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I'm thinking about how to construct, if possible, a sequence $(x_n)$ with the set of adherence values of $(x_n)$ equal to $\mathbb{N}$.

I already constructed a sequence $(y_n)$ with the set of adherence values equal to $[0,1]$, that being an enumeration of the rationals in $[0,1]$, since every real number is a limit of a sequence of rational numbers. So I'm thinking that it is possible to build such sequence with $\mathbb{N}$ being it's set of adherence values, since $\mathrm{card } ([0,1]) > \mathrm{card } (\mathbb{N})$.

Can anyone give me a hint? (Just a hint, not the answer please)

Thanks in advance.

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2 Answers 2

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HINT: For each $n\in\Bbb N$ let $\sigma_n$ be a sequence converging to $n$. Now combine the sequences $\sigma_n$ into a single sequence. A more specific hint is hidden below.

More specifically, consider the set $$\left\{n+\frac1m:n\in\Bbb N\text{ and }m\in\Bbb Z^+\right\}\,.$$

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  • $\begingroup$ That's really helpful, thank you! $\endgroup$ Commented Feb 8, 2021 at 21:55
  • $\begingroup$ @LeandroAbib: You’re welcome! $\endgroup$ Commented Feb 8, 2021 at 21:58
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Hint:

For every integer $n$ construct a sequence $u_n$ whose only limit is $n$. Let $p_n$ be the $n$-prime number. Consider $v_{p_n^m}=u_m$ its limit is $n$. Write $A_{p_n}=\{p_n^m\}$

Let $A=\cup A_{p_n}$ $A$ is numerable, there exists a bijection $f: \mathbb{N}\rightarrow A $.

Define $w_n=v_{p_n^m}$ where $f(n)=p_n^m$.

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  • $\begingroup$ The notation $p_n^m$ means what? $\endgroup$ Commented Feb 8, 2021 at 21:56
  • $\begingroup$ it means the product of $m$ copies of $p_n$. $\endgroup$ Commented Feb 8, 2021 at 21:58
  • $\begingroup$ Oh, I thought it was another index. Okay, I'll give it a go. Thanks! $\endgroup$ Commented Feb 8, 2021 at 22:00

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