\) with the usual order and its order topology. Decide which points are isolated.
The minimum \(1\) has immediate successor \(2\), so \(\{z\in X:z<2\}=\{1\}\) is open. The maximum \(4\) has immediate predecessor \(3\), so \(\{z\in X:3<z\}=\{4\}\) is open. The point \(2\) has immediate predecessor \(1\) and immediate successor \(3\), giving \((1,3)=\{2\}\). The point \(3\) has immediate predecessor \(2\) and immediate successor \(4\), giving \((2,4)=\{3\}\). Thus every point is isolated.
Check: the four displayed open sets are respectively \(\{1\}\), \(\{4\}\), \(\{2\}\), and \(\{3\}\). Each contains the stated point and no other element of \(X\), as required.
Worked Example: A Dense Order
Give \(\mathbb{Q}\) its usual order and order topology. Show that no rational number is isolated.
Fix \(q\in\mathbb{Q}\). There are rationals on both sides of \(q\), for example \(q-1<q<q+1\), so \(q\) is neither a minimum nor a maximum. If \(a\in\mathbb{Q}\) and \(a<q\), then \((a+q)/2\) is rational and satisfies \(a<(a+q)/2<q\); thus \(q\) has no immediate predecessor. If \(b\in\mathbb{Q}\) and \(q<b\), then \((q+b)/2\) is rational and satisfies \(q<(q+b)/2<b\); thus \(q\) has no immediate successor. The isolation criterion now shows that \(q\) is not isolated.
Check: the midpoint calculations stay in \(\mathbb{Q}\) because the sum of two rational numbers is rational and division by \(2\) preserves rationality. The argument applies to every possible candidate \(a<q\) and \(b>q\), not just to the particular rationals \(q-1\) and \(q+1\).
Worked Example: An Endpoint, an Adjacent Pair, and a Gap in the Order
Let \(X=(0,1]\cup\{2\}\subseteq\mathbb{R}\), ordered by the usual order, and equip \(X\) with its order topology. Determine which points are isolated.
For \(t\in(0,1)\), there is no immediate predecessor. Indeed, for any \(a\in X\) with \(a<t\), the number \((a+t)/2\) belongs to \((0,1)\subseteq X\) and satisfies \(a<(a+t)/2<t\). There is also no immediate successor. If \(t<b\le1\), then \((t+b)/2\in X\) lies strictly between \(t\) and \(b\). The only other possibility for \(b\in X\) with \(b>t\) is \(b=2\); in that case \((t+1)/2\in X\) and \(t<(t+1)/2<2\). Since \(t\) is not an endpoint and has neither immediate neighbor, it is not isolated.
The point \(1\) has no immediate predecessor: for every \(a\in X\) with \(a<1\), the point \((a+1)/2\in(0,1)\) lies strictly between \(a\) and \(1\). It does have immediate successor \(2\), because there is no element of \(X\) strictly between \(1\) and \(2\). But \(1\) is not the minimum of \(X\), and it has no immediate predecessor; the criterion shows that \(1\) is not isolated.
The point \(2\) is the maximum, and \(1\) is its immediate predecessor. Hence \(2\) is isolated; the open final ray \(\{z\in X:1<z\}\) is exactly \(\{2\}\).
Check: the calculation for \(t=0.8\) and \(b=2\) illustrates why an intermediate value must belong to \(X\). The midpoint \((0.8+2)/2=1.4\) is not in \(X\), so it does not show that \(2\) fails to be an immediate successor. Instead, \((0.8+1)/2=0.9\) belongs to \(X\), and \(0.8<0.9<2\). This correctly rules out \(2\) as an immediate successor of \(0.8\).
Using the Criterion Carefully
A useful consequence is that a dense linear order with at least two elements has no isolated points. Here “dense” means that whenever \(a<b\), there is \(z\in X\) with \(a<z<b\). Density rules out immediate predecessors and successors. It also rules out an isolated endpoint: if \(x\) is a minimum, choose \(y>x\); density supplies an element strictly between them, so \(x\) has no immediate successor. The maximum case is the same with the order reversed. A non-endpoint has elements on both sides, and density rules out an immediate neighbor on either side. The criterion therefore excludes isolation in every case.
Common pitfall. Do not infer that a point has no immediate successor by taking the midpoint of it and a larger element in an ambient space. The midpoint might not belong to \(X\), as \(1.4\) does not belong to \((0,1]\cup\{2\}\). To disprove that \(b\) is an immediate successor of \(t\), one must exhibit an actual \(z\in X\) with \(t<z<b\). In the mixed-order example, when \(b\le1\) the midpoint \((t+b)/2\) belongs to \(X\); when \(b=2\), the point \((t+1)/2\) does the job. Checking membership is part of checking the order relation.
The criterion is useful because it turns a topological question—whether a singleton is open—into an order question about neighboring elements. It also shows why endpoints alone do not guarantee isolation: an endpoint must have an immediate neighbor on its available side. Conversely, an interior point needs gaps on both sides, not just one adjacent pair.
Check Your Understanding
For each question, use the order topology associated with the stated order.
- In \(\mathbb{N}=\{1,2,\dots\}\) with the usual order, which points are isolated? Justify your answer using the criterion.
- In the two-element order \(X=\{a,b\}\) with \(a<b\), write an open set equal to \(\{a\}\) and one equal to \(\{b\}\).
- Suppose \(x\) is neither a minimum nor a maximum and has an immediate predecessor but no immediate successor. Is \(x\) isolated? Explain.
- Give an example of a maximum that is not isolated, and verify that it has no immediate predecessor.
- Why does the one-element ordered set need to be treated separately in the isolation criterion?