Closedness and Limit Points
A closed set may contain points that are approached by other points of the set, as well as isolated points that are not approached by any other set points. The key condition is not that every point of a closed set be a limit point. Instead, a closed set must contain every limit point it has. This links the definition of closedness to the derived set introduced in the previous tutorial.
Earlier in the course, the Theorem (Closed Sets Contain All Their Limit Points) established the equivalent criterion that \(F\) is closed if and only if \(F'\subseteq F\). We will use this characterization rather than re-prove it. It makes the role of limit points explicit: a set fails to be closed precisely when at least one of its limit points is missing from the set.
There is a useful strengthening of the definition of a limit point. It is not enough that every neighborhood contain a point of the set other than the center just once in some fixed neighborhood. The condition must hold at every radius, however small. In fact, every neighborhood of a limit point contains infinitely many points of the set.
Proof. Suppose instead that for some \(r>0\), the set \(B_r(x)\cap E\) is finite. Removing \(x\), if it belongs to this set, still leaves a finite set: $$ \bigl(B_r(x)\setminus\{x\}\bigr)\cap E=\{y_1,\ldots,y_m\} $$ for some nonnegative integer \(m\). Since \(x\in E'\), every punctured ball around \(x\) meets \(E\). In particular, this finite set cannot be empty, so \(m\geq1\). Each \(y_j\) differs from \(x\), and therefore \(|y_j-x|>0\). Let $$ s=\min\left\{r,\frac{|y_1-x|}{2},\ldots,\frac{|y_m-x|}{2}\right\}. $$ Every quantity in this minimum is positive, so \(s>0\). No \(y_j\) lies in \(B_s(x)\), because \(|y_j-x|\geq 2s>s\). Also, \(B_s(x)\subseteq B_r(x)\), so there are no other points of \(E\) in the punctured ball \(B_s(x)\setminus\{x\}\). This contradicts \(x\in E'\), which requires every punctured ball around \(x\) to meet \(E\). Thus \(B_r(x)\cap E\) must be infinite for every \(r>0\). \(\square\)
The converse also holds: if every ball around \(x\) contains infinitely many points of \(E\), then each contains a point of \(E\) other than \(x\), so \(x\in E'\). Thus a point is a limit point exactly when every neighborhood around it contains infinitely many points of the set. The distinction between “at least one” and “infinitely many” is useful when checking whether a point is isolated.
Worked Example: Two Limit Points in a Closed Set
Consider $$ F=\{0,2\}\cup\left\{-\frac1n:n\geq1\right\}\cup\left\{2+\frac1n:n\geq1\right\}. $$ We will show that \(F'=\{0,2\}\), and hence that \(F\) is closed.
First, let \(r>0\). Choose a positive integer \(n>1/r\). Then \(0<1/n<r\), so \(-1/n\) is a point of \(F\setminus\{0\}\) within distance \(r\) of \(0\). Therefore \(0\in F'\). Also \(2+1/n\in F\setminus\{2\}\) and \(|(2+1/n)-2|=1/n<r\), so \(2\in F'\).
Now let \(x\notin\{0,2\}\). Choose \(\delta>0\) with \(\delta<\tfrac12\min\{|x|,|x-2|\}\); then every point of \(B_\delta(x)\) is at distance greater than \(\tfrac12\min\{|x|,|x-2|\}\) from both \(0\) and \(2\). Since \(-1/n\to0\), all but finitely many of the points \(-1/n\) lie outside \(B_\delta(x)\). Since \(2+1/n\to2\), all but finitely many of the points \(2+1/n\) also lie outside \(B_\delta(x)\). Thus \(B_\delta(x)\cap F\) is finite. By the theorem just proved, \(x\) cannot be a limit point of \(F\): if it were, this ball would contain infinitely many points of \(F\). We have shown that \(F'=\{0,2\}\).
Both limit points belong to \(F\). The Theorem (Closed Sets Contain All Their Limit Points) therefore gives that \(F\) is closed. Notice that most of the listed points are isolated, yet the set is still closed: isolated points do not need to be limit points in order for a set to be closed.
Removing a Point from a Closed Set
A closed set can lose its closedness when a point is removed. Whether this happens depends on the point. Removing an isolated point leaves no missing limit point behind; removing a limit point can leave that point approached by the remaining set, even though it is no longer included.
