When a Set Point Has Its Own Neighborhood
The previous tutorial studied accumulation points: points that have other set points arbitrarily close to them. This tutorial considers the contrasting situation. A point can belong to a set yet have a small neighborhood containing no other points of that set. Such a point is called an isolated point.
Isolation is a local property. It does not mean that a point is far from the entire set, or that the set has only finitely many points. It means that one particular neighborhood separates the point from all the other members of the set.
The condition \(x\in E\) is essential: a point outside \(E\) cannot be an isolated point of \(E\). The equality \(B_r(x)\cap E=\{x\}\) says both that the neighborhood contains \(x\) and that it contains no other point of \(E\). The radius may depend on \(x\); there need not be one radius that works for every point.
Isolation and Accumulation
For a point already in \(E\), being isolated is exactly the failure to be an accumulation point. Recall that \(E'\) denotes the set of accumulation points of \(E\), and that an accumulation point has a different point of \(E\) in every ball around it.
Proof. Suppose \(x\) is isolated in \(E\). Then \(x\in E\), and there is an \(r>0\) such that \(B_r(x)\cap E=\{x\}\). This ball contains no point of \(E\) different from \(x\), so \(x\) fails the definition of an accumulation point. Thus \(x\notin E'\).
Conversely, suppose \(x\in E\) and \(x\notin E'\). Since \(x\) is not an accumulation point, the definition of accumulation point fails: there is some \(r>0\) for which no \(y\in E\) satisfies \(0<|y-x|<r\). Every point of \(E\cap B_r(x)\) must therefore equal \(x\). Because \(x\in E\) and \(x\in B_r(x)\), we have \(B_r(x)\cap E=\{x\}\). Hence \(x\) is isolated. \(\square\)
In particular, an isolated point is always a member of the set, whereas an accumulation point may or may not be. The earlier Closure Decomposes into Limit Points and Isolated Points theorem expresses this distinction by decomposing the closure into accumulation points and isolated members of the set.
Worked Example: Isolated Points in a Finite Set
Let \(E=\{-2,1,5\}\). For \(x=-2\), take \(r=1\). The ball \(B_1(-2)=(-3,-1)\) contains \(-2\), but neither \(1\) nor \(5\), so \(B_1(-2)\cap E=\{-2\}\).
For \(x=1\), take \(r=1\). The ball \((0,2)\) contains \(1\) and neither of the other two set points. For \(x=5\), take \(r=2\). The ball \((3,7)\) contains \(5\), but not \(1\) or \(-2\). Thus every point of \(E\) is isolated. These radii are examples, not unique choices: any sufficiently small positive radius also works.
Isolation Does Not Require a Finite Set
A set can have infinitely many isolated points, even when those points gather around another point. In that case, each point in the sequence can be separated from the other set points locally, while the sequence as a whole approaches a point that is not in the set. The size of a suitable isolating radius may shrink as the points get closer together.
Worked Example: Isolated Terms Approaching a Missing Point
Consider \(E=\{3-1/n:n\geq1\}\). Write \(x_n=3-1/n\). These points are distinct: if \(m>n\), then \(1/m<1/n\), so \(x_m>x_n\).
For each \(n\geq1\), choose $$ r_n=\frac{1}{2n(n+1)}. $$ The next point \(x_{n+1}\) is at distance $$ x_{n+1}-x_n=\left(3-\frac{1}{n+1}\right)-\left(3-\frac{1}{n}\right)=\frac{1}{n(n+1)}=2r_n. $$ If \(n>1\), the preceding point is at distance $$ x_n-x_{n-1}=\left(3-\frac{1}{n}\right)-\left(3-\frac{1}{n-1}\right)=\frac{1}{n(n-1)}. $$ Since \(n-1<n+1\), this preceding distance is greater than \(1/[n(n+1)]=2r_n\). The points \(x_k\) increase with \(k\), so every point other than the adjacent ones is at least as far away as the corresponding adjacent point. For \(n=1\), there is no preceding point, and all other terms lie above \(x_2\). It follows in every case that \(B_{r_n}(x_n)\cap E=\{x_n\}\). Thus every member of \(E\) is isolated.
The point \(3\) is not in \(E\), since \(1/n>0\) for every positive integer \(n\). Nevertheless, it is an accumulation point: given \(r>0\), choose \(n\) with \(1/n<r\). Then \(x_n\in E\), \(x_n\ne3\), and \(|x_n-3|=1/n<r\). The example shows that isolated points of a set can approach an accumulation point.
