Ordinary Intervals and a Deleted Sequence
To specify a topology from a basis, one must check that the proposed sets cover the space and that intersections can be refined around each of their points. For the K-topology, the family combines ordinary open intervals with intervals from which a particular sequence has been removed. The point \(0\) is a useful test case: it is not in the sequence, but every interval around it contains terms of the sequence.
Throughout, \(\mathbb{R}\) has its standard topology when we refer to ordinary open intervals. Set \(K=\{1/n:n\in\mathbb{N}\}\), where \(\mathbb{N}=\{1,2,\ldots\}\). Thus \(K=\{1,1/2,1/3,1/4,\ldots\}\), and \(0\notin K\).
In this definition, the deleted interval retains precisely those points of \((a,b)\) that are not in \(K\). For example, \((0,1)\setminus K\) omits \(1/n\) for every integer \(n\geq 2\); the point \(1\) is not in \((0,1)\) to begin with. The definition does not say that deleted intervals are ordinary open intervals. Rather, it proposes a collection of sets and requires a basis check before using them to generate a topology.
The Basis Criterion
Proof. First, each set \((a,b)\setminus K\) is nonempty: an ordinary nondegenerate interval is uncountable, whereas \(K\) is countable. Ordinary intervals are nonempty because \(a<b\). For every \(x\in\mathbb{R}\), the ordinary interval \((x-1,x+1)\) belongs to \(\mathcal{B}_K\) and contains \(x\). This verifies the covering condition.
For the intersection condition, take \(B_1,B_2\in\mathcal{B}_K\) and \(x\in B_1\cap B_2\). Write each \(B_i\) as \(I_i\) or \(I_i\setminus K\), where \(I_i\) is an ordinary open interval. The intersection \(J=I_1\cap I_2\) is an ordinary open interval containing \(x\): the intersection of two open intervals, when nonempty, is an open interval. If neither \(B_1\) nor \(B_2\) deletes \(K\), then \(J\in\mathcal{B}_K\), and \(x\in J\subseteq B_1\cap B_2\). If at least one deletes \(K\), membership of \(x\) in that set implies \(x\notin K\). Thus \(J\setminus K\in\mathcal{B}_K\), \(x\in J\setminus K\), and \(J\setminus K\subseteq B_1\cap B_2\). In both cases the required basis element exists.
For completeness, the criterion really does make the unions a topology. The empty set is the empty union, and \(\mathbb{R}\) is a union of basis elements because they cover \(\mathbb{R}\). Any union of unions of basis elements is again a union of basis elements. Finally, if \(U\) and \(V\) are unions of basis elements and \(x\in U\cap V\), choose \(B_1\subseteq U\) and \(B_2\subseteq V\) from the respective unions with \(x\in B_1\cap B_2\). The intersection condition supplies \(B_3\in\mathcal{B}_K\) with \(x\in B_3\subseteq U\cap V\). Doing this at each point shows that \(U\cap V\) is a union of basis elements. Hence the unions satisfy the topology axioms. \(\square\)
Computing Basis Refinements
Worked Example: Intersecting Two Deleted Intervals
Let \(B_1=(0,2/3)\setminus K\) and \(B_2=(1/3,1)\setminus K\). Compute their intersection and check membership of \(2/5\).
The ordinary intervals intersect in \((1/3,2/3)\), so \(B_1\cap B_2=(1/3,2/3)\setminus K\). To identify the deleted points in this interval, solve \(1/3<1/n<2/3\). Since \(n\) is a positive integer, the left inequality gives \(n<3\), and the right gives \(n>3/2\). The only possibility is \(n=2\). Thus the intersection is \((1/3,2/3)\setminus\{1/2\}\).
For the membership check, \(1/3<2/5\) because \(5<6\), and \(2/5<2/3\) because \(6<10\). Also \(2/5\notin K\): an equality \(2/5=1/n\) would give \(n=5/2\), which is not a positive integer. Hence \(2/5\in B_1\cap B_2\). The computed intersection is itself a basis element, namely \((1/3,2/3)\setminus K\), which checks the refinement requirement for this pair.
