5

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.

Charlie
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5 Answers5

15

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 $\epsilon = \frac{1}{2} (\alpha - \frac 1 n)$.

  • 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 $1/n < \epsilon$. –  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. –  Sep 14 '13 at 20:06
  • bongers im sorry if im asking too much can you elaborate a bit more in your last sentence where 0 < alpha < 1 thanks – Charlie Sep 14 '13 at 20:16
  • For $\alpha = 1$, it is more logical to set $\epsilon = 1/2$ as every $0 < \epsilon \leq 1/2$ will do. – Stand with Gaza Jul 02 '18 at 19:42
  • The $0 < \alpha < 1$ case is wrong. Consider $\alpha = 0.499$. Instead I think you should select the $1/n$ closest to $\alpha$ and set $\epsilon = |\alpha - 1/n|$. – Stand with Gaza Jul 03 '18 at 06:25
  • slightly off topic but would $(0,1)\cup (1,2)$ count as having three limit points? – PythonSage Jan 09 '20 at 04:42
6

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.

Dominic Michaelis
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3

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?

Dan Rust
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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
1

You can prove no other element in the set is a limit point because you can let 1/n be any of the elements and show that there is an r that gives you a neighborhood that doesn't have any of the other the elements in it. Because to be a limit point you have to have another element in the set in the neighborhood. for example, if you have 1/5 then you can make a neighborhood where r is really small then 1/4, and 1/6 wont be in the set and there is no other numbers in the neighborhood. So you would do 1/n-1 and 1/n and 1/n+1 and show that for a very small r like half the distance between 1/n-1 and 1/n then there wont be any elements. It must be for EVERY neighborhood. Correct me if I'm wrong.

Robert Duke
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0

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

copper.hat
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