Accumulation points of sequences – Serlo

WARNING: There are several types of accumulation points

In mathematics, there are two types of accumulation points: accumulation points of sequences and accumulation points of sets. Both are closely related to each other. Nevertheless, they need to be distinguished well within lectures and exercises.

This article concerns about accumulaation points of sequences. If we talk about an accumulation point here, we mean the accumulation point of a sequence.

Introductory example

Accumulation points are encountered when investigating the limits of sequences. Some sequences seem to converge to "multiple limits". Limits are always unique, so we need to replace the term limit by a new notion. The replacement is done by the term accumulation point.

For instance, let us take the sequence $a_{n}=(-1)^{n}{\tfrac {n}{n+1}}$. Its elements are

$\left((-1)^{n}{\tfrac {n}{n+1}}\right)_{n\in \mathbb {N} }=\left(-{\tfrac {1}{2}},\,{\tfrac {2}{3}},\,-{\tfrac {3}{4}},\,{\tfrac {4}{5}},\,-{\tfrac {5}{6}},\,{\tfrac {6}{7}},\,\ldots \right)$

Plotted in a diagram, they look as follows:

The first elements of the sequence (-1)^n*n/(n+1)

This sequence does not converge towards a unique limit, but rather splits in 2 parts: one is converging to $1$ ant the other towards $-1$ .

Later, we will mathematically say that the sequence has accumulation points: one at $1$ and one at $-1$ . Intuitively, it looks like

accumulation points are limits to which a part of a sequence converges.

How can we formulate this intuition mathematically? This will be our next question to be answered.

Definition of an accumulation point

We intend a mathematically precise definition of the expression:

"A part of the sequence ... converges to a limit ..."

In the last chapter, we introduced the concept of a subsequence in order to say what a part of a sequence is. So we replace "part of a sequence" by the mathematically precise expression "subsequence":

"There is a subsequence of ... converging to a limit ..."

And this is already the definition of an accumulation point:

Definition (accumulation point of a sequence)

A number $a$ is called accumulation point of a sequence $(a_{n})_{n\in \mathbb {N} }$, if there is a subsequence $\left(a_{n_{k}}\right)_{k\in \mathbb {N} }$ of $(a_{n})_{n\in \mathbb {N} }$ , which converges to $a$ .

Examples

The introductory example considered the sequence $a_{n}=(-1)^{n}{\tfrac {n}{n+1}}$ , which intuitively has the accumulation points $1$ and $-1$ . But do these accumulation points also satisfy the mathematical definition? For this, we need to find 2 subsequences of $a_{n}=(-1)^{n}{\tfrac {n}{n+1}}$: one converging to $1$ and one converging to $-1$. How can we construct these subsequences? Let us take a look at the sequence elements again:

$-{\tfrac {1}{2}},\,{\tfrac {2}{3}},\,-{\tfrac {3}{4}},\,{\tfrac {4}{5}},\,-{\tfrac {5}{6}},\,{\tfrac {6}{7}},\,\ldots$

or plotted in a diagram:

The sequence a_n=(-1)^n*n/(n+1) with 2 subsequences

It looks like all even-indexed elements tend towards $1$. This subsequence consists of only the positive elements:

${\cancel {-{\tfrac {1}{2}}}},\,{\color {OliveGreen}{\tfrac {2}{3}}},\,{\cancel {-{\tfrac {3}{4}}}},\,{\color {OliveGreen}{\tfrac {4}{5}}},\,{\cancel {-{\tfrac {5}{6}}}},\,{\color {OliveGreen}{\tfrac {6}{7}}},\,\ldots$

There even is an explicit form of this subsequence ${\color {OliveGreen}(c_{k})_{k\in \mathbb {N} }}$ :

$c_{k}=a_{2k}=(-1)^{2k}{\tfrac {2k}{2k+1}}={\tfrac {2k}{2k+1}}$

And the limit theorems indeed yield that it converges to $1$ :

${\begin{aligned}\lim _{k\rightarrow \infty }c_{k}&=\lim _{k\rightarrow \infty }{\frac {2k}{2k+1}}=\lim _{k\rightarrow \infty }{\frac {2k}{2k\left(1+{\frac {1}{2k}}\right)}}=\lim _{k\rightarrow \infty }{\frac {1}{1+{\frac {1}{2k}}}}\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{limit theorems}}\right.}\\[1em]&={\frac {\lim _{k\rightarrow \infty }1}{\lim _{k\rightarrow \infty }1+\lim _{k\rightarrow \infty }{\frac {1}{2k}}}}={\frac {1}{1+0}}=1\end{aligned}}$

