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I learned how to construct a Cantor Set, and I am asked to do the following.

"Construct a bounded set with exactly 3 limit points."

Since the Cantor set contains infinitely many points, I don't think something like it will not work. But this is the only thing that I have learned from the book that tells me anything about constructing a set that has a limit point.

I am also considering the interval $[0,1]$ and constructing a set so that the limit points are $\{0,1/2,1\}$.

If possible, I would like to see more than one simple examples because I am new to analysis and I have no teacher. It's very tough.

hyg17
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    Can you construct a bounded set with exactly one limit point? – Gerry Myerson May 05 '13 at 07:37
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    Or to make it a little bit clearer, to solve this problem, just answer the following 3 questions: 1. How would you construct a set contained within $[0; 1]$ that has $0$ as a limit point? 2. How would you construct a set contained within $[0; 1]$ that has $\dfrac{1}{2}$ as a limit point? 3. How would you construct a set contained within $[0; 1]$ that has $\left{0; \dfrac{1}{2}; 1 \right}$ as limit points? – user49685 May 05 '13 at 07:53
  • For example, Let E_0 = [0,1]. Now let E_1 = (1/2,1],... E_n = (1/(n+1),1]. I want to say that 1 is a limit point if the infinite intersection of E_i s are taken. – hyg17 May 05 '13 at 22:35
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    The solution is probably much easier than you think. – Giuseppe Negro May 07 '13 at 20:09
  • This StackExchange question is concerned with realizing any specified nonempty closed set as the set of subsequence limit points of some sequence, which should be of help. – Dave L. Renfro May 08 '13 at 21:08
  • Thanks for the advice, everyone. But I'm not really getting there.

    I want to say that taking the infinite subsets of [0,1/n),[2/n,2/(n+1)],(2/(n+1),1] leads to having 0,1/2 and 1 as a limit point, but I don't think it works like that. Furthermore, I don't know how to prove so.

    – hyg17 May 09 '13 at 07:52
  • But I'm not really getting there. O-K, then this should be of much greater help than the last StackExchange question I cited. – Dave L. Renfro May 09 '13 at 20:45

7 Answers7

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Construct a set with exactly one limit point and then add two distinct copies of it; let me be more clear: you'll surely agree with the fact that $\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}$ has only one limit point. Thus if we repeat twice a translation we'll get what we're looking for: $(\{0\} \cup \{{\frac{1}{n} : n \in \mathbb{N} ^{*}}\}) \sqcup (\{2\} \cup \{{\frac{1}{n}+2 : n \in \mathbb{N} ^{*}}\}) \sqcup (\{4\} \cup \{{\frac{1}{n} +4: n \in \mathbb{N} ^{*}}\})$.

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Hint: What are the limit points of $\left\{\frac{1}{n}\middle|\ n\in \mathbb{N}\right\}$?

Asaf Karagila
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Take a sequence $\{ a_n \}_{n \in \mathbb{N}}$ converging to $0$, another sequence $\{ b_n \}_{n \in \mathbb{N}}$ converging to $\frac{1}{2}$, and another sequence $\{ c_n \}_{n \in \mathbb{N}}$ converging to $1$. Then just take the set $$ S := \{ a_n,b_n,c_n; n \in \mathbb{N} \} $$ to be their union. By definition $0,\frac{1}{2}$ and $1$ will be limit points. For example, you can let $a_n := \frac{1}{n}$, $b_n := \frac{1}{2} - \frac{1}{2^n}$, $c_n := 1 - \frac{1}{n!}$. This choice also conforms to your wish that $S \subseteq [0,1]$.

Rookatu
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"Construct a bounded set with exactly 3 limit points."

consider this set: $$\{0.1,\space 0.2,\space0.3,\space\space0.11,\space0.22,\space0.33,\space\space0.111,\space0.222,\space0.333,\space\space0.1111,\space0.2222,\space0.3333,...\}$$

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Let $$A=\left\{\frac ab\;\bigg|\; a,b\in\mathbb N, a=1\text{ or }a=b-1\text{ or }b=2a\pm1\right\}$$

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Let $$X=\left\{z\in\mathbb{C}\::\:|z|=1-\frac{1}{n},n\in\mathbb{N}_{\geq 1}, \arg z\in\{0,\frac{2}{3}\pi,\frac{4}{3}\pi\}\right\}.$$

Dan Rust
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{1/n ; n belongs to N} U {(n+1)/n ; n belongs to N} U {(2n+1)/n ; n belongs to N}

this is an infinite set with 3 limit points {0,1,2}