# Convergence of sequence points to the point of accumulation

Wikipedia says:

Every finite interval or bounded interval that contains an infinite number of points must have at least one point of accumulation. If a bounded interval contains an infinite number of points and only one point of accumulation, then the sequence of points converge to the point of accumulation. [1]

How could i imagine a bounded interval of infinite points having a single limit point and that sequence of interval's points converge to that limit point.

Any examples?

Thank you.

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Every convergent sequence with infinitely many distinct terms gives an example. – Jonas Meyer Aug 15 '11 at 22:54
You should note that a sequence is a function whose domain is the nonnegative integers and whose range is the space. It is not a set of points. – ncmathsadist Aug 16 '11 at 0:00
@ncmathsadist: True enough. However, it's easy to show that, in this particular case, any sequence composed of distinct points in the set must converge to the accumulation point. – Ilmari Karonen Aug 16 '11 at 13:34

What about: $$\left \{\frac{1}{n}\Bigg| n\in\mathbb Z^+ \right \}\cup\{0\}$$ This set (or sequence $a_0=0,a_n=\frac{1}{n}$) is bounded in $[0,1]$ and has only limit point, namely $0$.

A slightly more complicated example would be: $$\left \{\frac{(-1)^n}{n}\Bigg| n\in\mathbb Z^+ \right \}\cup\{0\}$$ In the interval $[-1,1]$. Again the only limit point is $0$, but this time we're converging to it from "both sides".

I believe that your confusion arises from misreading the text. The claim is that if within a bounded interval (in these examples, we can take $[-1,1]$) our set (or sequence) have infinitely many points and only one accumulation point, then it is the limit of the sequence.

There is no claim that the interval itself has only this sequence, or only one limit/accumulation point.

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A bounded interval $I\subset{\mathbb R}$, say $I:=[{-1},1]$, contains all real numbers $x$ with $-1\leq x\leq1$, which is a huge number of points.

If you are given a ${\it set}\ A$ of "special points" in this interval, say $A:=\{x\in I\>|\>e^{2\pi i/x}=1\}$, then such a set may have accumulation points. An ${\it accumulation\ point\ \xi\ of\ the\ set}\ A$ is defined by the property that any punctuated neighborhood $\dot U$ of $\xi$ contains at least one point from $A$. If $A$ is an infinite bounded set then by Bolzano's theorem there is at least one such accumulation point which may or may not belong to $A$.

Sometimes the given set $A$ has a natural numbering, i.e., there is a bijective function $\phi:\ {\mathbb N}\to A, \ n\mapsto a_n$ which "produces" the points of $A$ one for one. The accumulation points of the set $A$ are also accumulation points of the sequence $(a_n)_{n\geq 0}\$ (see below). A point $\xi$ is called an ${\it accumulation\ point\ of\ the\ sequence}\ (x_n)_{n\geq0}$ if any neighborhood $U$ of $\xi$ contains $x_n$'s with arbitrary large $n$. But note that in a plot of the situation you might see nothing special: The sequence defined by $x_n:=(-1)^n$ has the two accumulation points $\pm1$ which is all you can see.

Now comes the question you raised about the situation when a given set $A$ has exactly one accumulation point $\xi$. Such a set is countable to begin with (I omit the proof), therefore this set can be bijectively produced by a sequence $(x_n)_{n\geq0}$. Maybe this set was defined as the image set of such a sequence to begin with. I claim that for any such sequence $\lim_{n\to\infty} x_n=\xi$.

Proof: Consider an $\epsilon>0$. If there were infinitely many $x_n\geq\xi+\epsilon$ then these $x_n$ would form an infinite bounded set without an accumulation point, and similarly it is impossible that infinitely many $x_n$ are $\leq\xi-\epsilon$. It follows that there is an $n_0$ with $x_n\in U_\epsilon(\xi)$ for all $n>n_0$.

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