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I came across this definition of cluster point. Let $S \subset R$ be a set. A number $x \in \mathbb R$ is called a cluster point of $S$ if for every $\varepsilon > 0$, the set $(x−ε, x+ε)∩S \setminus \{x\}$ is not empty.

For the set $\{1/n:n∈\mathbb N\}$ why 1 is not a cluster point? I came across this proof while googling but still cant understand it

http://mathonline.wikidot.com/cluster-points

Suppose that $0<c≤1$. By one of the Archimedean corollaries, since $1c>0$ then there exists a natural number $n_c\in\mathbb N$ such that $n_c−1≤1c<n_c$ and so $1n_c<c≤1n_c−1$. Choose $δ_0=\min\{∣c−1n_c∣,∣1n_c−1−c∣\}. Then $Vδ(c)={x∈\mathbb R:∣x−c∣<\min{∣c−1n_c∣,∣1n_c−1−c∣}}=∅$.

I take $n_c=10, c=1$, then delta is 0.888888889. I can find an $x=0.5$ such that the inequality satisfies. Am I doing it wrong?

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  • $\begingroup$ I strongly recommend you to visit meta.math.stackexchange.com/questions/5020/…, right now the question is hardly understandable. $\endgroup$ – BigbearZzz Nov 10 '15 at 1:37
  • $\begingroup$ 1 isn't a cluster point because it doesn't fit the definition for any epsilon less than 1/2. The nearest 1/n to 1 is an entire 1/2 unit away. 1/n s do not cluster around 1. They don't even come close to the 1. $\endgroup$ – fleablood Nov 10 '15 at 2:11
  • $\begingroup$ To be a cluster point, the condition must be true for all epsilon greater than 0. Not just a specific epsilon. delta = .88888889 is a very large delta. You have to show it also is true for delta =.00000001. can you? You have to show it is true for delta = $10^{-100}$. Can you? $\endgroup$ – fleablood Nov 10 '15 at 2:54
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A number x∈R is called a cluster point of S if for !!!!!EVERY!!!!! ε>0, the set (x−ε,x+ε)∩S∖{x} is not empty.

So if $\epsilon = 2/3$ then, yes, $(1 - \epsilon = 1/3, 1 + \epsilon = 1 2/3)\cap \{1/n\} / \{1\} =\{1/2\}$ is not empty. Fine but we need to show that for !!!!!EVERY!!!!!! epsilon.

Let $\epsilon \le 1/2$ then $(1 - \epsilon \ge 1/2, 1 + \epsilon )\cap \{1/n\} / \{1\}$ is empty. So one is not a cluster point.

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If we take $\epsilon$ to be $.25$ and we have $S = \{\frac{1}{n} : n \in \mathbb{N}\}$, then $(1-\epsilon,1+\epsilon) \cap S = (.75,1.25) \cap S = 1,$ so $1$ is not a cluster point.

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