Natural numbers =accumulation values in a sequence

Question

Is there such a sequence where exactly the natural numbers are accumulation values? Because I think not, but not sure how to prove it.

Answer 1

Consider the sequence:

$$1, 2, 1.1, 3, 2.1, 1.01, 4, 3.1, 2.01, 1.001, 5, 4.1, 3.01, 2.001, 1.0001, 6, \ldots$$

in which you are enumerating the natural numbers, but after each one you repeat all previous natural numbers $+\varepsilon$ for $\varepsilon \rightarrow 0$. In the example above, I've suggested one concrete approach: Specifically, to return to the earlier natural numbers but with smaller and smaller powers of $10$ added to them.

For this answer to be complete, you would need to include a justification of why (1) the natural numbers are all accumulation points in this sequence, and (2) why no other numbers are accumulation points in this sequence.

(Note: If you include $0 \in \mathbb{N}$, then the proposed sequence would need to be altered ever so slightly.)