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I have to give An example of set with a countably infinite set of accumulation points, and I say:

We can consider the set or real numbers and we take an arbitrary real number $x$ then the interval $(x-\epsilon,x+\epsilon)$ is infinite and since we can do this with any other real number,hence we have a set with a countably infinite set of accumulation points.

How can I say it more formally or better ?. Thank you :)

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The real interval $]x- \varepsilon, x+ \varepsilon[$ is not countably infinite ; it is uncountably infinite. This is not one of the examples you are looking for ; anything with non-empty interior will have uncountably many limit points. – Patrick Da Silva Sep 1 '14 at 0:17
thank You @PatrickDaSilva Then which could be a good example? that is not the interval (0,1) or the rational numbers :) thanks – user162343 Sep 1 '14 at 0:24

Let $S_{k}=\displaystyle\left\{k+\frac{1}{n}: n\in\mathbb{N}\right\}$ for every integer $k$. For example, $S_{0}=\displaystyle\left\{\frac{1}{n}: n\in\mathbb{N}\right\}$ has only one accumulation point, namely $x=0$. In general, the set $S_{k}$ has only one accumulation point, namely $x=k$.

How many accumulation points does the set $S=\displaystyle \bigcup_{n=0}^{\infty} S_{n}$ have?

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Well we can argue that since each $S_{k}$ has one point of acumulation the union has infinitely points because there are infinitely k isnt it ? – user162343 Sep 1 '14 at 0:29
@user162343: Yes, exactly. The union $S$ has countably infinite set of accumulation points: Indeed, $\{0, 1, 2, 3, …\}$ is the set of accumulation points for the set $S$. – Prism Sep 1 '14 at 0:34

Take any discrete subset $X \subseteq \mathbb R$, that is, for each $x \in X$, there exists $r > 0$ such that $]x-r,x+r[ \, \cap \, X = \{x\}$. For each $x \in X$, choose a sequence $x_n(x) \in ]x-r,x+r[$ such that $x_n(x) \to x$. Then the set $$ \bigcup_{x \in X} \{ x_n(x) \, | \, n \in \{1,2,\cdots\} \} $$ has limit set point $X$.

Now all you have to do is pick a discrete subset of your choice ($\mathbb N, \mathbb Z$, etc.) and such sequences ($x_n(x) = x + \frac r{2n}$ will always do the trick).

Hope that helps,

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