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Recently I am confused about the definition of infinite dimensional vector bundle, and would be grateful if someone could help me to clarify the following simple example:

Let $(e_i)_{i=1}^{\infty}$ be the standard orthonormal base for the classical $l^2$ space, fix the interval (-1,1) as base space and assign a family $(M_x)_{x\in(-1,1)}$ of infinite dimensional vector spaces over $(-1,1)$ by letting $M_x=l^2$ for $x\neq0$, and $M_0$ be the subspace of $l^2$ spanned by $(e_i)_{i=2}^{\infty}$.

Intuitively, $(M_x)_{x\in(-1,1)}$ is not a continuous family over any neighborhood of 0, or $(M_x)_{x\in(-1,1)}$ is not a vector bundle around 0.

To give a formal proof with the classical definition of vector bundles, I think think(not sure of it) one have to show that there dose not exist a sequence of $l^2$-valued continuous function $(f_i)$ around $0$ such that for every $x$ ,(1) $(f_i(x))_{i=1}^{\infty}$ is a linearly independent set (2) $span(f_i(x))_{i=1}^{\infty}=M_x$ (Here we mean the closed linear span) . However, I can not prove it and wonder if there are alternative proofs.

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Recently I am confused about the definition of infinite dimensional vector bundle, and would be grateful if someone could help me to clarify the following simple example:

Let $(e_i)_{i=1}^{\infty}$ be the standard orthonormal base for the classical $l^2$ space, fix the interval (-1,1) as base space and assign a family $(M_x)_{i=1}^{\infty}$ (M_x)_{x\in(-1,1)}$of infinite dimensional vector spaces over$(-1,1)$by letting$M_x=l^2$for$x\neq0$, and$M_0$be the subspace of$l^2$spanned by${e_i}_{i=2}^{\infty}$.(e_i)_{i=2}^{\infty}$.

Intuitively, $(M_x)_{i=1}^{\infty}$ (M_x)_{x\in(-1,1)}$is not a continuous family over any neighborhood of 0, or$(M_x)_{i=1}^{\infty}$(M_x)_{x\in(-1,1)}$ is not a vector bundle around 0.

To give a formal proof with the classical definition of vector bundles, I think one have to show that there dose not exist a sequence of $l^2$-valued continuous function $(f_i)$ around $0$ such that (1) for every $x$ ,(1) $(f_i(x))_{i=1}^{\infty}$ is a linearly independent family for every $x$ set (2) $span(f_i(x))_{i=1}^{\infty}=M_x$ (Here we mean the closed linear span) . However, I can not prove it.

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Recently I am confused about the definition of infinite dimensional vector bundle, and would be grateful if someone could help me to clarify the following simple example:

Let ${e_i}_{i=1}^{\infty}$ (e_i)_{i=1}^{\infty}$be the standard orthonormal base for the classical$l^2$space, fix the interval (-1,1) as base space and assign a family${M_x}$(M_x)_{i=1}^{\infty}$ of infinite dimensional vector spaces over $(-1,1)$ by letting $M_x=l^2$ for $x\neq0$, and $M_0$ be the subspace of $l^2$ spanned by ${e_i}_{i=2}^{\infty}$.

Intuitively, ${M_x}$ (M_x)_{i=1}^{\infty}$is not a continuous family over any neighborhood of 0, or${M_x}$(M_x)_{i=1}^{\infty}$ is not a vector bundle around 0.

To give a formal proof with the classical definition of vector bundles, I think one have to show that there dose not exist a sequence of $l^2$-valued continuous function ${f_i}$ (f_i)$around$0$such that (1)${f_i(x)}$(f_i(x))_{i=1}^{\infty}$ is a linearly independent family for every $x$ (2) $span{f_i(x)}=M_x$(Here span(f_i(x))_{i=1}^{\infty}=M_x\$ (Here we mean the closed linear span) . However, I can not prove it.

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