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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** ??2012-10-24

4 Answers 4

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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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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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    Of course; I just wanted to explicate stuff in case OP desired that $p \in S$.2012-10-24
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The sequence $(-1)^n\left(1+\frac{1}{n}\right)$ has exactly two A.Ps $1$ and $-1$.

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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    Prove that $a_{2n}\rightarrow a , a_{2n+1}\rightarrow −a ,a_{2n}\neq a ,a_{2n+1}\neq −a$, for (almost) all $n\in\mathbb N$. Also show that $a$ and $−a$ are the only numbers that can be the limit points of subsequences of $(a_n)_{n\in\mathbb N}$.2018-11-26
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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 sequence2012-10-24