So $1$ is an accumulation point of the sequence $a_{n}=(-1)^{n}{\tfrac {n}{n+1}}$, since $(c_{k})_{k\in \mathbb {N} }$ is a subsequence converging to $1$ . Analogously, we can show that $-1$ is an accumulation point. We have to sort out the negative elements:

${\color {Blue}-{\tfrac {1}{2}}},\,{\cancel {\tfrac {2}{3}}},\,{\color {Blue}-{\tfrac {3}{4}}},\,{\cancel {\tfrac {4}{5}}},\,{\color {Blue}-{\tfrac {5}{6}}},\,{\cancel {\tfrac {6}{7}}},\,\ldots$

Again, the limit theorems yield convergence to $-1$.

We may as well visualize that $1$ and $-1$ are the accumulation points of the sequence by plotting the sequence elements on the line of real numbers. They will "accumulate" near $1$ and $-1$ (which also is where the name "accumulation point" comes from):

1 and -1 are accumulation points of the sequence (a_n)

Question for understanding: What is the explicit form for the odd-indexed sequence elements?

The odd indices are $n_{k}=2k-1$. The explicit form of $(d_{k})_{k\in \mathbb {N} }$ hence reads:

$d_{k}=a_{n_{k}}=a_{2k-1}=(-1)^{2k-1}{\frac {2k-1}{(2k-1)+1}}={\frac {1-2k}{2k}}$

Question for understanding: Why does the subsequence of odd-indexed elements converge to $-1$ ?

This follows from the limit theorems:

${\begin{aligned}\lim _{k\rightarrow \infty }d_{k}&=\lim _{k\rightarrow \infty }{\frac {1-2k}{2k}}=\lim _{k\rightarrow \infty }{\frac {2k\left({\frac {1}{2k}}-1\right)}{2k}}=\lim _{k\rightarrow \infty }{\frac {{\frac {1}{2k}}-1}{1}}\\[1em]&{\color {OliveGreen}\left\downarrow \ {\text{limit theorems}}\right.}\\[1em]&={\frac {\lim _{k\rightarrow \infty }{\frac {1}{2k}}-\lim _{k\rightarrow \infty }1}{\lim _{k\rightarrow \infty }1}}={\frac {0-1}{1}}=-1\end{aligned}}$

Alternative definition of accumulation points

Neighbourhood definition

A limit of a sequence is characterized by almost all sequence elements being in each $\epsilon$-neighbourhood around the limit. I.e. all but finitely many elements must be there - no matter how small we choose $\epsilon$ . For accumulation points, there is a similar characterization (of which we still need to show that it is equivalent to the first one): A number is an accumulation point, if there are infinitely many elements in each neighbourhood. This is a much weaker assumption: if there are infinitely many elements inside the neighbourhood, there may still be infinitely many elements outside the neighbourhood.

Reminder: Mathematically, an $\epsilon$-neighbourhood of $h$ is an open interval $(h-\epsilon ,h+\epsilon )$ with $\epsilon >0$.

So $h$ is an accumulation point if and only if there are infinitely many elements inside the open interval $(h-\epsilon ,h+\epsilon )$ . Now, a sequence element $a_{n}$ is situated inside the open interval $(h-\epsilon ,h+\epsilon )$, if and only if the strict inequality $|a_{n}-h|<\epsilon$ holds. So there must be infinitely many indices $n\in \mathbb {N}$ with $|a_{n}-h|<\epsilon$ . The alternative definition of an accumulation point hence reads:

Definition (Neighbourhood definition of accumulation point)

A sequence $(a_{n})_{n\in \mathbb {N} }$ has an accumulation point at $h\in \mathbb {R}$, whenever there are infinitely many sequence elements of $(a_{n})_{n\in \mathbb {N} }$ in every neighbourhood of $h$. That means, for any $\epsilon >0$ there must be infinitely many indices $n\in \mathbb {N}$ with $|a_{n}-h|<\epsilon$ .

The definition of an accumulation point is just a weaker form of a limit: For a limit, almost all elements must be inside every $\epsilon$-neighbourhood of the corresponding number. Only finitely many elements may be situated on the outside. For an accumulation point, there may also be infinitely many elements on the outside, as long as there are also infinitely many elements on the inside. Hence, almost all is a special form of infinitely many and every limit is at the same time an accumulation point.

