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How to prove the following mind-blowing fact?

Let $X$ be a separable Banach space and let $\ell_1$ be the space of all absolutely summable scalar sequences. Then there exists such closed subspace $A\subset \ell_1$ that factor space $\ell_1/A$ and $X$ are isomorphic as normed spaces.

Edit: So what, this is like a classification up to isomorphism of all separable Banach spaces? Each separable Banach space corresponds to some closed subspace of $\ell_1$?

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  • $\begingroup$ Please look here (Theorem 2.3.1) springer.com/cda/content/document/cda_downloaddocument/… $\endgroup$ – MotylaNogaTomkaMazura Nov 20 '15 at 21:33
  • $\begingroup$ Your question is a bit garbled. First of all, $\ell_1$ is the space of absolutely summable scalar sequences, not the space of converging sequences. Second, $X$ has to be separable in order for $\ell_1/A\cong X$. (Obviously, nonseparable spaces won't be a quotient of $\ell_1$!) The proof is in the above link. Third, quotients are not subspaces. You have to transfer to the dual $\ell_\infty$, and then every separable space is a closed subspace. $\endgroup$ – Ben W Nov 20 '15 at 22:30
  • $\begingroup$ You might find this useful. $\endgroup$ – David Mitra Nov 21 '15 at 11:32
  • $\begingroup$ Not every Banach space is separable. $\endgroup$ – Asaf Karagila Nov 21 '15 at 12:20
  • $\begingroup$ @AsafKaragila, yes, thanks, it was lost in editing $\endgroup$ – Glinka Nov 21 '15 at 13:36
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Let $X$ be a Banach space and let $\{x_d\colon d\in D\}$ be a dense subset of the unit ball of $X$. Consider the space $\ell_1(D)$ of all absolutely summable sequences on $D$. We define a linear map $T\colon \ell_1(D) \to X$ by

$$T\Big((\lambda_d)_{d\in D}\Big) = \sum_{d\in D}\lambda_d x_d\qquad ((\lambda_d)_{d\in D} \in \ell_1(D)).$$ This is a well-defined linear map as the right-hand side converges absolutely for every $(\lambda_d)_{d\in D}\in \ell_1(D)$, hence it defines an element of $X$. For each $(\lambda_d)_{d\in D} \in \ell_1(D)$ we have $$\|T\big((\lambda_d)_{d\in D}\big)\|\leqslant \sum_{d\in D}\|\lambda_d x_d\|\leqslant \sum_{d\in D}|\lambda_d|=\|(\lambda_d)_{d\in D} \|.$$

Consequently, $T$ is a bounded (actually norm-one) linear operator. Since $\{x_d\colon d\in D\}$ is dense in the unit sphere of $X$, $T$ is surjective. By the first isomorphism theorem, $$X\cong \ell_1(D) / \ker T.$$

Note that separability of $X$ means that we may take $D=\mathbb{N}$.

Here's a reference to the literature.

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  • $\begingroup$ It is not so obvious that $T$ is surjective. It needs more explaining! $\endgroup$ – Yiorgos S. Smyrlis Dec 1 '15 at 16:54

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