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The definition in my text defines A.P. as: A point p is an A.P. of a set S if it is the limit of a sequence of points of S-{p}.

I am confused by my classnotes and information about A.P. in my textbook is not enough for me to understand the topic.

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So: "accumulation point" is for a set not for a sequence ?? –  GEdgar Oct 24 '12 at 21:24

4 Answers 4

up vote 3 down vote accepted

The sequence $(-1)^n\left(1+\frac{1}{n}\right)$ has exactly two A.Ps $1$ and $-1$.

Or more generally if $(a_n)_{n\in \mathbb{N}}$ is a sequence s.t. $a_n \to a \neq0 $ and the set $\{a_n:n \in \mathbb{N}\}$ is infinite then the sequence $((-1)^na_n)_{n\in \mathbb{N}}$ has exactly two accumulation points $a$ and $-a$.

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how can this be proved? –  Maximiliano Oct 24 '12 at 21:14
    
Hint: Prove that $a_{2n} \to a$ ,$a_{2n+1} \to -a$ ,$a_{2n} \neq a$ ,$a_{2n+1} \neq -a$ for (almost) all $n \in \mathbb{N}$ and that ${a,-a}$ can be the only limit points of subsequences of $(a_n)_{n \in \mathbb{N}}$. –  P.. Oct 24 '12 at 21:25

Exactly two accumulation points can be done using the sequence $$\frac{1}{3}, \frac{2}{3},\frac{1}{4},\frac{3}{4}, \frac{1}{5},\frac{4}{5},\frac{1}{6}, \frac{5}{6},\frac{1}{7}, \frac{6}{7},\dots,$$ and in many other ways. In this case the accumulation points are $0$ and $1$.

Here I intend the set $S$ of the definition to be $\{\frac{1}{3}, \frac{2}{3},\frac{1}{4},\frac{3}{4}, \frac{1}{5},\frac{4}{5},\frac{1}{6}, \frac{5}{6},\frac{1}{7}, \frac{6}{7},\dots\}$. We could work with a somewhat larger set for $S$, by, in addition to the points mentioned, throwing in all the integers, or something else that would not produce additional accumulation points.

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I presume your set is the unit interval? –  Lord_Farin Oct 24 '12 at 20:57
    
No, my set $S$ is the set $\{\frac{1}{3},\frac{2}{3},\frac{1}{4},\dots\}$. –  André Nicolas Oct 24 '12 at 21:05
    
Of course; I just wanted to explicate stuff in case OP desired that $p \in S$. –  Lord_Farin Oct 24 '12 at 21:05

This seems a little vague. What is your question? Do you want an example of the sequence or do you want more info. about accumulation points?

Intuitively, accumulations points are the points of the set S which are not isolated.

e.g. 1 is an A.P. of (0,1) but 2 is not an A.P of $(0,1) \cup$ {2} (because it's isolated.) To prove this, the set $X = $ ($(0,1) \cup$ {2} ) - {2} = $(0,1)$ but 2 is not the limit of a sequence in $(0,1)$ so 2 is not an A.P. of $(0,1) \cup$ {2}

from there its pretty easy to construct such a sequence.

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Yes an example of a sequence –  Maximiliano Oct 24 '12 at 20:58

Point $p$ is accumulation point of a set $S$, if every punctured neighborhood of the point $p$ contains points of $S.$ (In other words, if there exists sequence $\{x_n\} \subset S$ that converges to $p.$) Set $$S={\left \lbrace \sin{\dfrac{1}{n}}, \cos{\dfrac{1}{n}} \right \rbrace}_{n \in \mathbb{N}}=\left \lbrace \sin{1}, \cos{1},\sin{\dfrac{1}{2}}, \cos{\dfrac{1}{2}}, \ldots, \sin{\dfrac{1}{n}}, \cos{\dfrac{1}{n}} , \ldots \right \rbrace$$ has exactly two accumulation points.

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