Proof that limit points are unique I ams having problems to prove that the limits points of a sequence are unique. For example given the following sequence
\begin{equation}
x_n=(-1)^n+\frac{1}{n}
\end{equation}
To find the limit points, I establish these subsequences
\begin{equation}
y_n=x_{2n}=(-1)^{2n}+\frac{1}{2n}=1+\frac{1}{2n}
\end{equation}
\begin{equation}
z_n=x_{2n+1}=(-1)^{2n+1}+\frac{1}{2n+1}=-1+\frac{1}{2n+1}
\end{equation}
Where it can be seen that
\begin{equation}
y_n\rightarrow 1
\end{equation}
\begin{equation}
z_n\rightarrow -1
\end{equation}
My question is how can I prove that these are the only two limit points of the sequence.
I suppose that it is related to the fact that
\begin{equation}
Image(f(n))+Image(g(n))=\mathbb{N}
\end{equation}
Being
\begin{equation}
f(n)=2n\; \; \; \;  
g(n)=2n+1
\end{equation}
 A: If the sequence had a limit point $p$ other than $-1$ or $1$, then there would exist a subsequence $(x_{n_k})$  such that $x_{n_k}\to p$.
Note that the set $S:=\{n_k:k\in \mathbb N\}$ can't have an infinite intersection with the set of even natural numbers because if it did, then $(x_{n_k})$ and $(y_n)$ will share a subsequence $s$. Since every subsequence of a convergent sequence must converge, $s$ will also converge to $p$ and to $1$, which is impossible.
Similarly, the set $S$ can't have infinite intersection with the set of all odd numbers.
Therefore $S$ can't be an infinite set, which contradicts our assumption that $(x_{n_k})$ is a subsequence.
It follows by contradiction that no subsequence of $(x_n)$ can have a limit point other than $-1$ or $1$.
A: Suppose that $L$ is another number, not equal to $1$ or to $-1$. Then we can find neighborhoods of $1$, $-1$, and $L$, all mutually disjoint. Since the sequence is eventually inside the union of the neighborhoods of $1$ and $-1$, then it's eventually outside the neighborhood of $L$, so $L$ is not a limit point.
A: Take $a\in\Bbb R\setminus\{1,-1\}$. Then, take $\varepsilon=\min\{|a+1|,|a-1|\}$. Since $a\ne\pm1$, $\varepsilon>0$. Since $\lim_{n\to\infty}x_{2n}=1$ and $\lim_{n\to\infty}x_{2n-1}=-1$, there is some $N_e\in\Bbb N$ such that$$n\geqslant N_e\implies|x_{2n}-1|<\frac\varepsilon2$$and there is some $N_o\in\Bbb N$ such that$$n\geqslant N_o\implies|x_{2n-1}+1|<\frac\varepsilon2.$$But then, since each natural number is either even or odd, if $n$ is large enough, then you have $|x_n+1|,|x_n-1|<\frac\varepsilon2$. So, since $\varepsilon\leqslant|a+1|,|a-1|$, $|x_n-a|>\frac\varepsilon2$. Since the inequality $|x_n-a|>\frac\varepsilon2$ takes place for every $n$ is large enough, $a$ cannot be the limit of a subsequence of $(x_n)_{n\in\Bbb N}$.
A: Suppose $\ L \not\in \{-1,1\}\ $ is a limit point of $\ \{x_n\}_{n\in\mathbb{N}}\ $ and define $\ \delta := \min \left\{\ \frac{\vert L-1 \vert}{2} ,\ \frac{ \vert L-(-1) \vert }{2}\  \right\}.\ $
Show that $\ (L-\delta, L+\delta) \cap \{x_n\}_{n\in\mathbb{N} }\ $ is finite and conclude.
A: One more attempt:
Consider a point $a\not =\pm 1.$
There are open neighborhoods of $a, - 1,1$ that do not intersect each other: $U_a, V_{-1}, W_{1}$.
There is a $n_0$ s. t. for $n \ge n_0$ $x_n \in V_{-1}$ or $x_n  \in W_1$.
There are only finitely many indices $n <n_0$, s. t. $x_n$ not in $V_{-1}$ or $W_{1}.$
Hence:
$a$ is not a limit point of $x_n$.
Recall:
$a$ is a limit point of the sequence $x_n$ if for every open neighbourhood of $a$ there are  infinitely many indices $n$ s.t. $x_n$ is in this neighbourhood.
