Reading a Neighborhood from the Order
In an order topology, a neighborhood is built by placing endpoints on either side of a point. The useful question is often not just which intervals contain the point, but how to choose endpoints so that the resulting intervals fit inside any given neighborhood. This exercise-based discussion uses that method in several different orders, including orders with endpoints and limit points.
Let \(X\) be a linearly ordered set with its order topology. An open interval is \((a,b)=\{y\in X:a<y<b\}\). The basic open rays are \((-\infty,b)=\{y\in X:y<b\}\) and \((a,\infty)=\{y\in X:a<y\}\). The order topology is generated by these intervals and rays. In a one-element order \(X=\{x\}\), the basis must include \(X\) itself: there is no interval or proper ray containing \(x\). For the empty order the empty basis gives the topology \(\{\varnothing\}\).
The cuts record all possible endpoints around \(x\). A subset \(C_-\subseteq L_x\) is cofinal below \(x\) if for every \(a\in L_x\), some \(c\in C_-\) satisfies \(a\le c\). A subset \(C_+\subseteq R_x\) is coinitial above \(x\) if for every \(b\in R_x\), some \(d\in C_+\) satisfies \(d\le b\). Cofinal endpoints can be chosen as close to the point from below as needed; coinitial endpoints can be chosen as close from above as needed.
Proof. Each stated interval or ray is open in the order topology and contains \(x\), so every member of the collection is an open neighborhood. Let \(U\) be any open neighborhood of \(x\). By the interval-and-ray basis for the order topology, \(U\) contains a basic interval or ray \(B\) containing \(x\). Write its lower endpoint as \(a\) if it has one and its upper endpoint as \(b\) if it has one. If \(a\) exists, cofinality gives \(c\in C_-\) with \(a\le c<x\); if there is no lower endpoint, use the formal endpoint \(-\infty\). If \(b\) exists, coinitiality gives \(d\in C_+\) with \(x<d\le b\); if there is no upper endpoint, use \(+\infty\). Every point strictly between the chosen endpoints is then strictly between the endpoints of \(B\), so the selected basis neighborhood is contained in \(B\subseteq U\). This proves the neighborhood-base condition, including when one or both cuts are empty. \(\square\)
Here “cofinal” is about the order on a cut, not about distance. For example, there need not be a closest point below \(x\); a cofinal collection still supplies endpoints that pass any prescribed lower endpoint.
Worked Neighborhoods in Familiar Orders
Worked Example: A Neighborhood Base in the Real Line
Give a neighborhood base at \(2\) in \(\mathbb{R}\) with its standard topology. Take \(C_-=\{2-1/n:n\in\mathbb{N}\}\) and \(C_+=\{2+1/n:n\in\mathbb{N}\}\). The intervals \[ (2-1/n,2+1/n),\qquad n\in\mathbb{N}, \] form a neighborhood base at \(2\).
To check the cut conditions, take any \(a<2\). Since \(2-a>0\), choose \(n\) large enough that \(1/n\le 2-a\). Then \(a\le 2-1/n<2\). For any \(b>2\), choose \(n\) large enough that \(1/n\le b-2\). Then \(2<2+1/n\le b\). The cut theorem therefore applies.
For a direct check, let \(U\) be an open neighborhood of \(2\). It contains a basic interval \((a,b)\) with \(a<2<b\). Choose \(n\) so that \(1/n<\min(2-a,b-2)\). Then \(a<2-1/n\) and \(2+1/n<b\), giving \((2-1/n,2+1/n)\subseteq(a,b)\subseteq U\).
Check: substituting \(n=4\) gives \((2-1/4,2+1/4)=(7/4,9/4)\), which contains \(2\), as required. The point \(2\) is not isolated: every one of these intervals also contains \(2+1/(2n)\), a point different from \(2\).
Worked Example: Vertical Neighborhoods in the Dictionary Plane
Give a neighborhood base at \((a,b)\) in \(\mathbb{R}\times\mathbb{R}\) with the lexicographic order topology. The lexicographic order is \((u,v)<(u',v')\) if \(u<u'\), or if \(u=u'\) and \(v<v'\). For each \(n\in\mathbb{N}\), define \[ N_n=((a,b-1/n),(a,b+1/n)). \] These are order-open neighborhoods of \((a,b)\), and they form a neighborhood base.
