Limit point of sequence vs limit point of the set containing all point of the sequence

I need to show that there exist sequences s.t. for fix $\epsilon>0$ there exist $|z_n-\alpha|<\epsilon$ (1) holds for infinitely many $n\in N$ but s.t. $\alpha$ is not a limit point of the set containing all terms $z_n$.

Thus far I've basically constructed a sequence with several limit points but that does not converge and I guess that's way to go. What confuses me that if I have infinitely many $n \in N$ that satisfy (1) how can there be an epsilon neighborhood around $\alpha$ which contains no points of $z_n$? Is this the archimedean principle at work? My hunch is that im first choosing $N(\epsilon)$ and then in then in then in the set part choosing $\epsilon(N)$.

I'd love any answer that gets me any closer to understanding this and/or the archimedean principle at work here.

Thanks

/I

Solution 1:

Let $x_n=(-1)^n$ and $\alpha=1$. Then $S=\{x_n:n\in\Bbb N\}=\{-1,1\}$ is a finite set, so it has no limit points.

The point of this exercise is that cluster points of sequences and limit points of sets aren’t quite the same thing. A point $x$ is a cluster point of the sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ if some subsequence of $\sigma$ converges to $x$; $x$ is a limit point of the set $S=\{x_n:n\in\Bbb N\}$ if every open nbhd of $x$ contains a point of $S\setminus\{x\}$. If $x$ is a limit point of the set $S$, then every open nbhd of $x$ contains infinitely many different points of $S$; if $x$ is a cluster point of $\sigma$, then every open nbhd of $x$ contains infinitely many different terms of $\sigma$, but those terms might all be the same point, as in the example that I gave above.