Example (Neighbourhood definition of accumulation point)

We consider the sequence $(a_{n})_{n\in \mathbb {N} }=((-1)^{n})_{n\in \mathbb {N} }$, which is jumping between $1$ and $-1$. indeed, both are accumulation pints by the neighbourhood definition: For $1$ , we can choose $\epsilon >0$ as small as we wish - all even indexed elements $a_{2k}$ will be situated inside the interval $(1-\epsilon ,1+\epsilon )$. These are infinitely many. Mathematically, we may write $a_{2k}=(-1)^{2k}=1\in (1-\epsilon ,1+\epsilon )$ for all $k\in \mathbb {N}$ and any given $\epsilon >0$.

However, not almost all elements are situated in this neighbourhood (at least not for any $\epsilon >0$). For instance, with $\epsilon ={\tfrac {1}{2}}$ the odd-indexed elements $a_{2k-1}$ lie on the outside: $a_{2k-1}=(-1)^{2k-1}=-1\notin ({\tfrac {1}{2}},{\tfrac {3}{2}})=(1-\epsilon ,1+\epsilon )$ for all $k\in \mathbb {N}$. So $1$ is not a limit. Analogously, one can of course show that $-1$ is an accumulation point, since the infinitely many odd-indexed elements all are situated in $a_{2k-1}=(-1)^{2k-1}\in (-1-\epsilon ,-1+\epsilon )$ .

Proof of equivalence

We still didn't answer the question, why both definitions are equivalent. More precisely: why is an accumulation point as a limit of a subsequence also an accumulation point in the neighbourhood definition? The following theorem gives the answer:

Theorem (Equivalence of the accumulation point definitions)

Let $(a_{n})_{n\in \mathbb {N} }$ be a sequence. For a number $h\in \mathbb {R}$ the following two definitions are equivalent:

• There is a subsequence in $(a_{n})_{n\in \mathbb {N} }$ which converges to $h$ .
• In every neighbourhood of $h$ there are infinitely many sequence elements of $(a_{n})_{n\in \mathbb {N} }$.

Proof (Equivalence of the accumulation point definitions)

Step 1: We prove: If there is a subsequence converging to $h$ , then in each $\epsilon$-neighbourhood of $h$ , there are infinitely many sequence elements.

We know that there is a subsequence converging to $h$ . Let $U$ be any neighbourhood of $h$. By the definition of convergence, we know that almost all elements of the subsequence must be inside that neighbourhood $U$. Those are infinitely many elements. And all those infinitely many elements of the subsequence are also part of the sequence. So there are infinitely many sequence elements inside $U$ and we have an accumulation point by the neighbourhood definition.

Step 2: We prove: If in each $\epsilon$-neighbourhood of $h$ here are infinitely many sequence elements, then one may construct a subsequence converging to $h$ .

In each neighbourhood of $h$, there are infinitely many sequence elements. We now construct a subsequence out of them and prove that this subsequence converges to $h$ . The trick for the construction is to choose a sequence of neighbourhoods getting smaller and smaller:

$U_{n}=\left(h-{\tfrac {1}{n}},h+{\tfrac {1}{n}}\right)$

Those neighbourhoods are all intervals, namely:

${\begin{aligned}U_{1}&=\left(h-1,h+1\right)\\U_{2}&=\left(h-{\tfrac {1}{2}},h+{\tfrac {1}{2}}\right)\\U_{3}&=\left(h-{\tfrac {1}{3}},h+{\tfrac {1}{3}}\right)\\U_{4}&=\left(h-{\tfrac {1}{4}},h+{\tfrac {1}{4}}\right)\\&\ \vdots \end{aligned}}$

Inside each of those $U_{n}$ there are infinitely many sequence elements and they all have a distance of less than ${\tfrac {1}{n}}$ to $h$. So if we pick one element out of each neighbourhood and put it into a subsequence $(t_{n})_{n\in \mathbb {N} }$ , then the elements will move closer and closer to $h$, as $n$ increases. One just needs to pay attention, that an element is not picked twice. Mathematically, we recursively define:

• $t_{1}$ is chosen to be any element of $U_{1}$.
• When picking $t_{n}$ with $n>1$, then the subsequence elements $t_{1},t_{2},\ldots ,t_{n-1}$ have already been chosen. As the next subsequence element, we choose $t_{n}$ to be any element of $U_{n}$, with the index of $t_{n}$ inside $(a_{n})_{n\in \mathbb {N} }$ being greater than any of the indices of $t_{1},t_{2},\ldots ,t_{n-1}$ . this way, we assure that no element is picker twice and further, the element indices in $(a_{n})_{n\in \mathbb {N} }$ are increasing.

