Distance between convex hulls in a bounded closed convex set

Question

Let $X$ be an infinite-dimensional Banach space and $C\subseteq X$ be a bounded closed convex subset. Let $\{z_i\}_{i\in\mathbb{N}}$ be a sequence of linearly independent points in $C$ and for each $n\in\mathbb{N}$, define $V_{\leq n} = \operatorname{conv}\{z_i\}_{i\leq n}$ and $V_{\geq n} = \operatorname{conv}\{z_i\}_{i\geq n}$. For each $z\in V_{\geq 1}$, let $N(z)$ denote the least positive integer such that $z\notin V_{\geq N(z)}$. For each $n\in\mathbb{N}$, define:

$$ d_n: C\rightarrow [0, \infty), \hspace{0.3cm} c\mapsto \inf\big\{ \|c-z\|: z\in V_{\geq n} \big\} $$

and clearly each $d_n$ is continuous.

My question: if, for a fixed $m\in\mathbb{N}$, there exists $\lambda>0$ and $M\in\mathbb{N}$ such that $d_M(z_i) > \lambda$ for each $i\leq m$, is it true that for each $z\in V_{\leq m}$, there exists $n\geq \max(N(z), M)$ such that $d_n(z) > \lambda$?

What inspires this question is that, given $m\in\mathbb{N}$ and $\lambda>0$ such that $d_M(z_i)>\lambda$ for each $i\leq m$, I need to find $M'\in\mathbb{N}$ such that:

$$ \inf\big\{ \|z-z'\|: z\in V_{\leq m}, z'\in V_{\geq M'}\big\} > 0 $$

Notice that each $d_n$ is continuous. If my guess is true, then by the fact that $V_{\leq m}$ is compact the inequality above will be true. Any hints, as well as other methods to prove the inequality above without answering my question, will be appreciated.

Answer 1

$\newcommand\la\lambda$The answer is no. E.g., suppose that $X=\ell^\infty$, $z_1=e_1$, $z_2=-e_1+e_2$, and $z_k=e_2/2+e_k/k$ for $k\ge3$, where $(e_1,e_2,\dots)$ is the standard basis of $\ell^\infty$. Let $C$ be the unit ball in $\ell^\infty$. Then $C$ is a bounded closed convex subset of $X$ and the $z_k$'s are linearly independent points in $C$.

Let $m=2$, $M=3$, $\la=1/2$, and $z:=(z_1+z_2)/2=e_2/2$. Then $d_M(z_i)=1>\la$ for each $i\le m$. However, for each $n\ge M$ we have $d_n(z)=0\not>\la$. $\quad\Box$