Attained successive minima and adapted flag

(Attainment.) Fix $i$ and let $t_0$ be as in [F1]. Since ${\lambda }_i\le t_0<\mathrm{\infty }$ and ${\lambda }_i=\inf E_i$ for $E_i=\{t>0:\dim \mathrm{span}(tC\cap \mathrm{\Lambda })\ge i\}$, [A1] lets us select for each $m\ge 1$ a real $t_m\in E_i$ with ${\lambda }_i\le t_m<{\lambda }_i+1\mathrm{/}m$. Thus $t_m\to {\lambda }_i$ and $t_m<t_0+1$, because ${\lambda }_i\le t_0$ and $1\mathrm{/}m\le 1$.

For any $v\in \mathrm{int}({\lambda }_{i}C)$, since ${\lambda }_i>0$, the point $c:=v\mathrm{/}{\lambda }_i$ lies in $\mathrm{int}(C)$. The map $\epsilon \mapsto v\mathrm{/}({\lambda }_i-\epsilon )$ is continuous at $0$ with value $c$; because $C$ contains an open ball about $c$, there is ${\epsilon }_0\in (0,{\lambda }_i)$ such that $v\mathrm{/}({\lambda }_i-\epsilon )\in C$ for every $0<\epsilon <{\epsilon }_0$. Thus $v\in ({\lambda }_i-\epsilon )C$ for every such $\epsilon$.

By [F2] the sets $t_{m}C\cap \mathrm{\Lambda }$ are contained in the finite set $F:=(t_0+1)C\cap \mathrm{\Lambda }$, since $t_m<t_0+1$ implies $t_{m}C\subseteq (t_0+1)C$ by [F1]. The collection of subsets $\{t_{m}C\cap \mathrm{\Lambda }:m\ge 1\}$ of $F$ is therefore finite, so some subset $S\subseteq F$ occurs for infinitely many $m$; fix such an infinite subsequence.

Along that subsequence $t_m\to {\lambda }_i$ and $\dim \mathrm{span}S=\dim \mathrm{span}(t_{m}C\cap \mathrm{\Lambda })\ge i$.

Every $v\in S$ lies in $t_{m}C$ for all $m$ of the subsequence, that is $v\mathrm{/}{t}_m\in C$; since $v\mathrm{/}{t}_m\to v\mathrm{/}{\lambda }_i$ and $C$ is closed by [F3], also $v\mathrm{/}{\lambda }_i\in C$, so $v\in {\lambda }_{i}C$. Hence $S\subseteq {\lambda }_{i}C\cap \mathrm{\Lambda }$ and $\dim \mathrm{span}({\lambda }_{i}C\cap \mathrm{\Lambda })\ge \dim \mathrm{span}S\ge i$, which is clause 1.

(Adapted basis.) Fix an enumeration $v_1,\dots ,v_N$ of the finite set $F=(t_0+1)C\cap \mathrm{\Lambda }$ of step 2.1. Define $a_1$ to be the first $v_k$ with $v_k\in {\lambda }_{1}C\cap \mathrm{\Lambda }$ and $v_k\mathrm{\ne }0$, and for $k\ge 2$ define $a_k$ to be the first $v_j$ with $v_j\in {\lambda }_{k}C\cap \mathrm{\Lambda }$ and $v_j\mathrm{\notin }\mathrm{span}\{a_1,\dots ,a_{k-1}\}$. Each step succeeds: $\dim \mathrm{span}({\lambda }_{k}C\cap \mathrm{\Lambda })\ge k$ by step 4.1, while $\mathrm{span}\{a_1,\dots ,a_{k-1}\}\subseteq \mathrm{span}({\lambda }_{k}C\cap \mathrm{\Lambda })$ because $a_j\in {\lambda }_{j}C\subseteq {\lambda }_{k}C$ for $j\le k$ by [F1]. The recursion is a definition by the least index $j$ in a fixed finite list, so it selects nothing.

