I have the following problem which I can solve if there is some sort of converse to Riesz (which I wasn't able to prove):
If $X$ is an infinite dimentional normed vector space, prove there is a sequence $(x_n)_{n\in \mathbb{N}}$ such that $x_n\not \in \overline{\text{span}\{x_j| j>n\}}$ for every $n$.
I tried using Zorn like this: take a certain $0\not= x_1\in X$ with unitary norm and consider the set $\mathcal{S}$ of all closed subspaces such that $d(x_1,S)\geq \frac{1}{2}$. This set is partially ordered by inclusion. If we have a chain in $\{C_i| i \in I\}\subseteq \mathcal{S}$ we consider the following closed subspace:
$$C=\overline{\cup_{i\in I} C_i}$$
Clearly, if $c \in C$, we have that $\lVert c-c_i\rVert\leq \varepsilon$ for some $c_i\in C_i$. Thus:
$$\lVert c-x\rVert\geq \lVert x-c_i \rVert-\lVert c_i-c\rVert\geq \frac{1}{2}-\varepsilon$$
Because $\varepsilon>0$ is arbitrary, $d(x,C)\geq \frac{1}{2}$ and we have an upper bound to our chain. Hence, there is a maximal element in $\mathcal{S}$. Let us call this maximal element $X_1$. We repeat the process in $X_1$ to build a sequence such that $x_i\not \in X_i$ and all $\{x_j| j>i\}\subset X_i$ which is a closed subspace, which implies $\overline{\text{span}\{x_j| j>i\}}\subset X_i$.
There is a flaw in this argument which can be overcome using something similar to Riesz but not quite Riesz:
We need (given a point $x$ in a normed vector space of norm $1$) to find closed subspaces far from it, namely $d(x,C)\geq \frac{1}{2}$ and we need these vector spaces to be large in dimension and plentiful enough such that $\mathcal{S}$ has a maximal element of infinite dimension.
Can my argument be salvaged, or is there a more natural approach?
Best Answer
For a subspace $V\subseteq X$ denote $V'=V\setminus\{0\}$.
Let $\mathscr{F}=\{V':V\text{ is a subspace of }X\text{ of finite co-dimension}\}$.
By the axiom of choice, there is a function $\operatorname{ch}:\mathscr{F}\to X$ such that $\operatorname{ch}(V')\in V'$ for all $V'\in\mathscr{F}$. By replacing $\operatorname{ch}(V')$ with $\operatorname{ch}(V')/\|\operatorname{ch}(V')\|$ we may assume $\|\operatorname{ch}(V')\|=1$.
Let $x_0=\operatorname{ch}(X')$. By Hahn-Banach there is a bounded linear functional $\beta_0$ such that $\beta_0(x_0)=1$. Let $K_0=\ker\beta_0$.
Let $x_1=\operatorname{ch}(K_0')$. By Hanh-Banach there is a bounded linear functional $\beta_1$ such that $\beta_1(x_1)=1$. Let $K_1=K_0\cap\ker\beta_1$.
Let $x_2=\operatorname{ch}(K_1')$. By Hanh-Banach there is a bounded linear functional $\beta_2$ such that $\beta_2(x_2)=1$. Let $K_2=K_1\cap\ker\beta_2$.
etc., etc.
PS to make this precise I guess you need a choice function to choose the bounded linear functionals $\beta_i$ too...