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What are some examples of complex or real sequences having more than two accumulation points ?

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5  
$a_n=(-1)^n$ is the most simple example. –  Thomas Andrews Jan 16 '13 at 19:43

3 Answers 3

  1. $\langle 1,2,3,1,2,3,1,2,3,\dots\rangle$ and its congeners.

  2. Slightly more generally, let $\{a_0,\dots,a_{n-1}\}$ be a finite set of real numbers, and let $\langle\epsilon_n:n\in\Bbb N\rangle$ be any sequence converging to $0$. Let $x_k=a_{k\bmod n}+\epsilon_k$ for $k\in\Bbb N$; then $\langle x_k:k\in\Bbb N\rangle$ has $\{a_0,\dots,a_{n-1}\}$ as its set of cluster points.

  3. $\left\langle n\alpha-\lfloor n\alpha\rfloor:n\in\Bbb N\right\rangle$ for any irrational $\alpha$: every real number in $[0,1]$ is a cluster point.

  4. Any enumeration of $\Bbb Q$ in the form $\langle q_n:n\in\Bbb N\rangle$: every real number is a cluster point.

Added: Let $\langle X,d\rangle$ be a metric space, and let $\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$. If $\langle X,d\rangle$ converges to $x$, meaning that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x,x_n)<\epsilon$ for all $n\ge m_\epsilon$, then we say that $x$ is the limit of the sequence. A more informal (but perhaps more understandable) way to say this is that $x$ is the limit of the sequence if every open neighborhood of $x$ contains all but finitely many terms of the sequence.

If, on the other hand, $x$ merely has the property that each open neighborhood of $x$ contains infinitely many terms of the sequence, $x$ is a cluster point or accumulation point of the sequence. Take my first sequence above, for instance: $1$ is a cluster point, because if $V$ is any open interval around $1$, the infinitely many terms $x_0,x_3,x_6,x_9,\dots$ are all in $V$, since they’re all equal to $1$. (My $\Bbb N$ starts at $0$, not $1$.) Cluster points of a sequence are sometimes also called limit points of the sequence, but this is a confusing terminology that ought to be avoided. They can also be called subsequential limits of the sequence, because it can be proved (fairly easily) that $x$ is a cluster point of a sequence $\langle x_n:n\in\Bbb N\rangle$ if and only if some subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ of $\langle x_n:n\in\Bbb N\rangle$ converges to $x$.

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Thanks alot,I'm a bit confused about the concept of convergence and limit points. If a metric space is hausdorff ,then a sequence can't have more than one limit.What's the definition of a limit in that case ? Isn't it one of the accumulation points with certain property? –  nabil Jan 16 '13 at 19:52
    
@nabil If a limit exists in a Hausdorff space it is also the only accumulation point. (But note that a single accumulation point need not mean that a limit exists, e.g. $a_n = ((-1)^n+1)n$. –  Hagen von Eitzen Jan 16 '13 at 19:54
    
@nabil: Every metric space is Hausdorff, so a sequence in a metric space can converge to at most one point. I think that you’re getting confused by the unfortunate use of limit point to mean two different things. I’ll add something to my answer to address that. –  Brian M. Scott Jan 16 '13 at 20:01
    
@BrianM.Scott 4th example is beautiful. –  user140374 Apr 20 at 0:13

Let $z$ be a complex number of norm 1.

two muppets

If $\theta$ is a irrational multiple of $\pi$ the sequence $z^n$ is dense. And if $\theta$ is a rational multiple of $\theta$ the sequence $z^n$ is periodic.

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+1 --- for $\theta$ an irrational multiple of $\pi$, this is a particularly elegant example of a sequence with an uncountable number of limit (accumulation) points. –  Niel de Beaudrap Jan 17 '13 at 1:50
    
you are right!!! $\theta$ must be a irrational multiple of $\pi$ –  user52188 Jan 17 '13 at 1:53

$\mathbb{Q_{+}}$ has accumulation points everywhere in $\mathbb{R}$, see link for how to index the positive rational numbers. (Every irrational number is the limit of a sequence of rational numbers).

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