By construction $a_k\in {\lambda }_{k}C\cap \mathrm{\Lambda }$ and $a_k\mathrm{\notin }\mathrm{span}\{a_1,\dots ,a_{k-1}\}$ for every $k$, so $a_1,\dots ,a_n$ are linearly independent vectors of $\mathrm{\Lambda }$ with $a_k\in {\lambda }_{k}C$.

(Flag equality.) Fix $i$ and put $r:=\mathrm{\#}\{j:{\lambda }_j\le {\lambda }_i\}$, so $r\ge i$ and ${\lambda }_r\le {\lambda }_i<{\lambda }_{r+1}$ when $r<n$. For $j\le r$ one has ${\lambda }_j\le {\lambda }_i$, hence $a_j\in {\lambda }_{j}C\subseteq {\lambda }_{i}C$; the $a_1,\dots ,a_r$ are independent by step 6.1, so $\dim \mathrm{span}({\lambda }_{i}C\cap \mathrm{\Lambda })\ge r$. If the dimension exceeded $r$, then $\dim \mathrm{span}(tC\cap \mathrm{\Lambda })\ge r+1$ for $t={\lambda }_i$, whence ${\lambda }_{r+1}\le {\lambda }_i$ by definition of the infimum, contradicting ${\lambda }_{r+1}>{\lambda }_i$ (and for $r=n$ the dimension is at most $n=r$). Hence the dimension equals $r$ and, since $\mathrm{span}\{a_1,\dots ,a_r\}\subseteq \mathrm{span}({\lambda }_{i}C\cap \mathrm{\Lambda })$ is a subspace of the same dimension $r$, the two agree; that is $\mathrm{span}({\lambda }_{i}C\cap \mathrm{\Lambda })=\mathrm{span}\{a_j:{\lambda }_j\le {\lambda }_i\}$, which is clause 2.

(Interior flag.) Let $v\in \mathrm{\Lambda }\cap \mathrm{int}({\lambda }_{i}C)$ and put $p:=\mathrm{\#}\{j:{\lambda }_j<{\lambda }_i\}$, so ${\lambda }_p<{\lambda }_i={\lambda }_{p+1}$ when $p<n$ and ${\lambda }_j\le {\lambda }_p$ for $j\le p$. If $p=n$ then $\mathrm{span}\{a_j:{\lambda }_j<{\lambda }_i\}=\mathrm{span}\{a_1,\dots ,a_n\}={\mathbb{R}}^n$ contains $v$; so assume $p<n$.

Let ${\epsilon }_0$ be as in step 1.2. If $p=0$, choose any $0<\epsilon <{\epsilon }_0$. If $p>0$, choose $0<\epsilon <\min ({\epsilon }_0,{\min }_{{\lambda }_j<{\lambda }_i}({\lambda }_i-{\lambda }_j))$; the inner minimum is then over a nonempty finite set of positive numbers. In either case step 1.2 gives $v\in ({\lambda }_i-\epsilon )C$, and for each $j\le p$ convexity of $C$ with $0\in C$ gives $a_j\mathrm{/}({\lambda }_i-\epsilon )=\frac{{\lambda }_j}{{\lambda }_i-\epsilon }\cdot \frac{a_j}{{\lambda }_j}\in C$, since ${\lambda }_j\mathrm{/}({\lambda }_i-\epsilon )<1$.

Suppose $v\mathrm{\notin }\mathrm{span}\{a_j:{\lambda }_j<{\lambda }_i\}=\mathrm{span}\{a_1,\dots ,a_p\}$. By steps 1.2 and 7.3 the independent family $a_1,\dots ,a_p$ together with $v$ lies in $({\lambda }_i-\epsilon )C\cap \mathrm{\Lambda }$, so $\dim \mathrm{span}(({\lambda }_i-\epsilon )C\cap \mathrm{\Lambda })\ge p+1$; by definition of the infimum ${\lambda }_{p+1}\le {\lambda }_i-\epsilon <{\lambda }_i={\lambda }_{p+1}$, a contradiction. Therefore $v\in \mathrm{span}\{a_j:{\lambda }_j<{\lambda }_i\}$, which is clause 3.

Clauses 1, 2 and 3 are steps 4.1, 7.1 and 8.1 respectively.