What the Neighborhoods Tell Us
A point can belong to a set without being a limit point of it, and a point outside a set can still be a limit point. The distinction depends on the neighborhoods provided by the topology. We will use that local test to calculate closures, limit points, and isolated points in the usual topology on the real line and in three other familiar settings.
In a topological space \((X,\mathcal{T})\), an open neighborhood of \(x\) is an open set containing \(x\). The closure \(\overline{A}\) consists of the points whose every open neighborhood intersects \(A\). A point \(x\) is a limit point (or accumulation point) of \(A\) if every open neighborhood of \(x\) contains a point of \(A\) different from \(x\). The set of limit points is denoted \(A'\). A point \(a\in A\) is isolated in \(A\) if some open neighborhood \(U\) satisfies \(U\cap A=\{a\}\).
The neighborhood test gives \(\overline{A}=A\cup A'\). Indeed, every point of \(A\) is in its closure, and every limit point has neighborhoods meeting \(A\). Conversely, if \(x\in\overline{A}\) but \(x\notin A\), each neighborhood meeting \(A\) contains a point of \(A\) different from \(x\), so \(x\in A'\). We will use this relation to check the calculations below.
Finite Sets and a Convergent Sequence in the Usual Real Line
Unless another topology is specified, \(\mathbb{R}\) has its standard topology, whose basic open sets are open intervals. In this topology a finite set has no limit points: around any proposed point, one can avoid the finitely many members of the set other than possibly the point itself. The next calculation makes this neighborhood reasoning explicit.
Worked Example: A Finite Subset of the Real Line
Let \(A=\{-2,0,3\}\subseteq\mathbb{R}\). We calculate \(\overline{A}\), \(A'\), and the isolated points of \(A\).
For each \(a\in A\), the interval \((a-\tfrac12,a+\tfrac12)\) meets \(A\) only in \(\{a\}\): the distances between distinct members of \(A\) are \(2\), \(3\), and \(5\), each greater than \(\tfrac12\). Hence each member of \(A\) is isolated, and none is a limit point.
If \(x\notin A\), each of the three distances \(|x+2|\), \(|x|\), and \(|x-3|\) is positive. Let \(r\) be half their minimum. Then \(r>0\), and \((x-r,x+r)\) misses \(A\). So \(x\notin\overline{A}\), by the neighborhood test for closure.
Conclusion and check: \(\overline{A}=A\), \(A'=\varnothing\), and all three points of \(A\) are isolated. The closure calculation agrees with \(\overline{A}=A\cup A'\), since \(A\cup\varnothing=A\).
An infinite set may have limit points even when each of its own members is isolated. A basic example is a sequence of distinct real numbers approaching a point.
Worked Example: A Sequence Together with an Isolated Point
Let \(A=\{1/n:n\in\mathbb{N}\}\cup\{2\}\subseteq\mathbb{R}\), where \(\mathbb{N}=\{1,2,\ldots\}\). We claim that \(A'=\{0\}\) and \(\overline{A}=A\cup\{0\}\).
Every open neighborhood of \(0\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\). Choose \(n\in\mathbb{N}\) with \(n>1/\varepsilon\). Then \(0<1/n<\varepsilon\), so \(1/n\in A\) and \(1/n\ne0\). Thus \(0\in A'\).
Each \(1/k\) is isolated. For \(k=1\), the interval \((3/4,5/4)\) contains \(1\) and no other point of \(A\): the other sequence terms are at most \(1/2\), and \(2\) lies outside the interval. For \(k\ge2\), the nearest sequence values on either side of \(1/k\) are \(1/(k-1)\) and \(1/(k+1)\). Choose \(r>0\) smaller than half of each of the three positive distances \[ \left|\frac1{k-1}-\frac1k\right|,\qquad \left|\frac1k-\frac1{k+1}\right|,\qquad \left|2-\frac1k\right|. \] The interval \((1/k-r,1/k+r)\) contains no other sequence value: values with index less than \(k\) are at least \(1/(k-1)\), and those with index greater than \(k\) are at most \(1/(k+1)\). It also excludes \(2\). Hence \(1/k\) is isolated. The point \(2\) is isolated because \((3/2,5/2)\) meets \(A\) only in \(\{2\}\).
