Construct an example of a set of real numbers E that has no points of accumulation and yet has the property that for every ε > 0 there exist points x, y ∈ E so that 0 < |x − y| < ε.
so i know we need a convergent sequence to show that the difference between two elements can be as small as we like, also it should be a set not an interval to avoid accumulation points. but if a sequence converge to L, wouldnt L be an accumulation point too?
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$\begingroup$You can take the sequence $\left(\sqrt n\right)_{n\in\mathbb N}$, for instance. Can you prove that it works?
$\endgroup$ 0 $\begingroup$Consider $E := \mathbb{N}\setminus\{0\} \cup \{n + \frac{1}{n} \big| \ n \in \mathbb{N}\setminus\{0\} \}$
$E$ satisfies $\forall \epsilon > 0 \ \exists x,y\in E, \, |x-y|<\epsilon$.
There are no accumulation points in $E$ :
Let $x\in E$, there is $n \in \mathbb{N}$ such that $x = n$ or $x = n + \frac{1}{n}$. Now take $\epsilon < \frac{1}{n}$, and write $B(x,\epsilon)$ for the open ball centered on $x$ of radius $\epsilon$. Ona has $E \cap B(x,\epsilon) = \{x\}$ hence $x$ is not a point of accumulation.
More generally, all that is required for the existence of $x,y \in E$ satisfying $|x-y| < \epsilon$ is that there exist arbitrarily close points in $E$, but this does not mean that set $E$ is bounded. For $E \subseteq \mathbb R^+$, we can think of this as the existence of strictly increasing sequences of positive reals with no finite limit, but whose successive differences tend to $0$. We are already familiar with many examples of such sequences; indeed, many such sequences are found from the partial sums of certain infinite series: Let the sequence $\{s_n\}_{n \ge 0}$ be defined by $$s_n = \sum_{k=0}^n a_k,$$ where $\{a_k\}_{k \ge 0}$ is an infinite sequence of strictly decreasing positive reals satisfying $a_k > a_m > 0$ whenever $k < m$. If $\displaystyle \lim_{n \to \infty} s_n$ exists and is finite--that is to say, the infinite series converges to some value $L$--then $L$ is an accumulation point of the sequence of partial sums. But if it does not--and indeed, there are many such examples of divergent series--then as long as we have $a_k \to 0$ as $k \to \infty$, we will have a sequence $\{s_n\}$ that meets your criteria: no accumulation point but there exist successive terms whose absolute difference can be made arbitrarily small.
There also exist rather more "pathological" examples; e.g., $$s_n = 2^{\lceil \log_2 n \rceil} - 2^{-n},$$ which has the desired property of no accumulation point, yet arbitrarily small and arbitrarily large differences in successive terms.
$\endgroup$ 1 $\begingroup$Consider
$$E=\left\{\sum_{k=1}^n k^{-1}:n\in \Bbb N\right\}.$$
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