What a Neighborhood Can Look Like
In the K-topology, an ordinary interval is still open, but there are also basic open sets that remove a particular sequence from an interval. The key to testing a set is to check its points one at a time: a basis element containing the point must also fit inside the set. This check has a special feature at points of the sequence, because a deleted interval cannot contain any of them.
Throughout, \(\mathbb{R}\) has the K-topology described below, and intervals such as \((a,b)\) refer to ordinary real intervals. Set \(K=\{1/n:n\in\mathbb{N}\}\), where \(\mathbb{N}=\{1,2,\ldots\}\). Thus \(0\notin K\), and the points \(1/n\) approach \(0\) as \(n\) increases.
These sets do form a basis. They cover \(\mathbb{R}\), since every real number belongs to an ordinary open interval. To check intersections, the intersection of two ordinary intervals is an interval or the empty set. Intersecting an ordinary interval with a deleted interval gives an interval with \(K\) removed, or the empty set. Intersecting two deleted intervals also gives an interval with \(K\) removed, or the empty set. These are again basis elements (with the empty set allowed as an intersection). Thus the basis criterion applies.
Proof. Suppose \(U\) is open. By the definition of the topology generated by a basis, for each \(x\in U\) there is a basis element \(B\in\mathcal{B}\) with \(x\in B\subseteq U\). The basis element is either an ordinary interval or an interval with \(K\) removed, giving one of the two stated possibilities. Conversely, if one of these possibilities holds at every \(x\in U\), then every point of \(U\) lies in a basis element contained in \(U\). The basis criterion therefore says that \(U\) is open. For \(U=\varnothing\), the condition holds vacuously and \(\varnothing\) is open. \(\square\)
At \(x\in K\), no deleted interval can contain \(x\), since every set \((a,b)\setminus K\) omits all of \(K\). Consequently, an open set containing \(x\) must contain an ordinary interval around \(x\). At \(0\), which is not in \(K\), deleted intervals can be used: whenever \(a<0<b\), the basis element \((a,b)\setminus K\) contains \(0\). This distinction is the main tool in the calculations below.
Ordinary and Half-Open Intervals
Worked Example: Testing Interval Endpoints
Determine which of \((0,1)\), \([0,1)\), and \((0,1]\) are open in \(\mathbb{R}_K\).
The interval \((0,1)\). It is itself a basis element, so it is open. Equivalently, each \(x\in(0,1)\) belongs to the basis element \((0,1)\subseteq(0,1)\).
The interval \([0,1)\). The point \(0\) belongs to this set but is not in \(K\). Any basis element containing \(0\), whether an ordinary interval or a deleted interval, has the form of an interval around \(0\), possibly with \(K\) removed. In either case it contains negative real numbers: if \(a<0<b\), then \(a/2\in(a,b)\), and \(a/2\notin K\) because it is negative. Thus no basis element containing \(0\) is contained in \([0,1)\). This set is not open.
The interval \((0,1]\). The point \(1\in K\) belongs to this set. A deleted interval cannot contain \(1\), so any basis element containing \(1\) must be an ordinary interval \((a,b)\) with \(a<1<b\). Such an interval contains points greater than \(1\), and therefore cannot be contained in \((0,1]\). This set is not open.
Check: The two failures occur at the included endpoints: at \(0\), every available basis element extends to the left; at \(1\in K\), every available basis element extends to the right. Removing \(K\) from a basis interval does not change the endpoint test at \(0\), since there are still negative points in that deleted interval.
More generally, every ordinary open interval \((a,b)\) is open because it is a basis element. A half-open interval with an included endpoint fails the same local test at that endpoint: any basis element containing it extends past it. For an endpoint in \(K\), only ordinary intervals can serve; for an endpoint outside \(K\), deleted intervals still extend to both sides. This reasoning concerns openness, not whether a set is closed; those are separate properties.
Sets Obtained by Deleting the Sequence
Worked Example: An Interval with the Sequence Deleted
Let \(V=(-1,1)\setminus K\). Verify directly that \(V\) is open and check what happens at \(0\).
The set \(V\) is exactly the basis element \((-1,1)\setminus K\), so it is open. In particular, \(0\in V\), because \(0\notin K\). The basis element \((-1,1)\setminus K\) itself is an open neighborhood of \(0\) contained in \(V\). If \(x\in V\) is any other point, the same basis element contains \(x\) and remains contained in \(V\).
