Corson's rational metric space is not metacompact

Suppose $\mathcal{U}$ has a point-finite open refining cover $\mathcal{V}$. Since $\mathcal{V}$ is a set of the model, fix a finite support $E\subseteq A$ of $\mathcal{V}$ by [F1]. Enlarge $E$ by one atom if necessary, so that $E$ is nonempty, without destroying the support property. Let $D$ be the diameter of $E$.

By universality in [F2], choose $a\in A\setminus E$ such that $e<a$ and $d(a,e)=D+4$ for every $e\in E$. Fix an arbitrary integer $n\ge 1$. Since $\mathcal{V}$ covers $A$, choose $V\in \mathcal{V}$ with $a\in V$; by [L2], choose an integer $m\ge 1$ with $B(a,1\mathrm{/}m)\subseteq V$, and put $K:=nm$.

Extend $E\cup \{a\}$, using [F2], by points $a_0<\cdots <a_{3K}=a$ such that $d(a_i,e)=D+4$ for $e\in E$ and $d(a_i,a_j)=\mathrm{\mid }i-j\mathrm{\mid }\mathrm{/}K$ for $0\le i,j\le 3K$. These prescriptions form a finite ordered rational metric space: the old-to-new distances are constant and exceed the diameter of both $E$ and the new chain. Ultrahomogeneity then extends the partial isometry fixing $E$ and sending $a_i$ to $a_{i+1}$ for $0\le i<3K$ to an automorphism $\phi$. Since $E$ supports $\mathcal{V}$, every ${\phi }^j(V)$ belongs to $\mathcal{V}$.

The points $a_{3K-n+1},\dots ,a_{3K}$ lie in $B(a,1\mathrm{/}m)\subseteq V$, because their distances from $a=a_{3K}$ are all strictly less than $n\mathrm{/}K=1\mathrm{/}m$. Hence $a\in {\phi }^j(V)$ for every $0\le j<n$.

By [L1], $V$ has diameter at most $1$. Consequently, if $a_i\in V$, then $i\ge 2K$. Let $L$ be the least index with $a_L\in V$. Step 4.1 gives $2K\le L\le 3K-n+1$. For $0\le j<n$, one has $a_{L+j}\in {\phi }^j(V)$. Moreover, if $K\le i<L+j$, then $a_i\mathrm{\notin }{\phi }^j(V)$: otherwise $a_{i-j}={\phi }^{-j}(a_i)$ would lie in $V$, while $0\le i-j<L$, contradicting the minimality of $L$.

The members ${\phi }^j(V)$ for $0\le j<n$ are pairwise distinct. Indeed, for $j<j^{\mathrm{\prime }}<n$, the point $a_{L+j}$ belongs to ${\phi }^j(V)$, whereas step 5.1, applied with $i=L+j<L+j^{\mathrm{\prime }}$, shows that it does not belong to ${\phi }^{j^{\mathrm{\prime }}}(V)$. Thus, for every $n\ge 1$, the map $j\mapsto {\phi }^j(V)$ injects $n$ into $\{W\in \mathcal{V}:a\in W\}$. The set on the right is therefore not finite, contradicting point-finiteness at $a$. Hence $\mathcal{U}$ has no point-finite open refining cover, so the space is not metacompact and, a fortiori, not paracompact.