Two Topologies on the Same Ordered Subset
A subset of an ordered space has an order of its own: it inherits the comparisons between its points. There are then two natural ways to give the subset a topology. One is to put the order topology on that inherited order. The other is to give the subset the subspace topology from the original ordered space. These constructions need not agree. In particular, an ambient open set can cut out a very small piece of the subset, even when that piece is not open in the subset’s own order topology.
Let \((X,\mathcal{T})\) be a linearly ordered set with its order topology, and let \(A\subseteq X\). The order on \(A\) is the restriction of the order on \(X\). Write \(\mathcal{T}_{\mathrm{ord}}(A)\) for the order topology defined using this restricted order, and \(\mathcal{T}_{\mathrm{sub}}(A)\) for the subspace topology inherited from \((X,\mathcal{T})\). The latter consists of the sets \(A\cap U\) with \(U\in\mathcal{T}\).
The General Comparison
Proof. If \(A=\varnothing\), both topologies consist only of \(\varnothing\), and the inclusion holds. Suppose \(A\ne\varnothing\), and consider a basis set for \(\mathcal{T}_{\mathrm{ord}}(A)\). If it is an interval \((a,b)_A\), then \((a,b)_A=A\cap(a,b)_X\), where \((a,b)_X=\{x\in X:a<x<b\}\) is open in the order topology on \(X\). If it is a ray, then \((-\infty,b)_A=A\cap(-\infty,b)_X\) or \((a,\infty)_A=A\cap(a,\infty)_X\); the ambient rays are open, so these intersections are subspace-open. If the basis set is \(A\) itself, it equals \(A\cap X\), and \(X\) is open in its own topology. Thus every basis set for \(\mathcal{T}_{\mathrm{ord}}(A)\) is open in \(\mathcal{T}_{\mathrm{sub}}(A)\). Every order-open set in \(A\) is a union of such basis sets, so it is also subspace-open. \(\square\)
The theorem gives one direction only. The subspace topology may have extra open sets, produced by intersections with ambient open sets that are not open in the order topology on \(A\). The following criterion states exactly what must be checked to establish equality.
Proof. Suppose first that \(\mathcal{T}_{\mathrm{ord}}(A)=\mathcal{T}_{\mathrm{sub}}(A)\). Then \(A\cap U\) is open in the subspace topology and therefore in the order topology on \(A\). For each \(a\in A\cap U\), take \(V=A\cap U\); it has the required properties. Conversely, suppose the stated local condition holds. For each ambient open \(U\), every point \(a\in A\cap U\) belongs to an order-open set \(V\) contained in \(A\cap U\). Taking the union of all such \(V\) shows that \(A\cap U\) is order-open; if \(A\cap U=\varnothing\), it is open as well. Hence every subspace-open set is order-open, because every subspace-open set has the form \(A\cap U\). The theorem gives the reverse inclusion, so the topologies agree. \(\square\)
It is enough in this test to use ambient basis neighborhoods \(U\): every ambient open set is a union of basis sets, and a point in that union belongs to at least one of them. In examples, the test asks whether each ambient neighborhood, when restricted to \(A\), contains a neighborhood of the point in \(A\)’s own order topology.
Worked Example: The Rationals in the Real Line
Let \(X=\mathbb{R}\) with its standard topology and \(A=\mathbb{Q}\), with the order inherited from \(\mathbb{R}\). The order topology on \(\mathbb{Q}\) agrees with its subspace topology.
To check the nonautomatic inclusion, take any \(q\in\mathbb{Q}\) and any ambient open neighborhood \(U\) of \(q\). Since \(U\) is open in the standard topology on \(\mathbb{R}\), there is an \(\varepsilon>0\) such that \((q-\varepsilon,q+\varepsilon)\subseteq U\). There are rationals \(r,s\) with \(q-\varepsilon<r<q<s<q+\varepsilon\), by the density of \(\mathbb{Q}\) in \(\mathbb{R}\). Therefore the order-open interval \((r,s)_{\mathbb{Q}}\) contains \(q\) and satisfies \((r,s)_{\mathbb{Q}}\subseteq\mathbb{Q}\cap(q-\varepsilon,q+\varepsilon)\subseteq\mathbb{Q}\cap U\).
For completeness, the density used here follows from the Archimedean property: for any real numbers \(\alpha<\beta\), choose \(n\in\mathbb{N}\) with \(n(\beta-\alpha)>1\), and choose the integer \(m=\lfloor n\alpha\rfloor+1\). Then \(n\alpha<m\le n\alpha+1<n\beta\), so \(\alpha<m/n<\beta\). Apply this first to \((q-\varepsilon,q)\) and then to \((q,q+\varepsilon)\). The local test now gives \(\mathcal{T}_{\mathrm{sub}}(\mathbb{Q})\subseteq\mathcal{T}_{\mathrm{ord}}(\mathbb{Q})\); the general theorem gives the other inclusion.
Check: The interval endpoints satisfy \(r<q<s\), and the choice of their intervals puts both endpoints strictly between \(q-\varepsilon\) and \(q+\varepsilon\). Thus the interval used really is a neighborhood of \(q\) and is contained in the prescribed ambient neighborhood.
Worked Example: The Integers in the Real Line
Let \(X=\mathbb{R}\) with its standard topology and \(A=\mathbb{Z}\). Both the order topology on \(\mathbb{Z}\) and the subspace topology inherited from \(\mathbb{R}\) are discrete, meaning that every subset is open.