Indeed, let \(B\) be a basic open neighborhood of \((a,b)\). If \(B\) has a lower endpoint \((u,v)\), then \(u\le a\). If \(u<a\), every point with first coordinate \(a\) is above that endpoint, so there is no restriction on \(1/n\) from below. If \(u=a\), then \(v<b\); choosing \(n\) with \(1/n<b-v\) gives \(v<b-1/n\). Thus the lower endpoint of \(N_n\) lies above the lower endpoint of \(B\). On the upper side, an upper endpoint \((u',v')\) has \(u'\ge a\). If \(u'>a\), every point with first coordinate \(a\) lies below it; if \(u'=a\), choose \(n\) with \(1/n<v'-b\), so \(b+1/n<v'\). A sufficiently large \(n\) meets both restrictions that occur, and then \(N_n\subseteq B\). If \(B\) is a ray, only its existing endpoint imposes a restriction.
In particular, the entire vertical fiber \(\{a\}\times\mathbb{R}\) is open in the lexicographic order topology: at each of its points, an interval whose two endpoints have first coordinate \(a\) stays within that fiber. Yet no point of the fiber is isolated, since \(N_n\) contains, for example, \((a,b+1/(2n))\ne(a,b)\).
Check: for \(n=2\), the endpoints are \((a,b-1/2)\) and \((a,b+1/2)\). Their first coordinates agree, so every point strictly between them has first coordinate \(a\), and its second coordinate lies strictly between \(b-1/2\) and \(b+1/2\).
Isolated Points and Immediate Neighbors
Proof. Suppose first that every nonempty lower cut has a greatest element and every nonempty upper cut has a least element. If both cuts are nonempty, let \(a\) be the greatest element of \(L_x\) and \(b\) the least element of \(R_x\). There is no point strictly between \(a\) and \(x\), or between \(x\) and \(b\), so \((a,b)=\{x\}\), which is open. If \(L_x=\varnothing\) and \(R_x\ne\varnothing\), let \(b\) be the least element of \(R_x\); then \((-\infty,b)=\{x\}\). If \(R_x=\varnothing\) and \(L_x\ne\varnothing\), the greatest element \(a\) of \(L_x\) gives \((a,\infty)=\{x\}\). If both cuts are empty, \(X=\{x\}\), and the whole space is open.
Conversely, suppose \(\{x\}\) is open. If \(L_x\ne\varnothing\), an open basis element containing \(x\) and contained in \(\{x\}\) cannot be a ray extending to the left; it must have a lower endpoint \(a<x\), with no point strictly between \(a\) and \(x\). Thus \(a\) is the greatest element of \(L_x\). If \(R_x\ne\varnothing\), the same basis-element argument on the upper side gives an endpoint \(b>x\) with no point strictly between \(x\) and \(b\), so \(b\) is the least element of \(R_x\). Empty cuts impose no condition. This proves both directions. \(\square\)
The condition distinguishes an endpoint from an isolated point. Being the least point does not by itself make a point isolated: it also needs a least point above it, unless the order consists of that one point alone.
Worked Example: Open Sets in the Ordinal Interval \([0,\omega]\)
Give \([0,\omega]\) its order topology, where \(\omega\) is the first infinite ordinal and the points below it are \(0,1,2,\ldots\). Every finite point \(n\) is isolated. For \(n=0\), the basic initial ray \((-\infty,1)=\{0\}\) in this interval is open. For a positive finite \(n\), the interval \((n-1,n+1)=\{n\}\) is open: there is no integer strictly between \(n-1\) and \(n\), or between \(n\) and \(n+1\).
The point \(\omega\) is not isolated. Its neighborhoods have a base \[ (k,\omega]=\{n\in\mathbb{N}:n>k\}\cup\{\omega\},\qquad k\in\{0,1,2,\ldots\}. \] These are open in the order topology on the interval: each is the final ray above \(k\), including the endpoint \(\omega\). They form a base because any basic neighborhood of \(\omega\) must extend from some finite lower endpoint \(m\), and choosing \(k\ge m\) makes \((k,\omega]\) a subset of that neighborhood.
