Two Topologies on the Same Plane
The set \(\mathbb{R}^2\) carries more than one natural topology. In the product topology, neighborhoods of a point are built from open intervals in each coordinate. In the lexicographic order topology, neighborhoods come from intervals in the dictionary order: first compare first coordinates, and compare second coordinates only when the first coordinates agree. The two constructions give different collections of open sets. We will prove that every product-open set is open in the lexicographic topology, and then exhibit a lexicographically open set that is not product-open.
Throughout, both topologies are on the same set \(\mathbb{R}^2\). The product topology uses the standard topology on each copy of \(\mathbb{R}\). The lexicographic order is defined by \((x,y)<_{\mathrm{lex}}(u,v)\) when either \(x<u\), or \(x=u\) and \(y<v\). The order topology is the topology generated by the open intervals and open rays for this order.
For the product-topology description, these rectangles form a basis: every product-open set is a union of such rectangles. For the order topology, every interval of the form \((p,q)_{\mathrm{lex}}=\{z\in\mathbb{R}^2:p<_{\mathrm{lex}}z<_{\mathrm{lex}}q\}\) is open. We will use a particularly simple kind of order interval, whose endpoints have the same first coordinate.
A Vertical Slice Is a Lexicographic Interval
Proof. Write \(z=(x,y)\). If \(x<a\), then \(z<_{\mathrm{lex}}(a,b-\varepsilon)\), so \(z\) is not in the interval. If \(x>a\), then \(z>_{\mathrm{lex}}(a,b+\varepsilon)\), so again it is not in the interval. When \(x=a\), the defining inequalities for membership become \(b-\varepsilon<y<b+\varepsilon\). Thus membership in the interval is equivalent to \(x=a\) and \(b-\varepsilon<y<b+\varepsilon\), which is precisely membership in the displayed vertical slice. Since an open interval in the order topology is open, the slice is open. \(\square\)
This feature is the key to comparing the topologies. A product rectangle gives room to vary both coordinates around each of its points. A lexicographic interval with endpoints on the same vertical line varies only the second coordinate, but it is still an open neighborhood in the order topology. The existence of these thin open neighborhoods lets us fit a lexicographic neighborhood inside any product-open set.
Every Product-Open Set Is Lexicographically Open
Proof. Let \(O\) be product-open, and take any point \((a,b)\in O\). Because product rectangles form a basis, there are open intervals \(I,J\subseteq\mathbb{R}\) such that \((a,b)\in I\times J\subseteq O\). Since \(J\) is open and contains \(b\), there is an \(\varepsilon>0\) such that \((b-\varepsilon,b+\varepsilon)\subseteq J\). The proposition gives the lexicographically open neighborhood \(\big((a,b-\varepsilon),(a,b+\varepsilon)\big)_{\mathrm{lex}}=\{a\}\times(b-\varepsilon,b+\varepsilon)\) of \((a,b)\). This neighborhood is contained in \(I\times J\): its first coordinate is \(a\in I\), and all its second coordinates lie in \(J\). Therefore it is contained in \(O\). We have found a lexicographically open neighborhood inside \(O\) for every point of \(O\), so \(O\) is lexicographically open. If \(O\) is empty, it is open in both topologies by the topology axioms, so the conclusion holds in that case as well. \(\square\)
The argument proves the required inclusion of open sets directly: \(\mathcal{T}_{\mathrm{product}}\subseteq\mathcal{T}_{\mathrm{lex}}\). It does not say that a product rectangle itself is a lexicographic interval. Instead, the proof finds a suitable lexicographic neighborhood at each point of the rectangle. The distinction matters because an order interval whose endpoints have different first coordinates generally contains entire vertical fibers between them.
Worked Example: Finding a Lexicographic Neighborhood in a Rectangle
Let \(O=(-2,4)\times(-3,1)\) and \(p=(1,-1)\). We verify directly that a lexicographic open neighborhood of \(p\) lies inside \(O\).
Choose \(\varepsilon=\tfrac12\). The endpoints on the vertical line with first coordinate \(1\) are \((1,-1-\tfrac12)=(1,-\tfrac32)\) and \((1,-1+\tfrac12)=(1,-\tfrac12)\). The resulting lexicographic interval is \(\big((1,-\tfrac32),(1,-\tfrac12)\big)_{\mathrm{lex}}=\{1\}\times(-\tfrac32,-\tfrac12)\). Its first coordinate is \(1\), and \(-2<1<4\). Its second coordinates satisfy \(-3<-\tfrac32<y<-\tfrac12<1\). Hence the entire interval lies in \(O\), and it contains \(p\).
Check: the interval contains \((1,-1)\) because \(-\tfrac32<-1<-\tfrac12\). Every point in it has first coordinate \(1\in(-2,4)\) and second coordinate in \((-\tfrac32,-\tfrac12)\subset(-3,1)\), so it is indeed contained in \(O\).
Worked Example: A Point in an Open Disk
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2<1\}\), which is product-open, and consider \(p=(\tfrac35,0)\). We construct an explicit lexicographic neighborhood of \(p\) inside \(D\).
First, the product rectangle \(R=(\tfrac12,\tfrac7{10})\times(-\tfrac15,\tfrac15)\) contains \(p\): \(\tfrac12<\tfrac35<\tfrac7{10}\) and \(-\tfrac15<0<\tfrac15\). For any \((x,y)\in R\), we have \(0<x<\tfrac7{10}\) and \(|y|<\tfrac15\). Therefore \(x^2+y^2<(\tfrac7{10})^2+(\tfrac15)^2=\tfrac{49}{100}+\tfrac4{100}=\tfrac{53}{100}<1\). This proves \(R\subseteq D\).
