Turning Openness into a Calculation
In the lower limit topology, openness is tested from the left endpoint of a half-open interval: around each point, there must be a half-open interval extending to its right and staying inside the set. This makes endpoint choices important. A set can contain every point between two endpoints and still fail to be open if it includes a right endpoint with no room to its right.
Throughout, \(\mathbb{R}_\ell\) denotes the real numbers with the lower limit topology. Its basis is the collection of intervals \([a,b)\), where \(a,b\in\mathbb{R}\) and \(a<b\). Thus the topology consists of arbitrary unions of these intervals, including the empty union \(\varnothing\). A basis interval contains its left endpoint and excludes its right endpoint.
The test is local: it is enough to find one suitable interval at each point of \(U\). The interval may depend on the point. In particular, the condition does not require one fixed positive length to work everywhere in \(U\).
Proof. Suppose \(U\) is open. By the definition of the topology generated by the basis, \(U\) is a union of basis intervals. For any \(x\in U\), at least one interval in that union contains \(x\); write it as \([a,b)\). Then \(a\leq x<b\), and \([x,b)\subseteq[a,b)\subseteq U\). Conversely, suppose every \(x\in U\) has some \(b_x>x\) with \([x,b_x)\subseteq U\). Then \(U=\bigcup_{x\in U}[x,b_x)\): every interval in this union is contained in \(U\), and each \(x\in U\) belongs to its own interval. This is a union of basis intervals, so it is open. For \(U=\varnothing\), the same conclusion holds because it is the empty union. \(\square\)
Which Intervals Are Open?
Let \(a<b\). The basis test settles the common interval forms, but it helps to pay attention to what happens at each endpoint. At a point strictly to the left of \(b\), there is room to choose a short interval extending rightward and staying below \(b\). At \(b\), any basis interval containing the point extends strictly beyond \(b\).
| Set in \(\mathbb{R}_\ell\) | Open? | Reason |
|---|---|---|
| \([a,b)\) | Yes | It is a basis interval. |
| \((a,b)\) | Yes | Each point has room to the right before \(b\). |
| \([a,b]\) | No | At \(b\), every basis interval extends beyond the set. |
| \((a,b]\) | No | The same obstruction occurs at \(b\). |
For example, \((a,b)\) is a union of basis intervals: \[ (a,b)=\bigcup_{x\in(a,b)}[x,b). \] Every interval on the right lies in \((a,b)\), and each point \(x\in(a,b)\) belongs to the interval beginning at \(x\). The half-open interval \([a,b)\) is open by the definition of the basis.
In contrast, suppose \([a,b]\) were open. The basis test at \(b\) would give some \(c>b\) such that \([b,c)\subseteq[a,b]\). But \(c>b\), so there is a point strictly between \(b\) and \(c\); that point is in \([b,c)\) and not in \([a,b]\). This is impossible. The same argument rules out \((a,b]\). Notice that the issue is not whether \(b\) belongs to the set: it is whether a basis interval starting at \(b\) can remain inside it.
Worked Example: Classifying Four Intervals
Determine whether \((2,5)\), \([2,5)\), \([2,5]\), and \((2,5]\) are open in \(\mathbb{R}_\ell\).
The interval \((2,5)\). For any \(x\in(2,5)\), set \(b=(x+5)/2\). Since \(x<5\), we have \(x<b<5\), so \([x,b)\subseteq(2,5)\). The basis test shows that \((2,5)\) is open.
The interval \([2,5)\). It is a basis interval, since \(2<5\), and therefore is open.
The intervals \([2,5]\) and \((2,5]\). Both contain \(5\). Any basis interval containing \(5\) has the form \([c,d)\) with \(c\leq5<d\), and so it contains points greater than \(5\). Neither set contains such points. Thus neither set passes the basis test at \(5\), and neither is open.
Check: the two open sets have no right endpoint \(5\). The two sets that include \(5\) cannot contain any basis interval around \(5\).
Unbounded intervals can be handled in the same way. For example, \([a,\infty)\) is the union \(\bigcup_{n\in\mathbb{N}}[a,a+n)\). Also, \((-\infty,b)=\bigcup_{n\in\mathbb{N}}[b-n,b)\). These equalities follow by checking both containments: each displayed basis interval lies in the stated ray, and for any point in the ray an integer \(n\) can be chosen large enough that the point lies in the corresponding interval.
Calculating Unions of Basis Intervals
Basis intervals can overlap, leave gaps, or approach an endpoint without including it. To calculate a union, check whether each candidate point belongs to at least one interval. For a union \(\bigcup_{i\in J}[a_i,b_i)\), membership means that there is an index \(i\in J\) with \(a_i\leq x<b_i\). The strict inequality at the right endpoint is essential.
Proof. A point lies in the intersection exactly when it is at least \(a\) and \(c\), and strictly less than both \(b\) and \(d\). The first pair of inequalities is equivalent to \(x\geq\max(a,c)\), and the second is equivalent to \(x<\min(b,d)\). If \(\max(a,c)\geq\min(b,d)\), no point can satisfy both inequalities. If \(\max(a,c)<\min(b,d)\), the points satisfying them are precisely the interval stated. \(\square\)
Worked Example: A Countable Union Equal to an Interval
For \(n\in\mathbb{N}\), let \(U=\bigcup_{n\in\mathbb{N}}[1/n,1)\). Determine \(U\) and decide whether it is open in \(\mathbb{R}_\ell\).