Proof. Suppose first that \(p\) is isolated in \(F\). By definition, there is an \(r>0\) such that \(B_r(p)\cap F=\{p\}\). The set \(F\setminus\{p\}\) has no points in \(B_r(p)\). Consequently \(p\notin(F\setminus\{p\})'\). By monotonicity of the derived set, \(F\setminus\{p\}\subseteq F\) implies $$ (F\setminus\{p\})'\subseteq F'. $$ Since \(F\) is closed, the Theorem (Closed Sets Contain All Their Limit Points) gives \(F'\subseteq F\). Every point of \((F\setminus\{p\})'\) therefore belongs to \(F\), and it is not \(p\); hence it belongs to \(F\setminus\{p\}\). We have shown $$ (F\setminus\{p\})'\subseteq F\setminus\{p\}, $$ so the same closed-set criterion shows that \(F\setminus\{p\}\) is closed.
Conversely, suppose \(p\) is not isolated in \(F\). For every \(r>0\), the ball \(B_r(p)\) contains a point of \(F\) different from \(p\). Since \(p\) is the only point removed, that point belongs to \(F\setminus\{p\}\). Thus every punctured ball around \(p\) meets \(F\setminus\{p\}\), so \(p\in(F\setminus\{p\})'\). But \(p\notin F\setminus\{p\}\). The set \(F\setminus\{p\}\) does not contain all its limit points and is therefore not closed. This proves both directions. \(\square\)
Worked Example: Removing an Isolated Point or a Limit Point
Let $$ F=\{-3,5\}\cup\left\{5+\frac1n:n\geq1\right\}. $$ The points \(5+1/n\) approach \(5\), so \(5\in F'\). Every point of \(F\) other than \(5\) is isolated: \(-3\) is separated from the other points, and for each fixed \(n\), the point \(5+1/n\) has a positive gap from its neighboring terms \(5+1/(n+1)\) and, when \(n>1\), \(5+1/(n-1)\). Thus \(F'=\{5\}\). As \(5\in F\), the closed-set criterion shows that \(F\) is closed.
The point \(-3\) is isolated, so the Deleting a Point from a Closed Set Theorem implies that \(F\setminus\{-3\}\) is closed. In contrast, deleting \(5\) leaves all the points \(5+1/n\), which still approach \(5\). Thus \(5\) is a limit point of \(F\setminus\{5\}\), but it is not in that set. Consequently \(F\setminus\{5\}\) is not closed.
Limit Points Need Not Belong to the Set
A set is not automatically closed just because it contains infinitely many points. What matters is whether it includes the points that those points approach. An omitted limit point gives a direct certificate that a set is not closed. Conversely, checking that every limit point belongs to a set proves closedness by the established characterization.
Worked Example: A Missing Limit Point
Let \(A=\{2+1/n:n\geq1\}\). For any \(r>0\), choose a positive integer \(n>1/r\). Then \(2+1/n\in A\) and $$ \left|\left(2+\frac1n\right)-2\right|=\frac1n<r. $$ Since \(2+1/n\neq2\), every punctured ball around \(2\) meets \(A\), so \(2\in A'\). But \(2\notin A\). The set \(A\) therefore fails the criterion \(A'\subseteq A\) and is not closed.
This example also illustrates the infinitely-many-points characterization. Given any \(r>0\), all sufficiently large \(n\) satisfy \(1/n<r\), and the corresponding points \(2+1/n\) are distinct. Therefore every ball around \(2\) contains infinitely many points of \(A\), not just one point that happens to be close.
A common pitfall is to treat “closed” as meaning that the set contains all of its own points that are limit points, while overlooking limit points outside the set. The criterion \(F'\subseteq F\) has no such restriction: it tests every real number that is a limit point, whether or not that number was initially listed as an element. A second pitfall is to confuse isolated points with points that prevent closedness. An isolated point can be removed from a closed set without losing closedness; it is a non-isolated point, approached by the remaining set, whose removal makes the set fail the criterion.
The two tests developed here serve different purposes. The infinite-neighborhood test helps establish that a candidate really is a limit point, or rule out candidates when some neighborhood meets the set only finitely often. The closed-set criterion then turns the full collection of limit points into a test for closedness. Together they let us distinguish accumulation from mere membership and decide the effect of deleting a point.
Check whether every punctured ball around a candidate meets the set; equivalently, check whether every ball contains infinitely many set points.
A set is closed exactly when all of its limit points belong to it.
For a point in a closed set, removing it preserves closedness exactly when it was isolated in the original set.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- How many points of a set must every neighborhood of a limit point contain?
- What condition involving the derived set characterizes closedness?
- Can a closed set contain isolated points? Explain what happens to those points under the limit-point test.
- When does removing a point \(p\) from a closed set \(F\) leave a closed set?
- Why does a set fail to be closed if it has a limit point that is not one of its elements?