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Here is a way to define Banach manifolds given in Margalef-Differential topology

Definition 1

Let $E$ be a real Banach space and $\Lambda:=\{\lambda_1,...,\lambda_n\}$(possibly an empty set) be linearly independent continuous linear functionals on $E$. Define $E_\Lambda^+:=\{x\in E: \forall 1\leq i \leq n, \lambda_i(x)\geq 0\}$. We call this the $\Lambda$-quadrant of $E$.

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Definition 2

Let $X$ be a set and $E$ be a real Banach space and $\Lambda:=\{\lambda_1,...,\lambda_n\}$ be linearly independent continuous linear functionals on $E$. Let $U\subset X$ and $\phi:U\rightarrow E_\Lambda^+$ be an injection such that $\phi(U)$ is open in $E_\Lambda^+$. We call $(U,\phi,(E,\Lambda))$ a chart on $X$.

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Definition 3

Let $X$ be a set and $(U,\phi,(E,\Lambda)),(V,\psi,(F,\Gamma))$ be charts on $X$. Then they are $C^k$-compactible iff $\phi(U\cap V),\psi(U\cap V)$ are open and $\psi\circ \phi^{-1}:\phi(U\cap V)\rightarrow \psi(U\cap V)$, $\phi\circ \psi^{-1}:\psi(U\cap V)\rightarrow \phi(U\cap V)$ are of $C^k$

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Definition 4

Let $X$ be a set and $\mathscr{A}$ be a collection of charts. Then, $\mathscr{A}$ is a $C^k$-atlas iff the domains cover $X$ and any two of them are $C^k$-compatible

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Definition 5

Let $X$ be a set and $\mathscr{A},\mathscr{A}'$ be $C^k$-atlases. Then, they are equivalent iff there is a $C^k$-atlas containing both of them.

The author defines $C^k$ differentiable structures as the equivalence classes of atlases. However, is it a set and well-defined?

Alternatively, let $X$ be a set and $\mathscr{A}$ be a $C^k$-atlas on $X$. Then, does there exist a maximal atlas with respect to the $C^k$-compatible relation?

Unlike the union of Euclidean spaces form a set (i.e. $\bigcup_{n\in\mathbb{N}} \mathbb{R}^n$), the collection of all Banach spaces form a class. Hence, the usual argument for this problem on euclidean differentiable manifolds cannot be applied here.

How do we define a differentiable structure?

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    $\begingroup$ Why do you need to use different Banach spaces and linear functions for different charts on the same manifold? Surely you can fix the Banach space, and fix some sequence of linear functions for your corners, using only finitely many in each chart. $\endgroup$
    – Ben McKay
    Commented Feb 23, 2017 at 19:53
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    $\begingroup$ To be precise, the Banach model spaces are isomorphic on any connected component of $X$. In any case, you do not need the Zorn lemma or variations of it to ensure the existence of a maximal atlas containing a given atlas $\mathcal{A}$: it is the union of all atlases compatible with $\mathcal{A}$ (a proper class, so what? :) ) $\endgroup$ Commented Feb 23, 2017 at 22:32

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