Does there exist a real sequence such that every $n\in\mathbb N$ is its limit point?
I can not get anywhere with this.
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4The set of all rational numbers arranged in a sequence has every real number as a limit point. – Kavi Rama Murthy Jun 24 '21 at 12:05
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1Or something like $0,0,1,0,1,2,0,1,2,3,0,\ldots$ - which is $n$ minus the largest triangular number $\le n$ – Henry Jun 24 '21 at 12:07
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You can reason as follows:
If we want a single naturals as a limit point, we can use the constant sequence
$$1,1,1,1,1,1,1,1,1,1,\cdots$$
If we want two naturals as limit points, we can use the alternating sequence
$$1,2,1,2,1,2,1,2,1,2,\cdots$$
If we want a finite number of naturals as limit points (say $4$), we can use the round-robin sequence
$$1,2,3,4,1,2,3,4,1,2,\cdots$$
If we want all naturals, we lengthen the round-robin on each round,
$$1,1,2,1,2,3,1,2,3,4,\cdots$$
If you prefer "non-trivial" accumulation points, i.e. such that the integers that are never reached, you can add a perturbation of decreasing amplitude to every term, such as $\dfrac1{n+1}$.
