Proof We verify that if $d^E(x,y)=0$ then $x=y$ . This is clear when at least one of $x, y$ is in M. Assume $x, y\in E'(\mathcal {M})$ . Let $\varphi : M_x\to M_y$ be the map with $\varphi (x)=y$ and $\varphi (z)=z$ for all $z\in M$ . If $d^E(x,y)=0$ then $\mu (x,y)=\rho (x,y)=\lambda (x,y)=0$ , which implies that $\varphi $ is an isomorphism between $\mathcal {M}_x$ and $\mathcal {M}_y$ as continuous $\mathcal {L}$ -pre-structures. Thus $x=y$ .

Note that $\mu $ satisfies the triangle inequality because by Lemma 2.3 $\mathfrak {u}_{R,1}^*$ is subadditive for unary $R\in \mathcal {L}$ . For all $x, y, z\in E'(\mathcal {M})$ , $d(x, y)+d(y, z)\geq d(x, z)$ , because all of $\mu $ , $\rho $ and $\lambda $ satisfy the triangle inequality. Also, it is easy to verify that $\max \{\mu (x, y), \rho (x, y), \lambda (x, y)\}\leq \inf \left\{d^{M_x}(x, z)+d^{M_y}(y, z)\,:\, z\in M\right\}$ , thus $d^E$ also satisfies the triangle inequality.

Proof We verify that for any n-ary $R\in \mathcal {L}$ , $R^X$ is $1$ -Lipschitz with respect to $d^X_R$ .

Suppose first $n=1$ . It suffices to show that for any $x, y\in E'(\mathcal {M})$ ,

$$ \begin{align*}|R^X(x)-R^X(y)|\leq \mathfrak{u}_{R,1}(d^X(x,y)).\end{align*} $$

For this, note that by Lemma 2.3(3), we have

$$ \begin{align*} |R^X(x)-R^X(y)|&=|R^{M_x}(x)-R^{M_y}(y)| \\&\leq \mathfrak{u}_{R,1}(\mathfrak{u}_{R,1}^*(|R^{M_x}(x)-R^{M_y}(y)|)) \\&\leq \mathfrak{u}_{R,1}(\mu(x,y)) \\&\leq \mathfrak{u}_{R,1}(d^X(x,y)). \end{align*} $$

Next suppose $n\geq 2$ . The statement follows from Lemma 3.5 in all cases except when $\overline {u}\in M^n_x\setminus M^n, \overline {v}\in M_y^n\setminus M^n$ for distinct $x, y\in E'(\mathcal {M})$ .

Suppose $\overline {u}=(u_1,\dots , u_n),\overline {v}=(v_1,\dots , v_n)$ are given as above. Let

$$ \begin{align*} S_0&= \{ i\,:\, u_i, v_i\in M\}, \\S_1&= \{i\,:\, u_i=x, v_i\in M\}, \\S_2&= \{i\,:\, u_i\in M, v_i=y\}, \\S_3&= \{i\,:\, u_i=x, v_i=y\}. \end{align*} $$

Define $\overline {w}=(w_1,\dots , w_n)\in M_x^n$ by

$$ \begin{align*}w_i=\left\{\begin{array}{@{}ll}u_i, & \mbox{ if}\ i\in S_2\cup S_3,\\ v_i, & \mbox{ if}\ i\in S_0\cup S_1. \end{array}\right. \end{align*} $$

Then

$$ \begin{align*}& |R^X(\overline{u})-R^X(\overline{v})| \\& \quad \leq |R^X(\overline{u})-R^X(\overline{w})|+|R^X(\overline{w})-R^X(\overline{w}(y|x))|+|R^X(\overline{w}(y|x))-R^X(\overline{v})| \\& \quad \leq d^X_R(\overline{u},\overline{w})+\|w\|\rho(x, y)+d^X_R(\overline{w}(y|x),\overline{v}) \\& \quad \leq \displaystyle\sum_{i\in S_0\cup S_1}K_{R,i}d^X(u_i,v_i)+\sum_{i\in S_3}K_{R,i}d^X(u_i,v_i)+\sum_{i\in S_2} K_{R,i}d^X(u_i,v_i) \\& \quad = d^X_R(\overline{u},\overline{v}) \end{align*} $$

as required.

