'$ that has finite length.

To this end, fix a point $\ell \in \mathcal{M}^1(G)$.  There are several cases depending on if $\ell' \in \mathcal{M}^1_1$, $\ell' \in \mathcal{M}^1_2$ or $\ell' \in \mathcal{M}^1_{1,2}$.  

We first deal with the case that $\ell' \in \mathcal{M}^1_{1,2}$.  Without loss of generality we assume that $\mathfrak{h}_{G_1}((\ell')^1) = 1$.  In Proposition~\ref{prop:sep-finite length} we produced a path $\ell_t \colon\thinspace [0,1] \to \widehat{\mathcal{M}}^1(G)$ where $\ell_0 \in \mathcal{M}^1(G)$ and $\ell_1 \in \mathcal{M}_{1,2}$ is such that $\ell_1^1 = (\ell')^1$.  We can concatenate the path $\ell_t$ with a path $\tilde \ell_t \colon\thinspace [0,1] \to \mathcal{M}^1_{1,2}$ from $\ell_1$ to $\ell'$ in $\mathcal{M}^1_{1,2}$ as follows. 
We define the path by \[\tilde \ell_t = (\infty,\ell^1_1,(1-t) \cdot \ell_1^2 + t \cdot (\ell')^2)\] and we observe that $\tilde \ell_0 = \ell_1$ to $\tilde \ell_1 = \ell'$.  Note that by the convexity of entropy we have that \[\mathfrak{h}_{G_2}((1-t) \cdot \ell_1^2 + t \cdot (\ell')^2) \leq 1\] with equality only possible at the endpoints.  Hence the interior of the path $\tilde \ell_t$ lies in $\widehat{\mathcal{M}}^1(G)_{\rm reg}$.  Further, \[\norm{(\tilde \ell_t, \vardottilde{\ell}_t)}_{\mathfrak{h},G} = 0 \] as the length of edges in $G_2$ are changing linearly.  Hence there is a path of finite length from a length function in $\mathcal{M}^1(G)$ to any length function in $\mathcal{M}^1_{1,2}$.  

Next, we deal with the case that $\ell' \in \mathcal{M}_1$; the case of $\ell' \in \mathcal{M}_2$ is symmetric.  We will show that we can connect $\ell'$ to a length function in $\mathcal{M}_{1,2}$ with a concatenation of two paths that have finite length---in fact each has length $0$.  Let $\ell'' \in \mathcal{M}(G_2)$ have entropy less than $1$.  The paths $(\ell_1)_t$ and $(\ell_2)_t$ are as follows:
\begin{align*}
(\ell_1)_t & \colon\thinspace [0,\infty] \to \widehat{\mathcal{M}}^1(G) & (\ell_1)_t & = ((\ell')^0 + t,(\ell')^1,\infty \cdot \One) \\
(\ell_2)_t & \colon\thinspace [0,\infty] \to \widehat{\mathcal{M}}^1(G) & (\ell_2)_t & = (\infty,(\ell')^1,(\ell')^2 + t \cdot \One).
\end{align*}
The concatenation of $(\ell_1)_t|[0,\infty]$ and $(\ell_2)_t|[\infty,0]$ is a path from $\ell'$ to $\ell''$.  We observe that $(\ddot \ell_1)_t = 0$ and $(\ddot \ell_2)_t = 0$ as edge lengths are changing linearly.  Also, we observe that the interiors of these paths lie in $\widehat{\mathcal{M}}^1(G)_{\rm reg}$.  Hence $\norm{((\ell_k)_t,(\dot \ell_k)_t)}_{\mathfrak{h},G} = 0$ for $k = 1$, $2$ showing that the paths have finite---in fact zero---length.

This shows \eqref{item:sep-pseudo-metric}.

