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Construct a set of real numbers having exactly three limit points.

Progress: I know the set $\{ 1/n + k\}$ has one limit point $k$, but I am unable to justify why.

Please be a bit elaborate in your explanation because I need to understand this question if I face such a question in my exam.

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well i do know that the set { 1/n + k – Charlie Sep 14 '13 at 20:00
well i do know that the set { 1/n + k} has only k as a limit point can anybody explain why – Charlie Sep 14 '13 at 20:01

Consider the set $A = \{\frac 1 n : n \in \Bbb{Z}^+\} = \{1, 1/2, 1/3, 1/4, ...\}$. This has exactly one limit point, namely $0$.

Next consider the set $\{1 + \frac{1}{n} : n \in \Bbb{Z}^+\}$. This has exactly one limit point, namely $1$. Now consider the union of these two sets: Do you see why it has exactly two limit points? Do you see how to extend this to having $3$, or $n$, limit points?

To show that $A$ actually has $0$ as a limit point, just note that if given $\epsilon > 0$, there exists an $n$ for which $\frac{1}{n} < \epsilon$. So every neighborhood of $0$ intersects $A$ at a point.

Now suppose that $\alpha$ is a limit point of $A$. If $\alpha < 0$, set $\epsilon = \frac{-\alpha}{2}$, and note that $(\alpha - \epsilon, \alpha + \epsilon) \cap A = \emptyset$, a contradiction. Likewise, $\alpha > 1$ leads to a contradiction. If $\alpha = 1$, set $\epsilon = \frac{1}{3}$.

Finally, if $0 < \alpha < 1$, let $n$ be the least integer satisfying $\frac{1}{n} < \alpha$ and choose $\epsilon = \frac{1}{2} (\alpha - \frac 1 n)$.

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yes I do thanks but please can u justify for example why 0 is the only limit point of 1/n , i know the definition f a limit point but i dont know how to apply it – Charlie Sep 14 '13 at 19:59
@MEHYEDDINEZEIN To show that $0$ is a limit point, just note that every neighborhood of zero contains an element of the set. For if given $\epsilon > 0$, choose $n$ with $1/n < \epsilon$. – user61527 Sep 14 '13 at 20:01
ok then 0 is a limit point why is it the only limit point how can you prove it is the only limit point – Charlie Sep 14 '13 at 20:03
that proof is called the archemidean property i believe – Charlie Sep 14 '13 at 20:04
@MEHYEDDINEZEIN I've expanded my answer. – user61527 Sep 14 '13 at 20:06

Hint: The set $\{\frac{1}{n}\mid n\in\mathbb{N}^+\}$ has a single limit point. Can you see how to construct a set which has three limits points from this?

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yes i do, 1/n has 0 as a limit point, 1 + 1/n has 1 as a limit point point and k + 1/n has k as a limit oint but why is k the only limit point and no other – Charlie Sep 14 '13 at 19:58

Take a convergent non constant sequence $(a_n)_{n\in \mathbb{N}}$. Then the set $\{ a_n : n \in \mathbb{N}\}$ does have exactly one limit point. Now consider \[\bigcup_{i=1}^k \{a_n + i: n \in \mathbb{N}\}\] this does have exactly $k$ limit points.

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Take the sequence $x_{n+1} = x_n + 1 \mod 3$, with $x_1 =0$. Then take $a_n = x_n + \frac{1}{n}$.

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