Sequential Characterization of Limit Points

From Sequences in a Set to Limit Points

In the previous tutorial, sequences provided a way to test whether a set is closed: every convergent sequence in a closed set has its limit in the set. A related question is whether a particular point can be approached by points of a set other than the point itself. Such a point is called a limit point. Sequences give a precise test for this property, and the test works whether or not the point belongs to the set.

Definition: Let $E\subseteq\mathbb{R}$. A point $x\in\mathbb{R}$ is a limit point of $E$ if, for every $\varepsilon>0$, there exists $y\in E$ such that $0<|y-x|<\varepsilon$. Equivalently, every open interval around $x$ contains a point of $E$ different from $x$.

The strict inequality $0<|y-x|$ is essential: the point $x$ itself cannot be the witness. If $x\in E$, that membership alone does not make $x$ a limit point. Conversely, $x$ can be a limit point even when $x\notin E$. The condition asks whether other points of the set occur arbitrarily close to $x$.

A point $x\in E$ is called an isolated point of $E$ if some interval around $x$ contains no point of $E$ other than $x$. Thus a point in $E$ is either a limit point or an isolated point: if it is not a limit point, the negation of the definition supplies an $\varepsilon>0$ for which there is no $y\in E$ with $0<|y-x|<\varepsilon$. Points outside $E$ can also fail to be limit points.

The Sequential Characterization

The defining condition says that there are set points as close to $x$ as we ask. We can use it repeatedly, asking for a point within $1$, then within $1/2$, then within $1/3$, and so on. In fact, the selected points can be required to be distinct. That extra condition makes clear that a limit point is approached by endlessly many different points, not just by repeatedly naming one point.

Theorem (Sequential Characterization of Limit Points): Let $E\subseteq\mathbb{R}$ and $x\in\mathbb{R}$. Then $x$ is a limit point of $E$ if and only if there is a sequence $(x_n)$ of distinct points of $E\setminus\{x\}$ such that $x_n\to x$.

Proof. First suppose $x$ is a limit point of $E$. We choose the terms recursively. At stage $n$, there are only finitely many previously chosen points $x_0,\ldots,x_{n-1}$, all different from $x$. If $n>0$, let $d_n=\min\{|x_j-x|:0\leq j<n\}$. This minimum exists and is positive. Choose a positive radius $r_n$ that is at most $1/(n+1)$ and, when $n>0$, is less than $d_n$. For example, take $r_0=1$ and $r_n=\min\{1/(n+1),d_n/2\}$ for $n>0$.

Since $x$ is a limit point, there is an $x_n\in E$ with $0<|x_n-x|<r_n$. This point is not any of the earlier $x_j$: each earlier point has distance at least $d_n$ from $x$, while $x_n$ has distance less than $d_n/2$. Thus all the selected points are distinct and belong to $E\setminus\{x\}$. Also, $|x_n-x|<1/(n+1)$, which tends to zero. Therefore $x_n\to x$.

Conversely, suppose distinct points $x_n\in E\setminus\{x\}$ satisfy $x_n\to x$. Given any $\varepsilon>0$, convergence gives an index $N$ such that $|x_n-x|<\varepsilon$ for every $n\geq N$. In particular, $x_N\in E$ and $0<|x_N-x|<\varepsilon$, because $x_N\neq x$. This is exactly the limit-point condition. $\square$

The converse would still hold if the sequence were not required to have distinct terms. But distinctness in the forward construction is useful: it rules out a misleading idea that one set point, repeated indefinitely, can establish that $x$ is a limit point. The exclusion of $x$ from every term matters for the same reason. A constant sequence with value $x$ tells us nothing about whether other set points lie near $x$.

Worked Examples

Worked Example: Limit Points of an Open Interval

Let $E=(-1,2)$. We show that its limit points are exactly the points in $[-1,2]$. First take $x\in(-1,2)$, and let $\varepsilon>0$. Set $\delta=\min\{\varepsilon/2,(2-x)/2\}$. Then $\delta>0$, and $y=x+\delta$ satisfies $y>x>-1$ and $y\leq x+(2-x)/2=(x+2)/2<2$. Hence $y\in E$, $y\neq x$, and $|y-x|=\delta<\varepsilon$. So $x$ is a limit point.

At the left endpoint $x=-1$, take $y=-1+\min\{\varepsilon/2,1/2\}$. Then $-1<y\leq-1+1/2<2$, and $0<|y-(-1)|\leq\varepsilon/2<\varepsilon$. At the right endpoint $x=2$, take $y=2-\min\{\varepsilon/2,1/2\}$. This gives $-1<y<2$ and $0<|y-2|<\varepsilon$. Thus both endpoints are limit points, although neither belongs to $E$.

Finally, if $x<-1$, put $d=-1-x>0$. Every $y\in E$ satisfies $y>-1$, so $|y-x|=y-x>-1-x=d$. The interval of radius $d/2$ around $x$ contains no point of $E$. If $x>2$, put $d=x-2>0$; every $y\in E$ has $|y-x|=x-y>x-2=d$, so again a neighborhood misses $E$. No point outside $[-1,2]$ is a limit point. The limit-point set is therefore $[-1,2]$.