For each \(n\in\mathbb{Z}\), the order interval \((n-1,n+1)_{\mathbb{Z}}\) contains exactly one integer: \(n\). Indeed, an integer \(k\) satisfying \(n-1<k<n+1\) must equal \(n\). Thus every singleton \(\{n\}\) is open in the order topology on \(\mathbb{Z}\). Every subset of \(\mathbb{Z}\) is a union of such singletons and is therefore open.
In the subspace topology, the ambient open interval \((n-\tfrac12,n+\tfrac12)\) intersects \(\mathbb{Z}\) in exactly \(\{n\}\). Every singleton is consequently subspace-open, and again every subset is open. The two topologies agree.
Check: The only integer strictly between \(n-\tfrac12\) and \(n+\tfrac12\) is \(n\), since any other integer differs from \(n\) by at least \(1\). The strict endpoints therefore do not accidentally include a neighboring integer.
The Horizontal Axis in the Lexicographic Plane
Now let \(L=\mathbb{R}\times\mathbb{R}\) with the lexicographic order: \((x,y)<(x',y')\) if either \(x<x'\), or \(x=x'\) and \(y<y'\). Give \(L\) its order topology, and let \(H=\mathbb{R}\times\{0\}\) be its horizontal axis. The restricted order on \(H\) compares \((x,0)\) and \((x',0)\) exactly as the usual order compares \(x\) and \(x'\).
Worked Example: The Horizontal Axis
We compare the two topologies on \(H\). First, the map \(\phi:\mathbb{R}\to H\) defined by \(\phi(x)=(x,0)\) is an order isomorphism: it is a bijection, and \(x<x'\) holds exactly when \(\phi(x)<\phi(x')\). An order isomorphism carries open intervals and rays to open intervals and rays, and carries the whole space to the whole space. Hence it identifies the order topology on \(\mathbb{R}\) with \(\mathcal{T}_{\mathrm{ord}}(H)\). The order topology on \(\mathbb{R}\) is its standard topology, so the order topology on \(H\) is the usual real-line topology under this identification.
In contrast, fix any \(x\in\mathbb{R}\). The interval \(((x,-1),(x,1))_L\) is open in \(L\), and \[ ((x,-1),(x,1))_L\cap H=\{(x,0)\}. \] To verify the equality, points of \(L\) whose first coordinate is less than \(x\) lie below \((x,-1)\), and points whose first coordinate is greater than \(x\) lie above \((x,1)\). Among the points with first coordinate \(x\), the interval contains those with second coordinate strictly between \(-1\) and \(1\); its only point in \(H\) is \((x,0)\). Thus every singleton of \(H\) is open in the subspace topology. Every subset of \(H\), being a union of singletons, is subspace-open. The subspace topology on \(H\) is discrete.
The usual order topology on \(H\) is not discrete. If a singleton \(\{(x,0)\}\) were order-open, it would contain a basis neighborhood of \((x,0)\). Any interval or ray basis neighborhood in this order contains another point of \(H\), because between distinct real numbers there are real numbers and there are real numbers on either side of \(x\). The whole set \(H\), also a possible basis set, contains other points as well. No such basis neighborhood can fit inside the singleton. Therefore \(\{(x,0)\}\) is not order-open, and the order topology is strictly coarser than the subspace topology.
Check: The ambient interval uses endpoints with the same first coordinate as \((x,0)\), so its intersection with \(H\) is the singleton with first coordinate exactly \(x\), not a horizontal interval around it. This verifies both the discreteness claim and the source of the difference.
A Common Mistake and Why the Distinction Matters
Common mistake. It is incorrect to assume that the topology on a subset is automatically the order topology of its restricted order. For the horizontal axis, one might reason that \((x,0)\) has nearby points \((x',0)\) with \(x'\) near \(x\), and conclude that a neighborhood in the axis must contain such points. That reasoning describes neighborhoods in the order topology on \(H\), not in the subspace topology. The actual ambient interval calculation is \(((x,-1),(x,1))_L\cap H=\{(x,0)\}\), so the subspace topology makes that point isolated.
The correct general conclusion is the inclusion \(\mathcal{T}_{\mathrm{ord}}(A)\subseteq\mathcal{T}_{\mathrm{sub}}(A)\), not automatic equality. An ambient neighborhood may become unusually small when intersected with \(A\). The rationals and integers show two ways equality can still hold: the rationals have enough points between any two distinct points to supply order neighborhoods inside ambient neighborhoods, while the integers are discrete in both topologies. The horizontal axis shows what can fail when ambient neighborhoods isolate points that are not isolated in the restricted order.
This comparison is useful whenever a subset is treated both as an ordered set and as a subspace. Before transferring a continuity or openness claim between these viewpoints, identify which topology is being used. The local equality test gives a direct method: examine ambient neighborhoods at each point of the subset and check whether each restricted neighborhood contains an order-open neighborhood there.
Check Your Understanding
For each question, keep the order topology on the subset separate from the subspace topology inherited from the ambient space.
- State the general inclusion between the two topologies and explain why an interval in the order topology on \(A\) is subspace-open.
- In \(\mathbb{R}\) with its standard topology, compare the two topologies on \(A=\mathbb{Q}\). Which property of the rationals is used in the local test?
- Show that each singleton of \(\mathbb{Z}\) is open in the order topology on \(\mathbb{Z}\), and also in its subspace topology from \(\mathbb{R}\).
- For the lexicographic plane, compute \(((x,-2),(x,3))_L\cap H\). Does this show that the subspace topology on \(H\) is discrete?
- Why does a singleton in the horizontal axis fail to be open in the order topology on \(H\), even though it is open in the subspace topology?