nice and $a$ arbitrary, then there is a nice finite set $B$ containing $A\cup\{a\}$.*

*Proof of Claim:* Of course we may assume $a\notin A$. If there is no path from $a$ to $A$, clearly $A\cup\{a\}$ is nice. Hence we may also assume that there is some path $\gamma=(a=x_0,\ldots b)$ for some $b\in A$ and $\gamma\cap A=\{b\}$. It therefore suffices to prove the claim for the case where $a$ has a neighbour in $A$. If $a$ has two neighbours $x,y\in A$ then $(x,a,y)$ is a dense flag and $A\cup\{a\}$ is nice.

Now assume that $a\in V_i$ has a unique neighbour of type $V_{i+1}$ in $A$. (The other case then follows by self-duality.) If the $E_i$-connected component of $a$ does not intersect $A$, then again $A\cup\{a\}$ is nice. Otherwise there is some $E_i$-path $\gamma=(x_0=a,\ldots x_m=b)$ in $M_n$ with $\gamma\cap A=\{b\}$. If for some $V_{i-1}$-vertex $x_k$ of $\gamma$ there is an $E_{i-1}$-path to some $c\in A$, then the $E_i$-path from $c$ to $b$ extends $(x_k,\ldots x_m=b)$ and is entirely contained in $A$ since $A$ is nice. Since $\gamma\cap A=\{b\}$, no such $x_k$ exists implying that $A\cup\gamma$ is nice.

Let us say that $\gamma$ *changes direction in $x_i$* if $x_i\in V_j$ and either $x_{i-1},x_{i+1}\in 