Now use the lexicographic interval \(\big((\tfrac35,-\tfrac15),(\tfrac35,\tfrac15)\big)_{\mathrm{lex}}=\{\tfrac35\}\times(-\tfrac15,\tfrac15)\). It contains \(p\) and is contained in \(R\), because its first coordinate is \(\tfrac35\in(\tfrac12,\tfrac7{10})\). Thus it is a lexicographically open neighborhood of \(p\) inside the disk.
Check: a point \((\tfrac35,y)\) in this interval satisfies \(|y|<\tfrac15\), so \((\tfrac35)^2+y^2<\tfrac9{25}+\tfrac1{25}=\tfrac{10}{25}=\tfrac25<1\). Thus every point of the chosen interval is in \(D\), as required.
The Inclusion Is Strict
Proof. The preceding theorem proves that every product-open set is lexicographically open. To show that the inclusion is strict, consider \(V=\{2\}\times(0,3)\). By the proposition, \(V=\big((2,0),(2,3)\big)_{\mathrm{lex}}\), so \(V\) is open in the lexicographic order topology.
We show that \(V\) is not product-open. The point \(p=(2,\tfrac32)\) belongs to \(V\). Every product-open neighborhood of \(p\) contains a product rectangle \(I\times J\) with \(2\in I\) and \(\tfrac32\in J\). Since \(I\) is an open interval containing \(2\), it contains some \(x\ne2\); for example, if \(I=(2-\delta,2+\delta)\) for some \(\delta>0\), then \(2+\tfrac{\delta}{2}\in I\) and \(2+\tfrac{\delta}{2}\ne2\). The point \((2+\tfrac{\delta}{2},\tfrac32)\) then belongs to \(I\times J\), but not to \(V\), whose points must have first coordinate exactly \(2\). Thus no product-open neighborhood of \(p\) is contained in \(V\), so \(V\) is not product-open. Together with the inclusion already proved, this establishes strict fineness. \(\square\)
Worked Example: Testing a Vertical Segment
Take \(V=\{2\}\times(0,3)\) and \(p=(2,\tfrac32)\). In the lexicographic topology, \(\big((2,0),(2,3)\big)_{\mathrm{lex}}=\{2\}\times(0,3)=V\), so one open interval gives the whole set as a lexicographically open set.
For a product neighborhood test, let \(I\times J\) be any product rectangle containing \(p\). Because \(I\) is open and contains \(2\), there is a \(\delta>0\) with \((2-\delta,2+\delta)\subseteq I\). Set \(x=2+\tfrac{\delta}{2}\). Then \(2<x<2+\delta\), so \(x\in I\), and \(\tfrac32\in J\). Hence \((x,\tfrac32)\in I\times J\). But \(x\ne2\), so \((x,\tfrac32)\notin V\). This works for every product rectangle around \(p\), proving that none is contained in \(V\).
Check: \(0<\tfrac32<3\), so \(p\in V\). The point \((2+\tfrac{\delta}{2},\tfrac32)\) has the required second coordinate but not the required first coordinate, which verifies exactly why it lies outside \(V\).
A Common Mistake and the Main Comparison
Common pitfall. One might try to find a lexicographic interval crossing several first coordinates and assume it stays inside a product rectangle. For example, consider the rectangle \(Q=(0,2)\times(-1,1)\) and the lexicographic interval \(\big((0,0),(2,0)\big)_{\mathrm{lex}}\). The point \((1,100)\) belongs to that interval because \(0<1<2\), so it lies lexicographically between the endpoints regardless of its second coordinate. But \((1,100)\notin Q\), since \(100\notin(-1,1)\). Thus this interval is not contained in \(Q\). The correction is to use endpoints with the same first coordinate as the point under consideration. That interval varies only the second coordinate and can be chosen small enough to fit inside the product rectangle.
The comparison can be summarized as follows.
| Question | Product topology | Lexicographic order topology |
|---|---|---|
| Typical basic neighborhood | A rectangle \(I\times J\) | An order interval |
| Can a vertical slice \(\{a\}\times(b-\varepsilon,b+\varepsilon)\) be open? | No, when \(\varepsilon>0\) | Yes; it is an interval with endpoints sharing first coordinate \(a\) |
| Which topology has more open sets? | Every product-open set is lex-open | Strictly more: for example, \(\{2\}\times(0,3)\) is not product-open |
The example also explains why the two topologies should not be identified merely because they use the same set of points. In the product topology, a neighborhood of a point must allow variation in the first coordinate as well as the second. In the lexicographic topology, a vertical slice can itself be open, with no variation in the first coordinate. This local difference produces the proper inclusion of their collections of open sets.
Check Your Understanding
Use the definitions and the neighborhood argument above to answer the following questions.
- Write the lexicographic open interval with endpoints \((a,b-\varepsilon)\) and \((a,b+\varepsilon)\) as a subset of \(\mathbb{R}^2\).
- Why does every product-open set contain a lexicographically open neighborhood of each of its points?
- Show that \(\{0\}\times(1,4)\) is open in the lexicographic topology.
- Prove directly that \(\{0\}\times(1,4)\) is not open in the product topology.
- In the common-pitfall example, why does \((1,100)\) lie between \((0,0)\) and \((2,0)\) in lexicographic order?