Now let \(x\in\mathbb{R}\setminus(A\cup\{0\})\). In particular, \(x\ne0\). Choose \(N\ge2\) so that \(1/N<|x|/2\). For every \(n\ge N\), the reverse triangle inequality gives \[ \left|x-\frac1n\right|\ge |x|-\frac1n\ge |x|-\frac1N>\frac{|x|}{2}. \] The finitely many points \(2,1,1/2,\ldots,1/(N-1)\) all differ from \(x\). Take \(r>0\) smaller than \(|x|/4\) and smaller than half the distance from \(x\) to each of those finitely many points. Then \((x-r,x+r)\) misses those points, and it also misses every \(1/n\) with \(n\ge N\), since \(r<|x|/2\). Thus \(x\notin\overline{A}\).
Conclusion and check: the only limit point is \(0\), and every point of \(A\) is isolated. Therefore \(\overline{A}=A\cup A'=\{1/n:n\in\mathbb{N}\}\cup\{0,2\}\), as claimed.
The rationals behave differently: they have no isolated points, and every real number is a limit point. The calculation rests on the fact that every nonempty open interval contains a rational number. For completeness, if \((a,b)\) is such an interval, choose \(n\in\mathbb{N}\) with \(n(b-a)>2\). There is an integer \(k\) with \(na<k\le na+1\), and \(na+1<nb\). Consequently \(a<k/n<b\), so \(k/n\in(a,b)\cap\mathbb{Q}\).
Worked Example: The Rational Numbers
Let \(A=\mathbb{Q}\subseteq\mathbb{R}\). To show that every \(x\in\mathbb{R}\) is a limit point of \(A\), take any open neighborhood \(U\) of \(x\). There is an \(\varepsilon>0\) with \((x-\varepsilon,x+\varepsilon)\subseteq U\). The interval \((x+\varepsilon/3,x+2\varepsilon/3)\) is nonempty and lies inside \(U\), so the interval argument above gives a rational in it. This rational differs from \(x\), because it is strictly greater than \(x\). Therefore \(x\in\mathbb{Q}'\).
Every neighborhood of every real point thus meets \(\mathbb{Q}\), so every real point also lies in \(\overline{\mathbb{Q}}\). Since both sets are subsets of \(\mathbb{R}\), we obtain \(\mathbb{Q}'=\overline{\mathbb{Q}}=\mathbb{R}\). No rational is isolated: every neighborhood of a rational contains another rational by the same argument. The closure check is \(\mathbb{Q}\cup\mathbb{Q}'=\mathbb{Q}\cup\mathbb{R}=\mathbb{R}\).
Discrete and Indiscrete Topologies
In the discrete topology every subset is open. In the indiscrete topology on \(X\), the only open sets are \(\varnothing\) and \(X\). These two choices give opposite neighborhood behavior: discrete neighborhoods can separate a point from every other point, while every nonempty indiscrete neighborhood must be the whole space.
- if \(A=\varnothing\), then \(\overline{A}=A'=\varnothing\);
- if \(A=\{a\}\), then \(\overline{A}=X\), \(A'=X\setminus\{a\}\), and \(a\) is isolated in \(A\);
- if \(A\) has at least two points, then \(\overline{A}=A'=X\), and \(A\) has no isolated points.
Proof. In a discrete space, every set is both open and closed, so \(A\) is closed and \(\overline{A}=A\). For any \(x\in X\), the open neighborhood \(\{x\}\) contains no point different from \(x\), so \(x\notin A'\). For \(a\in A\), this same neighborhood meets \(A\) exactly in \(\{a\}\), proving that \(a\) is isolated.
Now suppose \(X\) is nonempty and indiscrete. If \(A=\varnothing\), no neighborhood meets \(A\), and no neighborhood contains a point of \(A\) other than its center; both sets are empty. If \(A\ne\varnothing\), the only open neighborhood of any \(x\in X\) is \(X\), which meets \(A\). Hence \(\overline{A}=X\). If \(A=\{a\}\), then for \(x=a\), \(X\) contains no member of \(A\setminus\{a\}\), while for each \(x\ne a\), \(X\) contains \(a\ne x\). Thus \(A'=X\setminus\{a\}\), and \(X\cap A=\{a\}\) shows \(a\) is isolated. If \(A\) has at least two points, then for every \(x\in X\), \(A\setminus\{x\}\) is nonempty, so \(X\) contains a member of \(A\) different from \(x\). Therefore \(A'=X\). No point of \(A\) is isolated, since the only neighborhood \(X\) meets \(A\) in at least two points. \(\square\)
Worked Example: The Same Subset in Two Topologies
Take \(X=\{a,b,c\}\) and \(A=\{a\}\). First give \(X\) the discrete topology. The singleton \(\{a\}\) is open and meets \(A\) only in \(a\), so \(a\) is isolated. Every point has its own singleton neighborhood, so no point is a limit point of \(A\). Since \(A\) is closed, \(\overline{A}=\{a\}\) and \(A'=\varnothing\).