Worked Example: A Common Point at Zero
Take \(B_1=(-1,1)\setminus K\) and \(B_2=(-1/2,1/2)\). Find a basis element containing \(0\) and contained in \(B_1\cap B_2\).
Since \(0\notin K\), we have \(0\in B_1\), and \(0\in B_2\) because \(-1/2<0<1/2\). The underlying interval intersection is \((-1/2,1/2)\). Because \(B_1\) deletes \(K\), the intersection is \[ B_1\cap B_2=(-1/2,1/2)\setminus K. \] This set is a member of \(\mathcal{B}_K\). It contains \(0\), and it is exactly the intersection, so it is in particular contained in the intersection.
There is no contradiction between \(0\notin K\) and the deletion of sequence terms arbitrarily close to \(0\). For instance, \(1/n\) lies in \((0,1/2)\) whenever \(n>2\), so \(B_1\cap B_2\) omits all those points while retaining \(0\). The basis condition asks for a basis element containing the common point; it does not require that this element contain every nearby real number.
Worked Example: One Deleted and One Ordinary Interval
Let \(B_1=(0,1)\setminus K\), \(B_2=(1/5,2/5)\), and \(x=2/7\). Check that \(x\) is in the intersection, then identify the intersection as a basis element.
First, \(0<2/7<1\). For the second interval, \(1/5<2/7\) follows from \(7<10\), and \(2/7<2/5\) follows from \(10<14\). Also \(2/7\notin K\), since \(2/7=1/n\) would require \(n=7/2\). Thus \(x\in B_1\cap B_2\).
The ordinary interval intersection is \((1/5,2/5)\). The terms \(1/n\) in it satisfy \(1/5<1/n<2/5\), equivalently \(5/2<n<5\). The positive integers satisfying these strict inequalities are \(n=3\) and \(n=4\). Therefore \[ B_1\cap B_2=(1/5,2/5)\setminus\{1/3,1/4\}=(1/5,2/5)\setminus K. \] The last equality holds because no other members of \(K\) lie in \((1/5,2/5)\). This is a basis element, and it contains \(2/7\) as checked above.
A Common Pitfall
Common mistake. One might claim that the intersection of two basis elements must be an ordinary open interval. For example, the incorrect calculation \[ \bigl((-1,1)\setminus K\bigr)\cap(-1,1)=(-1,1) \] drops the deletion. The right-hand side contains \(1/2\), but \(1/2\in K\), so \(1/2\notin(-1,1)\setminus K\). The calculation is false.
The correct calculation is \[ \bigl((-1,1)\setminus K\bigr)\cap(-1,1)=(-1,1)\setminus K. \] This set cannot be an ordinary open interval: it contains \(0\), so any ordinary open interval equal to it would contain some interval \((-\delta,\delta)\) with \(\delta>0\). Choose \(n\in\mathbb{N}\) large enough that \(1/n<\delta\); then \(1/n\) would belong to that interval, although \(1/n\) is excluded from \((-1,1)\setminus K\). The basis criterion does not require intersections themselves to be ordinary intervals. It requires a basis element around each point of the intersection, and a deleted interval can supply it.
The key point in the criterion is the membership of the particular point \(x\). If a common point lies in a deleted interval, it cannot belong to \(K\); this is exactly why deleting \(K\) from the smaller interval still leaves a basis element containing \(x\). This local check, rather than a claim that all intersections have one prescribed shape, is what establishes the K-topology.
Check Your Understanding
Use the basis criterion and the definition of \(K\) to answer these questions.
- Find \((0,1/2)\setminus K\) by listing the elements of \(K\) that lie strictly inside \((0,1/2)\).
- For \(B_1=(-1,2)\setminus K\) and \(B_2=(0,1)\), give a basis element equal to \(B_1\cap B_2\).
- Explain why \(0\) belongs to \((-\delta,\delta)\setminus K\) for every \(\delta>0\).
- For \(B_1=(1/4,1)\setminus K\) and \(B_2=(1/3,2)\), compute their intersection as an interval with the deleted points specified.
- State the two conditions that a collection must satisfy to be a basis.