Worked Example: A Finite Set Has No Limit Points

Consider $F=\{-4,1,6\}$. If $x\notin F$, all three distances $|x+4|$, $|x-1|$, and $|x-6|$ are positive. Their minimum $d$ is positive. Every point of $F$ is at distance at least $d$ from $x$, so the interval of radius $d/2$ around $x$ contains no point of $F$.

If $x\in F$, the distances from $x$ to the other two elements of $F$ are positive. Let $d$ be the smaller of those two distances. The interval of radius $d/2$ around $x$ contains no other element of $F$. Thus neither points outside $F$ nor points in $F$ satisfy the limit-point condition, and $F$ has no limit points. In particular, belonging to a set does not guarantee being a limit point.

Worked Example: Two Accumulation Locations

Let $$ E=\left\{5+\frac{1}{k+2}:k\in\mathbb{N}_0\right\}\cup \left\{-3-\frac{1}{k+2}:k\in\mathbb{N}_0\right\}. $$ For every $k\in\mathbb{N}_0$, the first expression is greater than $5$, and $\left|5+1/(k+2)-5\right|=1/(k+2)\to0$. The second expression is less than $-3$, and $\left|-3-1/(k+2)-(-3)\right|=1/(k+2)\to0$. Each sequence consists of distinct points of $E$ and avoids its respective limit. The Sequential Characterization shows that $5$ and $-3$ are limit points.

To see that there are no others, let $x$ be any limit point of $E$. By the characterization, there is a sequence of distinct points of $E\setminus\{x\}$ converging to $x$. Each term lies in one of the two displayed sets. At least one of those sets supplies infinitely many terms; otherwise their union would supply only finitely many terms. Take the subsequence from that set. A subsequence of a convergent sequence has the same limit $x$.

In the first set, distinct terms have distinct indices $k$. Along an infinite selection of distinct nonnegative integer indices, the indices eventually exceed every fixed bound: only finitely many indices are at most that bound. Hence $1/(k+2)\to0$ along the selected terms, so that subsequence converges to $5$. Uniqueness of limits gives $x=5$. In the second set the same reasoning shows that the selected subsequence converges to $-3$, so $x=-3$. Therefore the limit points of $E$ are exactly $\{-3,5\}$.

Limit Points and Closed Sets

The sequential characterization connects limit points directly to closedness. Recall the Sequential Characterization of Closed Sets from “Completeness and Cauchy Sequences”: a set is closed if and only if every convergent sequence of its points has its limit in the set. Combining that earlier result with the theorem above gives a useful equivalent test.

Theorem (Closed Sets Contain All Their Limit Points): A subset $E\subseteq\mathbb{R}$ is closed if and only if every limit point of $E$ belongs to $E$.

Proof. Suppose $E$ is closed, and let $x$ be a limit point of $E$. The Sequential Characterization of Limit Points provides a sequence in $E\setminus\{x\}$ converging to $x$. In particular, it is a convergent sequence whose terms all belong to $E$. The Sequential Characterization of Closed Sets then gives $x\in E$.

Conversely, suppose every limit point of $E$ belongs to $E$. Let $(x_n)$ be any convergent sequence with $x_n\in E$ for every $n$, and write $x_n\to x$. If $x\in E$, its limit is already in $E$. If $x\notin E$, then $x_n\neq x$ for every $n$. Given any $\varepsilon>0$, convergence supplies an $N$ such that $|x_n-x|<\varepsilon$ for all $n\geq N$. In particular $x_N\in E$ and $0<|x_N-x|<\varepsilon$. Thus $x$ is a limit point of $E$, so the hypothesis gives $x\in E$, a contradiction to $x\notin E$. Therefore every convergent sequence in $E$ has its limit in $E$, and the Sequential Characterization of Closed Sets implies that $E$ is closed. $\square$

Reading the Criterion Carefully

The sequence in the limit-point characterization must have its terms in $E\setminus\{x\}$. If a sequence merely lies in $E$ and converges to $x\in E$, it might be the constant sequence $x_n=x$; that sequence does not show that any other points of $E$ approach $x$. Requiring distinct terms is a convenient way to express the stronger idea that the set supplies endlessly many points near $x$.

A limit point need not be a member of the set. The endpoints in the open-interval example illustrate this: $-1$ and $2$ are approached by points of $(-1,2)$ but are not themselves in the interval. By contrast, a closed set contains all its limit points. This is why the last theorem is useful: it turns the question of closedness into a question about whether the set omits any points that its own elements approach arbitrarily closely.

Check Your Understanding

Use the definition and the sequential characterization to answer the following questions.

Why must the point $y$ in the definition of a limit point satisfy $y\neq x$?
What additional feature can be required of a sequence approaching a limit point, and why does that feature matter?
Can a point outside a set be a limit point? Give an example from this tutorial.
Why does the finite set $\{-4,1,6\}$ have no limit points?
How does the Sequential Characterization of Closed Sets help prove that a closed set contains all its limit points?