Non-convergence of a sequence implies 'consecutive' partial limits Let $n,d$ be fixed positive integers, and let $X^k=(x_1^k, \dots, x_n^k)$ be a bounded sequence, where $x_i^k \in \mathbb{R}^d$. (Every element of $X^k$ is an $n$-tuple of points in $\mathbb{R}^d$).
Claim: Suppose that $X^k$ has a finite number $N_0 \ge 2$ of accumulation points. Then there exist distinct accumulation points $Y=(y_1, \dots, y_n),Z=(z_1, \dots, z_n)$ and distinct subsequences $X^{k_j},X^{k_l}$ such that $X^{k_j} \to Y, X^{k_l} \to Z$ and $k_l=k_j+1$.
How to prove this claim?
I saw it mentioned in a paper without proof, so I assume there is an elementary solution.
 A: In your claim:

*

*$\{X^k\}_{k\in\mathbb N}$ must be bounded

*the number of accumulation points of $\{X^k\}_{k\in\mathbb N}$ must be greater or equal than 2

I define   $\ \forall k \in \mathbb N \ , \ U^k=X^{k+1}-X^k$.
$\{U^k\}_{k\in\mathbb N}$ is bounded and has therefore at least one point of accumulation $U$.
If $U\neq 0$, there exists a subsequence $\{U^{\varphi(k)}\}_{k\in\mathbb N}$ that converges to $U$. I extract from $\{X^{\varphi (k)}\}_{k\in\mathbb N}$ a subsequence $\{X^{\varphi \circ \psi(k)}\}_{k\in\mathbb N}$ that converges to $Y$. We have:
$\ \lim X^{\varphi \circ \psi(k)} = Y \ $ and $\ \lim X^{\varphi \circ \psi(k)+1} = Y +U$.
If the only point of accumulation of $\{U^k\}_{k\in\mathbb N}$ is $0$, then, $\{U^k\}_{k\in\mathbb N}$ converges to $0$. I note $d$ the minimal distance between the points of accumulation of  $\{X^k\}_{k\in\mathbb N}$ and $\varepsilon =d/3$. There exists $N\in \mathbb N$ such that $\forall p\geqslant N$, $X^p$ is in the union of $B(a,\varepsilon)$ where $a$ is a point of accumulation of  $\{X^k\}_{k\in\mathbb N}$ (otherwise, there would be another point of accumulation). There exists $M \geqslant N$ such that $\forall p\geqslant M \ , \|X^{p+1}-X^p\| < \varepsilon $. If we note $b$ the point of accumulation such that $X^M \in B(b,\varepsilon ) $, we can show by induction that $\forall p\geqslant M \ , X^{p} \in B(b,\varepsilon)$. But that would imply that $b$ is the only point of accumulation of  $\{X^k\}_{k\in\mathbb N}$. There is a contradiction.
And the claim is proved .
