Why 1 is not a cluster point of $\frac1n$ 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?

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.  
