The Basis That Defines the K-Topology
Two topologies on the same set can have different open sets, even when they use the same underlying points. To compare them, it is useful to work from their bases: a topology is finer when it has more open sets. The K-topology gives a particularly clear example. Its basis includes the usual open intervals, but it also includes intervals with a specified sequence of points removed.
Let \(K=\{1/n:n\in\mathbb{N}\}\subseteq\mathbb{R}\), where \(\mathbb{R}\) has its usual topology when no other topology is specified. Write \(\mathbb{R}_K\) for the real line with the K-topology.
For completeness, this collection satisfies the basis conditions. Every real number belongs to some interval \((a,b)\), so the basis covers \(\mathbb{R}\). For the intersection condition, take two basis elements whose intersection contains \(x\). Their underlying intervals intersect in an open interval \((c,d)\) containing \(x\). If neither element has \(K\) removed, their intersection is \((c,d)\); if at least one has \(K\) removed, their intersection is \((c,d)\setminus K\). In either case the intersection is itself a basis element containing \(x\).
The usual topology on \(\mathbb{R}\) has the open intervals \((a,b)\) as a basis. Thus the two bases have some members in common. The comparison question is whether the extra sets \((a,b)\setminus K\) add genuinely new open sets, or whether they can all be built from usual-open sets.
Every Usual-Open Set Is K-Open
Proof. Let \(U\) be usual-open. By the interval-basis description of the usual topology, for each \(x\in U\) there are \(a,b\in\mathbb{R}\) such that \(a<x<b\) and \((a,b)\subseteq U\). Every interval \((a,b)\) belongs to \(\mathcal{B}_K\), so it is K-open. Taking the union of these intervals over all \(x\in U\) expresses \(U\) as a union of K-open basis elements. Hence \(U\) is K-open. If \(U=\varnothing\), it is K-open as the empty union. \(\square\)
This argument uses the bases in the direction that matters: each usual basis interval is also a K-basis element. It does not require every K-basis element to be usual-open. In particular, proving containment does not yet tell us whether the two topologies are equal.
Worked Example: An Interval to the Left of Zero
Consider the K-basis element \((-2,0)\setminus K\). Every element \(1/n\) of \(K\) is positive, since \(n\in\mathbb{N}\) implies \(n\geq 1\) and therefore \(1/n>0\). Thus no element of \(K\) lies in \((-2,0)\), and $$ (-2,0)\setminus K=(-2,0). $$ The result is a usual-open interval and, by the theorem, is K-open. This example shows that writing “\(\setminus K\)” does not by itself make a basis element different from a usual interval: the removed set may not meet the interval at all.
Check: the left endpoint is \(-2\) and the right endpoint is \(0\), so every point of \((-2,0)\) is negative. Since each \(1/n\) is positive, the intersection \((-2,0)\cap K\) is empty, as required.
A Basic Open Set That Is Not Usual-Open
The important difference appears in an interval containing zero. Although \(0\notin K\), the points \(1/n\) lie arbitrarily close to \(0\). Removing \(K\) from an interval around zero leaves zero in the set while removing points close to it. The next proposition makes this failure of usual openness precise.
Proof. The set is a member of \(\mathcal{B}_K\), so it is K-open by the definition of the K-topology. Also \(0\in(a,b)\), and \(0\notin K\), so \(0\in(a,b)\setminus K\). Suppose this set were usual-open. Then, because it contains \(0\), there would be some \(\delta>0\) such that \((-\delta,\delta)\subseteq(a,b)\setminus K\). Choose \(n\in\mathbb{N}\) with \(n>1/\delta\), which is possible because the positive integers are unbounded. Then \(0<1/n<\delta\), so \(1/n\in(-\delta,\delta)\). But \(1/n\in K\), and hence \(1/n\notin(a,b)\setminus K\). This contradicts the proposed containment. Therefore \((a,b)\setminus K\) is not usual-open. \(\square\)
Worked Example: A Deleted Interval Around Zero
Take the K-basis element \(G=(-1,1)\setminus K\). Since \(-1<0<1\), the proposition applies: \(G\) is K-open but not usual-open. To see the obstruction with a specific neighborhood test, suppose a usual interval \((-\delta,\delta)\) around zero were contained in \(G\), where \(\delta>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\delta\). Then \(1/n<\delta\), and \(1/n>0>-\delta\), so \(1/n\in(-\delta,\delta)\). At the same time, \(1/n\in K\), so it is excluded from \(G\). No such interval can be contained in \(G\).
