# Set of adherence values of a sequence equal to $\mathbb{N}$

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)

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

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

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 $$f(n)=p_n^m$$.

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