Proof Let N be the largest arity of all $R\in \mathcal {L}$ . Let K be the least of all $K_{R,i}$ for n-ary $R\in \mathcal {L}$ where $n\geq 2$ and $1\leq i\leq n$ . Let J be the largest value of all $K_{R,i}/K$ for n-ary $R\in \mathcal {L}$ where $n\geq 2$ and $1\leq i\leq n$ .

Let D be a countable dense subset of M. Let

$$ \begin{align*}V=\{ d^M(x,y), R^M(\overline{u})\,:\, x, y\in D, \overline{u}\in D^n, R\in\mathcal{L} \mbox{ is}\ n\text{-}\mbox{ary}\},\end{align*} $$

and let G be the additive subgroup of $\mathbb {R}$ generated by $V\cup \mathbb {Q}$ . Then G is a countable dense subset of $\mathbb {R}$ .

We say that a continuous $\mathcal {L}$ -pre-structure $\mathcal {A}$ is G-valued if

$$ \begin{align*}\{d^A(x, y), R^A(\overline{u})\,:\, x, y\in A, \overline{u}\in A^n, R\in\mathcal{L} \mbox{ is}\ n\text{-}\mbox{ary}\}\subseteq G.\end{align*} $$

Let

$$ \begin{align*}\begin{array}{rcl} B&\!\!\!\!\!=\!\!\!\!\!&\{ x\in E'(\mathcal{M})\,:\, \mbox{ there is a finite support}\ F\subseteq D\ \mbox{ for}\ \mathcal{M}_x \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{ such that}\ \mathcal{F}_x\ \mbox{is}\ G\text{-}\mbox{valued}\}. \end{array}\end{align*} $$

Then $B\subseteq E(\mathcal {M},\omega )\cap E'(\mathcal {M})$ is countable. It suffices to show that B is dense in $E(\mathcal {M},\omega )\cap E'(\mathcal {M})$ .

Let $x\in E(\mathcal {M},\omega )\cap E'(\mathcal {M})$ and $1>\epsilon >0$ . Suppose $A=\{a_0, \dots , a_m\}\subseteq M$ is a finite support of $\mathcal {M}_x$ such that

$$ \begin{align*}d^{M_x}(x, a_0)\geq d^{M_x}(x, a_1)\geq \cdots\geq d^{M_x}(x, a_m)>0.\end{align*} $$

Let

$$ \begin{align*}\delta=\min\{ d^{M_x}(y,z)\,:\, y\neq z\in A_x=A\cup\{x\}\}>0.\end{align*} $$

Choose

$$ \begin{align*}0<\epsilon_0<\displaystyle\frac{\epsilon\min\{1,\delta\}\min\{1,K\}}{(9m+9)N}.\end{align*} $$

Then

$$ \begin{align*}\epsilon_0<\min\left\{\displaystyle\frac{\min\{\epsilon, \delta\}}{9m+9}, \frac{\epsilon\delta}{(9m+9)N}, \frac{K\epsilon\delta}{4}\right\}.\end{align*} $$

Since D is dense, we can find distinct $b_0,\dots , b_m\in D$ such that

$$ \begin{align*}d^M(a_i, b_i)<\displaystyle\frac{\epsilon_0}{2NJ}\end{align*} $$

for all $0\leq i\leq m$ . Let $F=\{b_0,\dots , b_m\}\subseteq D$ .

Define $g: F\to G$ by

$$ \begin{align*}g(b_i)=d^{M_x}(x, a_i)+(3i+1)\epsilon_0+\frac{\epsilon}{2}+\epsilon_i'\end{align*} $$

for $0\leq i\leq m$ , where we choose $\epsilon _i'\in (0,\epsilon _0)$ so that $g(b_i)\in G$ . Then the following computations demonstrate that for all $0\leq i<j\leq m$ ,

$$ \begin{align*}|g(b_i)-g(b_j)|\leq d^M(b_i, b_j)\leq g(b_i)+g(b_j).\end{align*} $$

