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I can answer this question when the cardinality $\lambda$ of Hamel basis of $X$ is not less than $\mathfrak{c}$. For any infinite dimensional Banach space its cardinality and cardinality of its Hamel basis are equal. See theorem 3.5 from The cardinality of Hamel bases of Banach spaces by Lorenz Halbeisen , Norbert Hungerbühler. Consider Banach space ...

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I presume you are reading Ted Odell's survey paper on ordinal indices in Banach spaces :) . Anyway, here are some details that may be of assistance. Since $(B_{X^\ast},w^\ast)$ is compact metric it is second countable, and so is every topological subspace of $(B_{X^\ast},w^\ast)$. Besides this fact regarding second countability we need two results whose ...

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Let $(u,v)\neq(0,0)$. $D_{(0,0)}f(u,v)=\displaystyle\lim_{\epsilon\to0}\dfrac{f\big((0,0)+\epsilon((u,v))\big)-f(0,0)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{f(\epsilon u,\epsilon v)}{\epsilon}=\lim_{\epsilon\to0}\dfrac{\epsilon^4 u^3v}{\epsilon(\epsilon^4 u^4+\epsilon^2 ... 1 Being able to extend an isometry to an isometry is rare. I don't think any nontrivial Banach space has this property. Here is a proof for$\ell^1$and$\ell^\infty$, the two spaces of main interest to you. In both cases$e_1,e_2,\dots$is the standard set of$0-1$vectors, e.g.$e_2=(0,1,0,0,\dots)$. The case of$\ell^1$Let$A=\{0,e_1,e_2\}$. Define ... 1 Yes there is one here. As I'm not from maths background so won't be able to explain the paper. There is another nice work Approximate proof for Carathéodory's theorem which gives a epsilon-approx being INDEPENDENT of the dimension of the set 1 Suppose that$x,y\in F$, but$x+y\notin F$.$F$is closed, so$x+y$has an open nbhd$U$such that$U\cap F=\varnothing$. Addition is a continuous function from$Z\times Z$to$Z$, so there are$O,O'\in\mathcal{O}$such that$x\in O$,$y\in O'$, and$O+O'\subseteq U\subseteq Z\setminus F\$.

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