Check: zero is in \((-1,1)\), and zero is not any \(1/n\), so \(0\in G\). Every usual-open set containing zero must contain some interval around zero, but the argument just given shows each such interval contains a point removed from \(G\). This verifies the failure of usual openness directly from the definition.
Proof. The theorem shows that every usual-open set is K-open, so the usual topology is contained in the K-topology. The set \((-1,1)\setminus K\) is K-open but not usual-open by the proposition, so the containment is strict. \(\square\)
“Strictly finer” refers to inclusion of the collections of open sets, not to a change in the underlying set: both topologies have underlying set \(\mathbb{R}\). Nor does the conclusion say that every K-basis element fails to be usual-open. The next example gives a deleted basis element that is usual-open as well.
Worked Example: Deleting the Points Inside a Positive Interval
Consider \(H=(1/3,2)\setminus K\). To find exactly which points are removed, solve \(1/3<1/n<2\) for \(n\in\mathbb{N}\). The inequality \(1/n<2\) holds for every \(n\geq1\), since \(1/n\leq1<2\). The inequality \(1/3<1/n\) is equivalent, after multiplying by the positive number \(3n\), to \(n<3\). Thus the possible integers are \(n=1,2\), and the points removed are \(1\) and \(1/2\). The endpoint \(1/3\) is not in the interval, and it is not removed from the interval.
Consequently, $$ H=(1/3,1/2)\cup(1/2,1)\cup(1,2). $$ Each of these three intervals is usual-open, so their union is usual-open. It is K-open too, since \(H\) is a K-basis element. This contrasts with the interval around zero: deleting \(K\) does not always destroy usual openness.
Check: the removed points satisfy \(1/2\in(1/3,2)\) and \(1\in(1/3,2)\). The next point of \(K\) is \(1/3\), which is the excluded left endpoint; for \(n\geq3\), \(1/n\leq1/3\), so no other point of \(K\) lies in the interval. The displayed union therefore removes exactly the required points.
Why the Location of Zero Matters
The examples are explained by the placement of \(K\): its points approach zero, while away from zero they do not accumulate. The proposition exploits the first fact. The positive-interval example is consistent with the second: around any point \(x>0\), only finitely many members of \(K\) can lie above \(x/2\), because \(1/n\geq x/2\) implies \(n\leq2/x\). This is why removing \(K\) from a positive interval can leave an ordinary open set. The distinction is local: what happens near zero controls whether a deleted interval containing zero can be usual-open.
Common pitfall. A tempting but incorrect argument says that \((a,b)\setminus K\) is usual-open because it is an interval with some points removed. For \(G=(-1,1)\setminus K\), that reasoning overlooks the removed points \(1/n\) arbitrarily close to the retained point \(0\). In particular, \(0\in G\), but every usual interval around zero contains some \(1/n\notin G\). The correction is to test openness at every point of the proposed set, especially at zero; being a K-basis element establishes K-openness, not usual openness.
For this comparison, the reliable procedure is to keep the two bases distinct. To prove usual-open sets are K-open, express them as unions of ordinary intervals, which are also K-basis elements. To disprove usual openness of a particular K-open set, find one of its points—here zero—at which no usual interval fits inside the set. Together these arguments establish both the inclusion and its strictness without confusing the two notions of openness.
Check Your Understanding
For each question, distinguish carefully between being K-open and being usual-open.
- Why does the fact that every usual interval belongs to \(\mathcal{B}_K\) imply that every usual-open set is K-open?
- Show directly that \((-3,0)\setminus K\) is usual-open.
- For \((-2,2)\setminus K\), identify a point that prevents the set from being usual-open and explain why.
- Find all points of \(K\) lying in \((1/4,1)\), and write \((1/4,1)\setminus K\) as a union of usual-open intervals.
- In the proof of strictness, where is the fact that \(1/n\) approaches zero used?