Indeed, suppose $0\leq i<j\leq m$ . We have

$$ \begin{align*} g(b_j)-g(b_i)&= d^{M_x}(x, a_j)-d^{M_x}(x, a_i)+3(j-i)\epsilon_0+\epsilon_j'-\epsilon_i' \\&< (3m+1)\epsilon_0 \\&= 3(m+1)\epsilon_0-2\epsilon_0 \\&< \delta-(d^M(a_i, b_i)+d^M(a_j,b_j)) \\&\leq d^M(a_i, a_j)-d^M(a_i,b_i)-d^M(a_j, b_j) \\&\leq d^M(b_i,b_j),\\g(b_i)-g(b_j)&= d^{M_x}(x, a_i)-d^{M_x}(x, a_j)-3(j-i)\epsilon_0+\epsilon_i'-\epsilon_j' \\&\leq d^{M_x}(x, a_i)-d^{M_x}(x, a_j)-2\epsilon_0 \\&< d^M(a_i,a_j)-d^M(a_i, b_i)-d^M(a_j, b_j) \\&\leq d^M(b_i,b_j), \end{align*} $$

and

$$ \begin{align*} d^M(b_i,b_j)& \leq d^{M_x}(x, a_i)+d^{M_x}(x, a_j)+d^M(a_i, b_i)+d^M(a_j, b_j) \\& < d^{M_x}(x, a_i)+d^{M_x}(x, a_j)+2\epsilon_0 \\& < g(b_i)+g(b_j). \end{align*} $$

We can thus define a one-point extension of the finite metric space $(F,d^F)$ by a point u such that for all $0\leq i\leq m$ , $d^{F_u}(u, b_i)=g(b_i)$ . By defining

$$ \begin{align*}d^{M_u}(u, y)=\inf\{ g(b_i)+d^M(b_i, y)\,:\, 0\leq i\leq m\}\end{align*} $$

for any $y\in M$ , we extend the metric to $M_u=M\cup \{u\}$ .

Next we define a continuous $\mathcal {L}$ -structure $\mathcal {F}_u$ with domain $F_u=F\cup \{u\}$ by defining the values of $R^{F_u}$ for all $R\in \mathcal {L}$ .

Suppose first $R\in \mathcal {L}$ is unary. Since G is dense in $\mathbb {R}$ , we may choose $R^{F_u}(u)\in G\cap [0,1]$ such that

$$ \begin{align*}|R^{F_u}(u)-R^{M_x}(x)|<\mathfrak{u}_{R,1}\left(\frac{\epsilon}{2}\right).\end{align*} $$

We verify that (UC_ℒ ) holds for R in $F_u$ . In fact, for any $0\leq i\leq m$ , if $g(b_i)\in I_{R,1}$ , then by the superadditivity of $\mathfrak {u}_{R,1}$ , we have

$$ \begin{align*} & |R^M(b_i)-R^{F_u}(u)| \\&\quad \leq |R^M(b_i)-R^M(a_i)|+|R^M(a_i)-R^{M_x}(x)|+|R^{M_x}(x)-R^{F_u}(u)| \\&\quad \leq \mathfrak{u}_{R,1}(d^M(b_i,a_i))+\mathfrak{u}_{R,1}(d^{M_x}(x, a_i))+\mathfrak{u}_{R,1}(\frac{\epsilon}{2}) \\&\quad \leq \mathfrak{u}_{R,1}(\epsilon_0)+\mathfrak{u}_{R,1}(d^{M_x}(x, a_i))+\mathfrak{u}_{R,1}(\frac{\epsilon}{2}) \\&\quad \leq \mathfrak{u}_{R,1}(g(b_i))=\mathfrak{u}_{R,1}(d^{F_u}(u,b_i)). \end{align*} $$

If $g(b_i)\not \in I_{R,1}$ then

$$ \begin{align*}|R^M(b_i)-R^{F_u}(u)|\leq 1\leq \mathfrak{u}_{R,1}(g(b_i))=\mathfrak{u}_{R,1}(d^{F_u}(u,b_i)).\end{align*} $$

Next suppose $R\in \mathcal {L}$ is n-ary for $n\geq 2$ . If $\overline {v}\in F^n$ , then let $R^{F_u}(\overline {v})=R^M(\overline {v})$ . Next we define $R^{F_u}(\overline {v})$ for $\overline {v}=(v_1,\dots , v_n)\in F_u^n\setminus F^n$ . For this, let $\overline {v}'=(v_1',\dots , v_n')\in A_x^n\setminus A^n$ be defined by

$$ \begin{align*}v_i'=\left\{\begin{array}{@{}ll} a_j, & \mbox{ if}\ v_i=b_j, \\ x, & \mbox{ if}\ v_i=u.\end{array}\right.\end{align*} $$

