Can somebody provide me with an example for a sequence which has as limits points the set $Z$.
Because $Z$ is a closed set than it can be the set of all the limits points.
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4$[0,-1,0,1,-2,-1,0,1,2,-3,-2,-1,0,1,2,3,\dots]$ – JMoravitz Jan 20 '21 at 18:56
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Any enumerationof the rationals. Or do you also mean that it should only have the integers as limit points? – Vercassivelaunos Jan 20 '21 at 18:57
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@Vercassivelaunos the set of limit points is the whole real line, not EXCLUSIVELY the integers. – Tito Eliatron Jan 20 '21 at 18:58
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@TitoEliatron: Yes, I noticed that the text sounds more like it should exclusively be the integers, while the title might allow for other limit points. I asked for clarification the moment before I saw your comment – Vercassivelaunos Jan 20 '21 at 19:02
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Basically if $\varphi$ is any bijection between $\mathbb N^2$ and $\mathbb N$ then calling $\varphi^{-1}(n)=(n_1,n_2)$
We get that $x_n=n_1+\dfrac 1{n_2}$ will have all natural numbers as accumulations points.
It is not too hard to adapt it to negative numbers (split for instance odd/even $\mapsto$ negative/positive).
For instance $\varphi^{-1}(n)=\bigg(\min(n-\lfloor\sqrt{n}\rfloor^2,\lfloor\sqrt{n}\rfloor)\quad,\quad2\lfloor\sqrt{n}\rfloor-\max(n-\lfloor\sqrt{n}\rfloor^2,\lfloor\sqrt{n}\rfloor)\bigg)$
--> see this post How to show that $\max(a,b)^2+\max(a,b)+a-b$ is bijective?
But there are others like Cantor-pairing function and such, it just happens I had this one above at hand...
zwim
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