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

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.

• Not if $X$ is finitedimensional. Do you want $B$ to have the same norm as $X$? Commented Oct 24, 2017 at 23:43
• Take the set of finite complex sequences with the $\ell^2$ norm or any norm you like. Banach spaces are completion of normed vector spaces. Subspace means a subset which is a vector space (closed under multiplication by scalar and finite additions). Closed subspace of a Banach space means sub-Banach space. Commented Oct 24, 2017 at 23:48
• It's not clear to me that this subspace will be Banach in $X$. Could you write this with more details as answer? Thanks Commented Oct 25, 2017 at 1:10
• I'll follow your argument: Let $X$ a normed space, there exists $B$ Banach space such that $X$ is a subspace dense of $B$. Now, how I get $S$ infinite dimensional subspace of $X$ such that $S$ is Banach in $X$? (Note that if $S$ is banach and infinite dimensional, we will have that the cardinality of S's basis is not countable). Commented Oct 25, 2017 at 1:13
• If you take $X=C[a,b]$ with the $L^2$ norm, then my answer here shows that there is no infinite-dimensional complete subspace: math.stackexchange.com/questions/2268098/…
– user138530
Commented Oct 25, 2017 at 5:26

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.