There Are At Most Countably Many Isolated Points
Although a set may have infinitely many isolated points, there is a limit on how many it can have in \(\mathbb{R}\): its isolated points are at most countable. The key is to place a rational-endpoint interval around each isolated point, small enough to contain no other point of the set. A single such interval cannot serve two different isolated points.
Proof. Let \(I\) be the set of isolated points of \(E\). For each \(x\in I\), choose an isolating radius \(r_x>0\), so \(B_{r_x}(x)\cap E=\{x\}\). By density of the rationals, there are rationals \(a_x,b_x\) such that $$ x-r_x<a_x<x<b_x<x+r_x. $$ The interval \((a_x,b_x)\) contains \(x\) and is contained in \(B_{r_x}(x)\). Consequently, $$ (a_x,b_x)\cap E=\{x\}. $$
Associate to \(x\) the rational pair \((a_x,b_x)\). If two isolated points \(x,y\in I\) are associated with the same pair \((a,b)\), then \(x,y\in(a,b)\). But the interval chosen for \(x\) satisfies \((a,b)\cap E=\{x\}\), and \(y\in E\cap(a,b)\); hence \(y=x\). Thus distinct isolated points are assigned distinct rational pairs. The set of rational pairs is countable, since \(\mathbb{Q}\) is countable and pairs of positive integer indices can be enumerated, for example by increasing sum of their indices. Therefore \(I\) is at most countable. \(\square\)
The rational interval is a label for an isolated point, not necessarily the only possible label. A choice of one such interval for each point gives an injection into a countable set. The argument works even when \(E\) itself is uncountable.
Worked Example: One Isolated Point in an Uncountable Set
Let \(E=[0,1]\cup\{2\}\). The point \(2\) is isolated: \(B_{1/2}(2)=(3/2,5/2)\) contains no point of \([0,1]\), and its intersection with \(E\) is \(\{2\}\).
No point \(x\in[0,1]\) is isolated. If \(x<1\), then for any \(r>0\), let $$ \delta=\min\left\{\frac r2,\frac{1-x}{2}\right\}. $$ This number is positive, and \(y=x+\delta\) satisfies \(x<y\leq(1+x)/2<1\), so \(y\in[0,1]\), \(y\ne x\), and \(|y-x|=\delta<r\). If \(x=1\), use \(\delta=\min\{r/2,1/2\}\) and \(y=1-\delta\). Then \(y\in[0,1]\), \(y\ne1\), and \(|y-1|=\delta<r\). Every ball around every point of \([0,1]\) therefore contains another point of \(E\). The only isolated point of \(E\) is \(2\), despite \(E\) being uncountable.
Worked Example: The Rationals Have No Isolated Points
Let \(E=\mathbb{Q}\), and fix any \(q\in\mathbb{Q}\). Given \(r>0\), density of the rationals gives a rational \(y\) in the nonempty interval \((q+r/3,q+2r/3)\). This interval lies to the right of \(q\), so \(y\ne q\), and $$ 0<|y-q|<\frac{2r}{3}<r. $$ Thus every ball around \(q\) contains a rational other than \(q\). Since this holds for every \(q\in\mathbb{Q}\), the set of isolated points of \(\mathbb{Q}\) is empty.
Using the Isolation Test Carefully
To prove that a point is isolated, start with a candidate that belongs to the set and find a positive radius whose ball excludes every other set point. For an infinite set, it is not enough to check only a few nearby examples: the argument must account for all its other members. To prove that a point of \(E\) is not isolated, show instead that every positive-radius ball contains some other point of \(E\).
A common pitfall is to confuse “isolated in \(E\)” with “surrounded by an empty interval.” An isolating ball is not empty: it contains the point \(x\) itself. Another is to assume that an isolated point cannot be close to an accumulation point. In the sequence example, the isolating radii become smaller as the set points approach \(3\); each term remains isolated, but there is no uniform positive radius that isolates all of them at once.
The countability theorem is useful precisely because it does not require the set to be countable or closed. It says that no subset of the real line can have uncountably many points each separated from its own set by a neighborhood. It does not say that the full set is countable, nor that a set with no isolated points must be empty. Those conclusions would go beyond what the isolation test establishes.
Check Your Understanding
Use the isolation definition, the examples, and the countability theorem to answer the following questions.
- What two conditions must hold for \(x\) to be an isolated point of \(E\)?
- Why is an isolating radius for \(x_n=3-1/n\) allowed to depend on \(n\)?
- How does an isolating interval with rational endpoints help prove countability?
- Can an uncountable set have an isolated point? Give the example from this tutorial.
- Why does the existence of a sequence of isolated points approaching \(3\) not make \(3\) an isolated point of that sequence set?
- What must be shown to prove that a member of \(E\) is not isolated?