Every interval \([1/n,1)\) is contained in \((0,1)\): its points satisfy \(1/n\leq x<1\), and \(1/n>0\). Thus \(U\subseteq(0,1)\).
For the reverse inclusion, take any \(x\in(0,1)\). Since \(x>0\), the Archimedean property gives \(n\in\mathbb{N}\) with \(n\geq1/x\). Multiplying by the positive number \(x/n\), or equivalently taking reciprocals of positive quantities, gives \(1/n\leq x\). Together with \(x<1\), this shows \(x\in[1/n,1)\subseteq U\). Therefore \(U=(0,1)\). As a union of basis intervals, \(U\) is open.
Check: \(0\) is not in any interval because \(1/n>0\); \(1\) is not in any interval because every right endpoint is excluded. Every \(x\) with \(0<x<1\) was shown to lie in at least one interval.
Worked Example: A Countable Union Covering the Real Line
Let \(V=\bigcup_{q\in\mathbb{Q}}[q,q+1)\), where \(\mathbb{Q}\) is the set of rational numbers. Calculate \(V\).
First, every interval in the union is a subset of \(\mathbb{R}\), so \(V\subseteq\mathbb{R}\). Now fix any \(x\in\mathbb{R}\). By the density of the rational numbers, there is a rational \(q\) with \(x-1<q<x\). The first inequality gives \(x<q+1\), and the second gives \(q\leq x\). Hence \(q\leq x<q+1\), which means \(x\in[q,q+1)\). Since \(q\) is among the indices in the union, \(x\in V\). Thus \(V=\mathbb{R}\), which is open.
Check: the covering inequality is strict at the right endpoint, \(x<q+1\), and weak at the left endpoint, \(q\leq x\). These are exactly the membership conditions for \([q,q+1)\).
Countable Sets Need Not Be Open
A countable family of basis intervals can have a large union, as the preceding example shows. That fact does not mean that a countable set of points is open. To be open, each point of the set must have an entire basis interval to its right contained in the set. The next examples apply that test directly.
Worked Example: A Reciprocal Sequence Is Not Open
Let \(S=\{1/n:n\in\mathbb{N}\}\). Determine whether \(S\) is open in \(\mathbb{R}_\ell\).
The point \(1=1/1\) belongs to \(S\). If \(S\) were open, the basis test would provide some \(b>1\) such that \([1,b)\subseteq S\). Choose \(y=(1+b)/2\). Since \(1<b\), we have \(1<y<b\), so \(y\in[1,b)\). But every element of \(S\) is at most \(1\), whereas \(y>1\). Therefore \(y\notin S\), contradicting \([1,b)\subseteq S\). Hence \(S\) is not open.
Check: the failure occurs at the member \(1\), not at a missing limit point. Any interval beginning at \(1\) and extending a positive distance to the right contains points larger than \(1\).
Worked Example: The Rational Numbers Are Not Open
Determine whether \(\mathbb{Q}\) is open in \(\mathbb{R}_\ell\).
Take any \(q\in\mathbb{Q}\). If \(\mathbb{Q}\) were open, there would be some \(\varepsilon>0\) with \([q,q+\varepsilon)\subseteq\mathbb{Q}\). Choose a rational number \(r\) with \(0<r<\varepsilon/\sqrt{2}\), using the density of the rationals. Then \(0<r\sqrt{2}<\varepsilon\). The number \(q+r\sqrt{2}\) lies in \([q,q+\varepsilon)\), but it is irrational: \(r\) is a nonzero rational and \(\sqrt{2}\) is irrational, so \(r\sqrt{2}\), and then its sum with the rational number \(q\), is irrational. This contradicts the assumed containment. Thus \(\mathbb{Q}\) is not open.
Check: the constructed point is strictly to the right of \(q\) and strictly less than \(q+\varepsilon\), so it satisfies both endpoint conditions for membership in \([q,q+\varepsilon)\).
A Common Endpoint Mistake
Common pitfall. One might try to represent \([0,1]\) as \(\bigcup_{x\in[0,1]}[x,1)\), reasoning that intervals starting at every point should cover the whole closed interval. But each interval \([x,1)\) excludes \(1\). Their union is exactly \([0,1)\), not \([0,1]\): every point \(y<1\) in \([0,1]\) belongs to the interval starting at \(x=y\), while \(1\) belongs to none of them. The proposed equality is therefore a wrong calculation. Moreover, \([0,1]\) is not open, since no basis interval containing \(1\) stays inside \([0,1]\).
When calculating a union, test endpoints separately rather than inferring them from the points between. The basis test also gives a reliable procedure: choose a point in the proposed set, identify an interval \([x,b)\) contained in it, and check both \(x\leq y\) and \(y<b\) for any point \(y\) whose membership is at issue.
Check Your Understanding
For each question, keep track of whether the right endpoint of a basis interval is included.
- Is \((3,7]\) open in \(\mathbb{R}_\ell\)? Justify your answer at the endpoint \(7\).
- Calculate \(\bigcup_{n\in\mathbb{N}}[2,2+n)\).
- Is \(\mathbb{Z}\) open in \(\mathbb{R}_\ell\)? Test the basis condition at \(0\).
- Show that \(\bigcup_{q\in\mathbb{Q}}[q,q+2)\) contains every real number.
- For \(a<b\), explain why \((a,b)\) is open even though it is not itself a basis interval.