Now give the same underlying set the indiscrete topology. The only neighborhood of each point is \(X\), so \(\overline{A}=X\). At \(a\), that neighborhood contains no member of \(A\setminus\{a\}\), but at \(b\) and \(c\) it contains \(a\), which differs from the center. Hence \(A'=\{b,c\}\); \(a\) is isolated because \(X\cap A=\{a\}\). The check \(\overline{A}=A\cup A'=\{a\}\cup\{b,c\}=X\) agrees with the neighborhood calculation.
The Cofinite Topology
On a set \(X\), the cofinite topology consists of \(\varnothing\) and all subsets whose complements in \(X\) are finite. Its closed sets are \(X\) and the finite subsets of \(X\). Unlike the discrete and indiscrete cases, the size of \(A\) determines the answer when \(X\) is infinite.
- if \(A\) is finite, then \(\overline{A}=A\), \(A'=\varnothing\), and every point of \(A\) is isolated;
- if \(A\) is infinite, then \(\overline{A}=X\), \(A'=X\), and \(A\) has no isolated points.
Proof. If \(A\) is finite, it is closed, so \(\overline{A}=A\). For any \(x\in X\), the set \(U=X\setminus(A\setminus\{x\})\) is open because its complement is finite, and it contains \(x\). It contains no point of \(A\) other than possibly \(x\), so \(x\notin A'\). If \(a\in A\), the same \(U\) satisfies \(U\cap A=\{a\}\), so \(a\) is isolated.
If \(A\) is infinite, no finite closed set can contain it. The only closed set containing \(A\) is therefore \(X\), giving \(\overline{A}=X\). To test limit points, fix \(x\in X\) and an open neighborhood \(U\) of \(x\). The set \(X\setminus U\) is finite, and \(A\setminus\{x\}\) is infinite. It cannot be contained in the finite set \(X\setminus U\), so some point of \(A\setminus\{x\}\) lies in \(U\). Thus \(x\in A'\) for every \(x\), and \(A'=X\). In particular, every neighborhood of a point of \(A\) contains another point of \(A\), so no point of \(A\) is isolated. \(\square\)
Worked Example: Finite and Infinite Sets in a Cofinite Space
Let \(X=\mathbb{Z}\) with the cofinite topology. For \(A=\{1,2,3\}\), the set \(A\) is finite and hence closed. For each \(x\in\mathbb{Z}\), the set \(U=\mathbb{Z}\setminus(A\setminus\{x\})\) is open and contains \(x\). It meets \(A\) in at most \(\{x\}\), so \(A'=\varnothing\), \(\overline{A}=A\), and each member of \(A\) is isolated.
Now take \(B=\{2n:n\in\mathbb{N}\}\), the positive even integers. This set is infinite. For any \(x\in\mathbb{Z}\), every open neighborhood \(U\) has finite complement. Since \(B\setminus\{x\}\) is infinite, it cannot all lie in that finite complement; consequently \(U\) contains a point of \(B\) different from \(x\). Thus \(B'=X\). The closure is also \(X\), either because every neighborhood meets \(B\) or because no finite closed set contains \(B\). No point of \(B\) is isolated. The check \(\overline{B}=B\cup B'=B\cup X=X\) confirms the closure calculation.
A Common Pitfall
Common mistake: “A finite set has no limit points, so a singleton has no limit points in any topology.” This is false. For example, in the indiscrete space \(X=\{a,b,c\}\), let \(A=\{a\}\). The neighborhood of \(b\) is necessarily \(X\), and it contains \(a\ne b\), so \(b\in A'\). The same reasoning gives \(c\in A'\). In fact \(A'=\{b,c\}\), even though \(A\) is finite. The correction is to test the neighborhoods in the topology at hand: finiteness guarantees the stated conclusion in the discrete and cofinite cases, but not in the indiscrete case.
Check Your Understanding
For each question, identify the topology before applying the neighborhood test.
- In the usual topology on \(\mathbb{R}\), find the closure, derived set, and isolated points of \(\{0,2,4\}\).
- In the usual topology, find the derived set and closure of \(\{1/n:n\in\mathbb{N}\}\cup\{-1\}\).
- In an indiscrete space \(X\), calculate \(A'\) when \(A\) has exactly two points. What if \(X\) itself has exactly one point and \(A=X\)?
- In a discrete space, can a nonempty subset have a limit point? Explain using an open neighborhood.
- In a cofinite topology, compare the derived sets of a finite subset and an infinite subset. Which result changes if the underlying set \(X\) is finite?