The previous argument shows that for any $\ell \in \mathcal{M}^1_1$, there is an $\ell' \in \mathcal{M}^1_{1,2}$ such that $d_{\mathfrak{h},G}(\ell,\ell') = 0$.  Likewise the analogous statement holds for $\ell \in \mathcal{M}^1_2$.  Given, $\ell, \ell' \in \mathcal{M}_{1,2}$, we will show that $d_{\mathfrak{h},G}(\ell,\ell') = 0$, completing the proof of~\eqref{item:sep-short cut}.  There are four cases depending on the entropies of the length functions $\ell^1$, $\ell^2$, $(\ell')^1$ and $(\ell')^2$.  

The first case we consider is when $\mathfrak{h}_{G_1}(\ell^1) = 1$ and $\mathfrak{h}_{G_2}((\ell')^2) = 1$.  In this case we consider the concatenation of the two paths $(\ell_1)_t$ and $(\ell_2)_t$ that are defined as follows:
\begin{align*}
(\ell_1)_t & \colon\thinspace [0,1] \to \widehat{\mathcal{M}}^1(G) & (\ell_1)_t & = (\infty,\ell^1,(1-t) \cdot \ell^2 + t\cdot (\ell')^2) \\
(\ell_2)_t & \colon\thinspace [0,1] \to \widehat{\mathcal{M}}^1(G) & (\ell_2)_t & = (\infty,(1-t) \cdot \ell^1 + t \cdot (\ell')^1,(\ell')^2).
\end{align*}
As above, the interiors of these paths lie in $\widehat{\mathcal{M}}^1(G)_{\rm reg}$ and they have length $0$ since the lengths of edges are changing linearly.  This completes this case.

The case where $\mathfrak{h}_{G_2}(\ell^2) = 1$ and $\mathfrak{h}_{G_1}((\ell')^1) = 1$ is similar.

Next we consider the case where $\mathfrak{h}_{G_1}(\ell^1) = 1$ and $\mathfrak{h}_{G_1}((\ell')^1) = 1$.  Fix a length function $\ell'' \in \mathcal{M}^1_{1,2}$ with $\mathfrak{h}_{G_2}((\ell'')^2) = 1$.  By the first argument, we can connect both $\ell$ and $\ell'$ to $\ell''$ with paths of length $0$.  Concatenating these two paths shows that this case holds as well.

The case where $\mathfrak{h}_{G_2}(\ell^2) = 1$ and $\mathfrak{h}_{G_2}((\ell')^2) = 1$ is similar.

This completes the proof of~\eqref{item:sep-short cut}. 

We observe that any path in $\widehat{\mathcal{M}}^1(G)$ is close to a path in $\mathcal{M}^1(G)$.  Thus by Proposition~\ref{prop:sep-norm extends}, we have that the inclusion $(\mathcal{M}^1(G),d_{\mathfrak{h},G}) \to (\widehat{\mathcal{M}}^1(G),d_{\mathfrak{h},G})$ is an isometric embedding, hence~\eqref{item:sep-isometric embedding} holds.
\end{proof}


\section{\texorpdfstring{$\mathcal{X}^1(\mathcal{R}_r,{\rm id})$}{X\textasciicircum1(R\_r,id)} Has Bounded Diameter in \texorpdfstring{$\mathcal{X}^1(\mathbb{F}_r)$}{X\textasciicircum1(F\_r)}}\label{sec:rose bounded}

In this section, we make use of the collapsing phenomona witnessed in the previous section to show that even though $(\mathcal{M}^1(\mathcal{R}_r),d_{\mathfrak{h},\mathcal{R}_r})$ has infinite diameter (Proposition~\ref{prop:infinite length}), the subspace $(\mathcal{X}^1(\mathcal{R}_r,{\rm id}),d_\mathfrak{h}) \subset (\mathcal{X}^1(\mathbb{F}_r),d_\mathfrak{h})$ has bounded diameter.  The idea is as follows.  Using the natural bijection $\mathcal{M}^1(\mathcal{R}_r) \leftrightarrow \mathcal{X}^1(\mathcal{R}_r,{\rm id})$, since $\widehat{\mathcal{M}}^1(\mathcal{R}_r)$ is the completion of $(\