Check: The deletion is set subtraction: \(V\) contains every point of \((-1,1)\) except the points \(1/n\) that lie in that interval. For example, \(1/2\in K\) and \(0<1/2<1\), so \(1/2\notin V\); whereas \(0\notin K\), so \(0\in V\).
A set formed by deleting only part of \(K\) may also be open, but its included points in \(K\) need their own ordinary intervals. The following example checks this point as well as the behavior at \(0\).
Worked Example: Deleting a Tail but Keeping One Sequence Point
Let \(W=\mathbb{R}\setminus\{1/n:n\ge2\}\). Determine whether \(W\) is open.
First, \(W=(\mathbb{R}\setminus K)\cup\{1\}\): the only point of \(K\) not deleted is \(1=1/1\). At \(0\in W\), the basis element \((-1,1)\setminus K\) is contained in \(W\). Indeed, it excludes every point of \(K\), so it cannot contain any of the deleted points \(1/n\) for \(n\ge2\).
At \(1\in W\), use the ordinary interval \((3/4,5/4)\). Every deleted point satisfies \(1/n\le1/2\) when \(n\ge2\), so none lies in this interval. Hence \((3/4,5/4)\subseteq W\), and this interval is a basis element containing \(1\).
Every other point of \(W\) lies in \(\mathbb{R}\setminus K\). To verify the required neighborhoods there, consider \(x\in\mathbb{R}\setminus K\). If \(x<0\), the interval \((x-|x|/2,x+|x|/2)\) is entirely negative and so avoids \(K\). If \(x=0\), use \((-1,1)\setminus K\). If \(x>0\), choose an integer \(N\ge2\) with \(1/N<x/2\). The points \(1/n\) with \(n\ge N\) are all less than \(x/2\). There are only finitely many remaining points \(1,1/2,\ldots,1/(N-1)\), none equal to \(x\). Choose a positive radius smaller than \(x/2\) and smaller than the distance from \(x\) to each of those finitely many points. The interval with that radius around \(x\) avoids all of \(K\), so it is a basis element contained in \(\mathbb{R}\setminus K\). This verifies the needed neighborhood at every point of \(W\).
Thus \(W\) is open. As a further check, \(K\) is closed in \(\mathbb{R}_K\), since its complement \(W\) is open. But \(K\) is not open: \(1\in K\), and every basis element containing \(1\) must be an ordinary interval. Every ordinary interval containing \(1\) also contains points not in \(K\), so none is contained in \(K\).
A Common Neighborhood Mistake
Common pitfall. It is incorrect to use a deleted interval as a neighborhood of a point of \(K\). For example, one might claim that \((1/3-\varepsilon,1/3+\varepsilon)\setminus K\) is a basis neighborhood of \(1/3\). This is a wrong test: \(1/3=1/n\) for \(n=3\), so \(1/3\in K\) and \(1/3\notin(1/3-\varepsilon,1/3+\varepsilon)\setminus K\), for every \(\varepsilon>0\). The set does not even contain the point whose neighborhood is being tested. The correction is to use an ordinary interval around \(1/3\). In particular, if an open set contains \(1/3\), it must contain some ordinary interval \((a,b)\) with \(a<1/3<b\).
The pointwise test prevents two opposite errors: rejecting a set just because it contains \(0\), or accepting a deleted interval as a neighborhood of a point it has removed. At any point, first check membership; then choose a basis element that contains that point and stays inside the proposed set. Near \(0\), deletion can make such a choice possible. At a point of \(K\), the deleted basis elements are unavailable, and an ordinary interval must do the work.
Check Your Understanding
For each set, test the points that are included endpoints or points of \(K\) using the basis criterion.
- Is \((-2,3)\setminus K\) open in \(\mathbb{R}_K\)? Explain how the basis criterion applies.
- Is \([1/4,2)\) open? Identify a point where the local test succeeds or fails.
- Is \(\mathbb{R}\setminus\{1/n:n\ge3\}\) open? Check the included point \(1/2\) separately.
- Explain why no deleted interval \((a,b)\setminus K\) can be a basis neighborhood of \(1/5\).
- Is \(K\) open, closed, both, or neither in \(\mathbb{R}_K\)? Justify both parts of your answer.