# Convergent sequences and accumulation points

I am very confused about the relationship between convergent sequences and accumulation points.

Basically, my question boils down to: if $(a_n)_{n\in N}$ is a convergent sequence in $\mathbb{R}$ then,

(1) Does the set $\{a_n\}$ have exactly one accumulation point?

(2) If so, does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?

I'm tempted to say no to (1), but I'm afraid that I'm missing something. My counter-example to (1) is $\{a_n\} = \{ 4, 3, 2, 1, 0,0,0,...\}$ (i.e. inserting $0$s after the 4th element). Then the set has no accumulation point and it converges to 0. Is that correct?

If I'm right about (1), let's assume that $\{a_n\}$ has exactly one accumulation point. Then does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?

Lastly, is it possible that $\{a_n\}$ could have more than one accumulation point?

Update:

I define accumulation point as: Let $a$ be an accumulation point of $A$. Then $\forall \ \epsilon >0$, $B_{\epsilon}(a) \setminus \{a\}$ contains an element of $A$.

• What is your definition of accumulation point? – wjm Dec 4 '17 at 5:44
• @Raptor updated. – user1691278 Dec 4 '17 at 5:57
• The situation is quite simple. If a sequence $(a_n)_{n\in\Bbb N}$ is convergent, then every accumulation point of the set $\{\, a_n\mid n\in\Bbb N\,\}$ is equal to the limit of the sequence (easy proof using an extracted sequence). This implies there is at most one accumulation point, but not that such an accumulation point exists. – Marc van Leeuwen Dec 4 '17 at 9:52

I am calling $x$ an accumulation point of the set $A$ iff $(B(x,\epsilon) \cap A)\setminus \{x\} \neq \emptyset$ for all $\epsilon>0$.

Suppose $a_n \to a$.

The set $\{a_n\}_n$ can have at most one accumulation point which would have to be $a$. If $b \neq a$, then there is some $\epsilon>0$ such that $B(b,\epsilon)$ contains a finite number of points hence $b$ cannot be an accumulation point.

Note that the sequence $a_n = 1$ has $\{a_n\}_n = \{1\}$ which has no accumulation points.

In general, the set $\{a_n\}_n$ will have $a$ as an accumulation point iff for all $N$ there is some $n \ge N$ such that $a_n \neq a$.

If $\{a_{n}\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\in B_{1}(a)-\{a\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\{a,a_{1},...,a_{n_{1}}\}$, proceed in this way we have $a_{n_{k}}\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\rightarrow a$.

Here I use the following definition:

$a$ is an accumulation point for $A$ if for every $\delta>0$, $(B_{\delta}(a)-\{a\})\cap A\ne\emptyset$.

And note that in the topology of ${\bf{R}}$, being such an accumulation point also implies that $(B_{\delta}(a)-\{a\})\cap A$ contains infinitely many points.

• If $\{a_{n}\}$ has two distinct accumulation points, say, $a$, $b$, one can find subsequences $(a_{n_{k}})$, $(a_{n_{l}})$ such that $a_{n_{k}}\rightarrow a$ and $a_{n_{l}}\rightarrow b$, a contradiction. – user284331 Dec 4 '17 at 5:54
• Your counterexample $(a_n)$ is not a convergent sequence though... – user1691278 Dec 4 '17 at 6:04
• I have edited, let's see if there is any mistake. – user284331 Dec 4 '17 at 6:13

The usual definition of an accumulation point (for a subset of $\mathbb{R}$) is as follows:

Let $A \subseteq \mathbb{R}$. We say that $a$ is an accumulation point of $A$ if for all $r > 0$ the set $B(a,r) \cap A \setminus \{a\}$ is nonempty. That is, every ball centered at $a$ contains a point of $A$ other than $a$ itself.

Even if the sequence $(a_n)$ converges, the set $\{ a_n \}$ needn't have any accumulation points. For example, any constant set $\{a, a, a, \dotsc, \} = \{a\}$ does not have any accumulation points (as no ball centered at $a$ contains any point of the set other than $a$, but the sequence $(a,a,a,\dotsc)$ converges to $a$. On the other hand, if $\{a_n\}$ has an accumulation point, and $(a_n)$ is convergent, then the accumulation point is necessarily the limit.

• Thank you. Does that mean it cannot have more than one accumulation point since limits are unique? – user1691278 Dec 4 '17 at 6:05
• In $\mathbb{R}$ with the usual topology, a convergent sequence can have at most one accumulation point, yes. As noted above, it needn't have even that. – Xander Henderson Dec 4 '17 at 14:38

Here's a somewhat more general proof of this:

Let $X$ be a Hausdorff topological space, and let $\left\{x_n\right\}_{n\in\mathbb N}$ be a sequence in $X$ that converges to $x\in X$. Suppose that $y\in X$ is an accumulation point of $\left\{x_n\right\}_{n\in\mathbb N}$ such that $x\neq y$. Then, since $X$ is Hausdorff, there are disjoint open neighborhoods $U$ and $V$ of $x$ and $y$, respectively. Since $\left\{x_n\right\}_{n\in\mathbb N}$ converges to $x$, there is an $N\in\mathbb N$ such that $x_n\in U$ whenever $n\geq N$. However, this implies that there are at most $N-1$ elements of $\left\{x_n\right\}_{n\in\mathbb N}$ in $V$ different from $y$. Denote this set of finite elements by $\left\{y_n\right\}_{n=1}^m$. Again, since $X$ is Hausdorff, there are disjoint open neighborhoods $U_n$ and $V_n$ of $y_n$ and $y$, respectively. Then $\bigcap_{n=1}^m V_n$ is an open neighborhood of $y$ containing no elements of $\left\{x_n\right\}_{n\in\mathbb N}$, which is a contradiction.

• In the context of this question, it is rather confusing to use braces to designate the sequence, especially when you then immediately use the very same notation for the set. OP rightly uses parentheses for the sequence, and braces for the set. – Marc van Leeuwen Dec 4 '17 at 9:56

Yes, you are right. If $x_n$ is a convergent sequence, then

• $\{x_n\}$ has an accumulations point if and only if $\{x_n\}$ is infinite.
• $\{x_n\}$ has at most one accumulation point and if it has one, then it coincides with the limit.

None of these statements is true for non-convergent squences.