Since $t_{n}\in U_{n}$ , there is $|h-t_{n}|<{\tfrac {1}{n}}$ (as for all $x\in U_{n}$ there is by definition $|h-x|<{\tfrac {1}{n}}$). This proves that $\lim _{n\to \infty }t_{n}=h$ , so the subsequence converges to $h$ and $h$ is an accumulation point by the subsequence definition.

Comparison: accumulation points vs. limits

Accumulation points are generalized limits

From both definitions of an accumulation point, we can directly imply that limits are a special kind of accumulation point: If we have a limit, then in each $\epsilon$-neighbourhood there are almost all points, i.e. infinitely many. This argument already proves the following theorem:

Theorem (Limits are also accumulation points)

For any convergent sequence, the limit is at the same time an accumulation point.

Question: Why does this theorem also hold for the subsequence-definition of an accumulation point?

If a sequence converges towards $h$, then the sequence can be considered a (trivial) subsequence of itself, so $h$ must be an accumulation point.

But on the converse, not every accumulation point is a limit. A counterexample is given in the introduction: A sequence like $a_{n}=(-1)^{n}{\frac {n}{n+1}}$ can have multiple accumulation points. But there may at most be one limit. Hence, the introductory example cannot have a limit.

So we can see the notion of an accumulation point as a generalization of a limit: Each limit is an accumulation point. But limits are unique. So if a sequence has 0 or more than one accumulation point, then it cannot have a limit:

Theorem

Every sequence with 0 or more than one accumulation point diverges.

Question: And what if there is 1 accumulation point? Does the sequence always have a limit then?

No, it doesn't. The reason is that a part of the sequence may converge to the accumulation point $h$ and another part may go to $\pm \infty$. A simple example, where this happens in the sequence

$a_{n}={\begin{cases}n,&n{\text{ even}}\\0,&n{\text{ odd}}\end{cases}}$

$h=0$ is an accumulation point (limit of all odd elements). But the even elements tend to $+\infty$, so the sequence diverges.

Accumulation points are not unique

The introductory example already had multiple accumulation points. So we have a counterexample that shows that accumulation points are not unique. This makes sense, since we introduced accumulation points to describe sequences which "seem to have multiple limits" (i.e. they have multiple accumulation points). It is even possible to get more than two accumulation points. Actually, infinitely many. Actually, one can even find sequences containing each real number as an accumulation point. Maybe, you can find one?

Exercise

Find a sequence $(a_{n})_{n\in \mathbb {N} }$, which has each real number $x\in \mathbb {R}$ as an accumulation point.

Proof

The rational numbers $\mathbb {Q}$ are dense in $\mathbb {R}$. And we know that $\mathbb {Q}$ is countable, so there is a bijective function $f:\mathbb {N} \to \mathbb {Q}$. This function can also be viewed as a sequence $a_{n}=f(n)$.

Now, consider any $x\in \mathbb {R}$ and any neighbourhood $U$ of $x$. By definition, there must be an open interval $(x-\epsilon ,x+\epsilon )$ inside this neighbourhood for some $\epsilon >0$ . This interval contains infinitely many rational numbers $a_{n}\in \mathbb {Q}$. I.e. infinitely many sequence elements. So $x$ is an accumulation point of $(a_{n})_{n\in \mathbb {N} }$. and this statement holds for any real number $x\in \mathbb {R}$

Properties of accumulation points and limits

limit	accumulation point
Every subsequence converges towards the limit.	At least one subsequence converges towards an accumulation point.
In every $\epsilon$-neighbourhood, there are almost all sequence elements.	In every $\epsilon$-neighbourhood, there are infinitely many sequence elements.
A sequence has at most one limit.	A sequence can have arbitrarily many accumulation points.
Every limit is an accumulation point.	Some accumulation points are not a limit.

Limes superior and limes inferior

→ Main article: Lim sup and Lim inf

There are two special accumulation points of a sequence, that you might encounter within some mathematical lectures. For a sequence $(a_{n})_{n\mathbb {N} }$ bounded from above, the limes superior is the greatest accumulation point. It is denoted $\limsup _{n\to \infty }a_{n}$. Conversely, the limes inferior $\liminf _{n\to \infty }a_{n}$ is the smallest accumulation point of a sequence (bounded from below). If the sequence is unbounded from above or below, we write $\liminf _{n\to \infty }a_{n}=+\infty$ or $\liminf _{n\to \infty }a_{n}=-\infty$, respectively. That means divergent sequences (!) can also have a limes superior or a limes inferior. Further details are given in the article „Lim sup and Lim inf“.