Prove that the algebraic dimension of an infinite dimensional Banach space is atleast $\mathfrak{c}$.

Question

Question: Prove that the algebraic dimension of an infinite dimensional Banach space is atleast $\mathfrak{c}$.

Proof: $(X, \|•\|) $ be a infinite dimensional Banach space.

Then $\dim(X) \ge \aleph_{0}$

Now, we can prove that $\dim(X) \neq \aleph_{0}$

Suppose, $\dim(X) =\aleph_{0}$ and ${\scr{B}}=\{e_1, e_2, ...\}$ be a Hamel basis.

Then, $X=\cup_{n\in \Bbb{N}} {span \{e_1, e_2,...,e_n\}}$

a) $span {\{e_1, e_2,..., e_n\}}$ closed set (being finite dimensional linear subspace of $X$ )

b) $span {\{e_1, e_2,..., e_n\}}$ contains no ball ( being a proper linear subspace of $X$ )

This implies $X$ can be written as countable union of n.w.dense subsets and is of meagre set. But this is contradiction, as $X$ is Baire space and hence $2$nd category in itself.

Hence, $\dim(X) >\aleph_{0} $ and by Continumm Hypothesis , it follows that $\dim(X) \ge \mathfrak{c}$

Alternative proof : Any infinite dimensional Banach space contains an isomorphic copy of $\ell_{\infty}$ space . i.e there exists a continuous injective linear map from $\ell_{\infty}$ to $X$.

Choose, $x_1\in X$ such that $\|x_1\|=1 $

Let, $X_1=span\{x_1\}$ and choose $x_2\in X\setminus X_1$ such that $\|x_2\|=\frac{1}{2}$

Let, $X_2=span\{x_1,x_2\}$ and choose $x_3\in X\setminus X_2$ such that $\|x_3\|=\frac{1}{2^2}$

Continuing in this way, we obtain a sequence $(x_n) \subset X$ such that $x_n \notin span\{x_1, x_2, ..., x_{(n-1) } \}$ forall $n>1$ and $\|x_n\|=\frac{1}{2^{n-1}}$

Now, define $T: \ell_{\infty} \to X$ by

$T(c) =\sum_{n\in\Bbb{N}} c_n x_n$ for $c=(c_1, c_2,...) \in\ell_{\infty}$

The above map is well defined as $ \sum_{n\in\Bbb{N}} c_n x_n$ is convergent as

$\begin{align} \sum_{n\in\Bbb{N}} \|c_n x_n\|&= \sum_{n\in\Bbb{N}} |c_n|\| x_n\| \\ &\le \|c\|_{\infty} \sum_{n\in\Bbb{N}} x_n\\ &=\|c\|_{\infty}\sum_{n\in\Bbb{N}} \frac{1}{2^{n-1}}\\&=2\|c\|_{\infty}\end{align} $

$T$ is a continuous linear map and $T$ is injective as $\ker(T)=\{c\in\ell_{\infty} : Tc=0 \}=\{0\}$

Hence, $\ell_{\infty}\cong T(\ell_{\infty}) \subseteq X$

Please verify.

1. Isn't Independentness of $\{x_n\}$ enough to prove $T$ is injective?

2. Does those above two arguments sound good?