Consequently, a subset \(U\subseteq[0,\omega]\) is open if and only if either \(\omega\notin U\), or \(U\) contains all finite integers larger than some \(k\). If \(\omega\notin U\), each point of \(U\) has its open singleton as a neighborhood, so \(U\) is a union of open singletons. If \(\omega\in U\) and \(U\) is open, it contains a basic neighborhood \((k,\omega]\). Conversely, if it contains such a tail, then \(\omega\) has an open neighborhood inside \(U\), and all its finite points are isolated.
Common pitfall: It is wrong to identify \(\{\omega\}\) with \((k,\omega]\) for some finite \(k\). For example, \((3,\omega]\) contains \(4\), since \(3<4\le\omega\), so it is not \(\{\omega\}\). In fact every neighborhood of \(\omega\) contains a finite tail, and \(\omega\) is not isolated.
A Limit-Block Neighborhood on the Long Ray
For a final cut calculation, let \(\omega_1\) be the first uncountable ordinal and consider the long ray \[ \mathcal{L}=[0,\omega_1)\times[0,1) \] with the lexicographic order topology. Thus \((\alpha,s)<(\beta,t)\) when \(\alpha<\beta\), or when \(\alpha=\beta\) and \(s<t\). Examine \(p=(\omega,0)\), where \(\omega\) is the first infinite ordinal.
Worked Example: A Neighborhood Base at \((\omega,0)\)
For \(n\in\mathbb{N}\) and \(0<\varepsilon<1\), set \[ N_{n,\varepsilon}=((n,1/2),(\omega,\varepsilon)). \] The point \(p\) lies in this interval because \((n,1/2)<(\omega,0)<(\omega,\varepsilon)\). These intervals form a neighborhood base at \(p\).
To verify the base claim, take any basic interval \((l,r)\) containing \(p\). The first coordinate of \(l\) is some finite ordinal \(m\), since \(l<(\omega,0)\). Choose \(n>m\); then \((n,1/2)>l\). If \(r=(\omega,t)\) for some \(t>0\), choose \(\varepsilon=t/2\), so \(0<\varepsilon<t\) and \((\omega,\varepsilon)<r\). If the first coordinate of \(r\) is greater than \(\omega\), choose \(\varepsilon=1/2\); then \((\omega,1/2)<r\). In either case \(N_{n,\varepsilon}\subseteq(l,r)\). For a neighborhood given by a ray, the same choices are made only for the endpoint that exists.
For instance, \(N_{2,1/4}=((2,1/2),(\omega,1/4))\) contains \((3,0)\) and \((\omega,1/8)\): each is strictly above \((2,1/2)\), and each is strictly below \((\omega,1/4)\). The point \(p\) is not isolated, since \((\omega,\varepsilon/2)\) belongs to \(N_{n,\varepsilon}\) and differs from \(p\) for every allowed \(n,\varepsilon\).
Using Cuts as a Practical Test
For order-topology exercises, first identify the lower and upper cuts at the point of interest. Then determine whether the order supplies convenient cofinal and coinitial endpoint sets. These choices yield a neighborhood base; the isolation criterion asks whether the cuts instead have adjacent endpoints. For a proposed open set, test each of its points: an interior point needs a basic interval inside the set, while an endpoint needs a basic ray or interval suited to its one-sided cut. At a limit point such as \(\omega\), checking only the endpoint itself is not enough; the entire tail required by its neighborhoods matters.
Check Your Understanding
Use the order topology and the cut descriptions above to answer each question.
- Give a neighborhood base at \(3\) in \(\mathbb{R}\) using endpoints of the form \(3-1/n\) and \(3+1/n\), and explain why it is a base.
- In the lexicographic order on \(\mathbb{R}\times\mathbb{R}\), is \(\{a\}\times(0,1)\) open? Justify your answer from neighborhoods of its points.
- Is \(\{0\}\) open in \([0,\omega]\)? Is \(\{\omega\}\) open? Give a basis neighborhood or explain why none can fit inside the proposed singleton.
- Describe the condition on a subset \(U\subseteq[0,\omega]\) for \(U\) to be open when \(\omega\in U\).
- At \(p=(\omega,0)\) in the long ray, explain why endpoints \((n,1/2)\) with \(n\in\mathbb{N}\) can be chosen above any specified lower endpoint of a basic neighborhood.