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If, for all $k\in\mathbb{N}$, $(x_{i}^{k})_{i=1}^{\infty}\in X^{\mathbb{N}}$ is normalized and $M$-basic and if, in addition, for all $k\leq i_{1}<i_{2}<\ldots$ the diagonal sequence $(x_{i_{k} }^{k})_{k=1}^{\infty}\in X^{\mathbb{N}}$ is $M$-basic, then $(x_{i}^{k})_{i=1,k\in\mathbb{N}}^{\infty}$ is said to be an $M$-basic array. Here is a definition of an asymptotic model of a Banach space $X$.

Let $(x^{k}_{i})_{i=1,k\in\mathbb{N}}^{\infty}$ be an $M$-basic array in $X$ and let $(v_{i})_{i=1}^{\infty}$ be a $1$-spreading normalized basis for a Banach space $(V,\|\cdot\|_{V})$. If there exist positive real numbers $\varepsilon_{N}\downarrow 0$ such that for all \begin{equation} \left\vert \left\Vert\sum_{k=1}^{N}\lambda_{k}x^{k}_{i_{k}}\right\Vert-\left\Vert\sum_{i=1}^{N}\lambda_{i}v_{i}\right\Vert_{V}\right\vert<\varepsilon_{N} \end{equation} for all $N\leq i_{1}<\ldots <i_{N}$ and for all scalars $(\lambda_{i})_{i=1}^{N}\in[-1,1]^{N}$, then $(v_{i})_{i=1}^{\infty}$ is said to be an asymptotic model of $X$ generated by $(x^{k}_{i})_{i=1,k\in\mathbb{N}}^{\infty}$.

My question is the following: does any sub-array of the form $(x_{i}^{k_{j}})_{i=1,j\in\mathbb{N}}^{\infty}$ where $k_{1}<k_{2}<\ldots$ generate the same asymptotic model $(v_{i})_{i=1}^{\infty}$ with respect to the same decreasing sequence $\varepsilon_{N}\downarrow 0$?

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The definition of the asymptotic model is not correct. The asymptotic model doesn't have to be spreading. You are allowed to pass to subsequences in each row but not in columns. With that the answer to your question is negative. Take any array that generates an asymptotic model which is not spreading. Then the subarrays of the form $(x_i^{k_j})$ will generate a subsequence of $(v_j)$. (To see this `fill in missing rows with zeroes', that is, take zeros as coefficients $\lambda_j$'s.)

See the original paper of Halbeisen and Odell where the notion is introduced and many examples are discussed.

However, if you are asking what if we start with an asymptotic model which happened to be 1-spreading, then yes, the subarray you described will generate the same asymptotic model. Again the proof is the same, fill in missing rows by zeroes.

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  • $\begingroup$ Thank you for your answer! To clarify: 1) the definition that I gave of an asymptotic model would be fine, provided that the $1$-spreading hypothesis is omitted. 2) As I understand it, we should extract a subsequence of $(v_{i})_{i=1}^{\infty}$ like this: $\lim_{i_{1}\to\infty}\|\lambda_{k_{1}}x_{i_{k}}^{k_{1}}\|= \lim_{i_{1}\to\infty}\left\Vert\sum_{k=1}^{k_{1}}\lambda_{k}x_{i_{k}}^{k}\right\Vert=\left\Vert\sum_{i=1}^{k_{1}}\lambda_{i}v_{i}\right\Vert_{V}=\|\lambda_{k_{1}}v_{k_{1}}\|_{V}$ where $k_{1}\leq i_{1}<i_{2}<\ldots<i_{k_{1}}$ and $\lambda_{k}=0$ for $k<k_{1}$ $\endgroup$
    – JWP_HTX
    Nov 18, 2020 at 14:58
  • $\begingroup$ continued from above....then if $(v_{i})_{i=1}^{\infty}$ were $1$-spreading to being with, it would be $1$-equivalent to its subsequence $v_{k_{1}},v_{k_{2}},\ldots$, in which case the sub-array $(x_{i}^{k_{j}})_{i=1,j\in\mathbb{N}}^{\infty}$ generates the same asymptotic model. $\endgroup$
    – JWP_HTX
    Nov 18, 2020 at 15:02

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