Does there exist a infinite dimensional Banach subspace in every normed space?

Question

We know that every normed space contains a separable subspace.

Let $X$ a normed space. Suppose that Hamel's basis of $X$ is uncountable, $X$ isn't a reflexive and Banach space (see below). Does there exist an infinite dimensional Banach subspace $B$ of $X$?

Remark:

If $X$ is Banach take any infinite dimensional closed subspace in $X$, and we have done. (Let $(x_{n})_{n}$ linearly independent sequence, so $S=\overline{ \langle (x_{n})_{n} \rangle}$ is a closed subspace in $X$, then $S$ is a Banach subspace of $X$).

We have to show the question when $X$ isn't a Banach space.

In particular, if $X$ isn't Banach, then $X$ isn't reflexive. (Suppose $X$ isn't Banach. If $X$ is reflexive , $J(X)=X''$, and $X''$ is Banach, then $X$ is Banach.)

We have to show the question when $X$ isn't a reflexive and Banach space.

If $X$ is normed space with Hamel's basis countable then $X$ can't have a subspace that is Banach in $X$. In fact, let $S$ a Banach subspace of $X$, then Hamel's basis of $S$ is uncountable, but Hamel's basis of $X$ is countable.

An example of normed space with countable basis:

"Consider $c_{00}$, the space of the sequences $x=(x_{n})$ of real numbers which have only finitely many non-zero elements, with the norm $ \|x\|=\sup _{n}|x_{n}|$ . Its standard basis, consisting of the sequences having only one non-zero element, which is equal to 1, is a countable Hamel basis." (https://en.wikipedia.org/wiki/Basis_(linear_algebra))

We have to show the question when Hamel's basis of $X$ is uncountable, and $X$ isn't a reflexive and Banach space.

Answer 1

Let $M$ be an uncountable set, e.g. $M = [0,1]$, equipped with the counting measure. $$ X_1 = \{x \in \ell^2(M) \, | \, x_i \ne 0 \quad \text{for finitely many} \quad i \in M\} $$ and $$ X_2 = \{x \in \ell^2(\mathbb{Z}) \, | \, x_i \ne 0 \quad \text{for finitely many} \quad i < 0\} $$ both with the $\ell^2$ norm.

Then $X_1$ is a normed space with uncountable Hamel basis for which every complete subspace is finite dimensional.

And $X_2$ is a normed space with uncountable Hamel basis which contains the Banach space $\ell^2(\mathbb{N})$ as a subspace.

In sum: There are normed spaces with uncountable Hamel basis that contain no infinite-dimensional Banach spaces as subspaces, and there is a simple tensor construction to produce such normed spaces that do contain infinite-dimensional Banach spaces.