We define $R^{F_u}: F_u^n\setminus F^n\to G\cap [0,1]$ such that:

1. (a) for all $\overline {v}\in F_u^n\setminus F^n$ , $|R^{F_u}(\overline {v})-R^{A_x}(\overline {v}')|<K\displaystyle \frac {\epsilon }{2}$ , and

2. (b) for all $\overline {v}, \overline {w}\in F_u^n\setminus F^n$ ,

   $$ \begin{align*}|R^{F_u}(\overline{v})-R^{F_u}(\overline{w})|\leq \left( 1-\displaystyle\frac{\epsilon_0}{\delta}\right) |R^{A_x}(\overline{v}')-R^{A_x}(\overline{w}')|.\end{align*} $$

Note that our choice of $\epsilon _0$ guarantees that

$$ \begin{align*}\displaystyle\frac{\epsilon_0}{\delta}<\frac{\epsilon}{(9m+9)N}\ll 1 \mbox{ and } \displaystyle\frac{2\epsilon_0}{\delta}<K\frac{\epsilon}{2}.\end{align*} $$

To define $R^{F_u}$ we enumerate all elements of $F_u^n\setminus F^n$ as

$$ \begin{align*}\overline{v}^0, \overline{v}^1,\dots, \overline{v}^{\ell}\end{align*} $$

such that

$$ \begin{align*}R^{A_x}({\overline{v}^0}')\leq R^{A_x}({\overline{v}^1}')\leq\cdots \leq R^{A_x}({\overline{v}^{\ell}}').\end{align*} $$

If $R^{A_x}({\overline {v}^i}')=0$ we define $R^{F_u}(\overline {v})=0$ . We also make the commitment that if $R^{A_x}({\overline {v}^i}')=R^{A_x}({\overline {v}^j}')$ then $R^{F_u}(\overline {v}^i)=R^{F_u}(\overline {v}^j)$ . With this commitment we assume without loss of generality that

$$ \begin{align*}0<R^{A_x}({\overline{v}^0}')< R^{A_x}({\overline{v}^1}')<\cdots < R^{A_x}({\overline{v}^{\ell}}').\end{align*} $$

We will then make our definition by induction on $i=0,\dots , \ell $ with the following inductive hypotheses:

1. (H1) $R^{F_u}(\overline {v}^i)\in G$ and

   $$ \begin{align*}\left(1-\displaystyle\frac{2\epsilon_0}{\delta}\right) R^{A_x}({\overline{v}^i}')< R^{F_u}(\overline{v}^i)<\left(1-\displaystyle\frac{\epsilon_0}{\delta}\right) R^{A_x}({\overline{v}^i}');\end{align*} $$

2. (H2) for all $j<i$ ,

   $$ \begin{align*}R^{F_u}(\overline{v}^i)\leq R^{F_u}(\overline{v}^j)+\left(1-\displaystyle\frac{\epsilon_0}{\delta}\right)(R^{A_x}({\overline{v}^i}')-R^{A_x}({\overline{v}^j}')).\end{align*} $$

For $i=0$ (H2) is vacuous and (H1) can be accomplished by the density of G. In general, suppose we have defined $R^{F_u}(\overline {v}^j)$ for $j\leq i$ to satisfy (H1) and (H2). Note that

$$ \begin{align*}\left(1-\displaystyle\frac{2\epsilon_0}{\delta}\right) R^{A_x}({\overline{v}^{i+1}}')<R^{F_u}(\overline{v}^i)+\left(1-\displaystyle\frac{\epsilon_0}{\delta}\right)(R^{A_x}({\overline{v}^{i+1}}')-R^{A_x}({\overline{v}^i}')).\end{align*} $$

We can then define $R^{F_u}(\overline {v}^{i+1})\in G$ to be a value in between the above two quantities. To see that the inductive hypotheses are maintained, note that by this definition (H2) and the left half of (H1) are immediate. For the right half of (H1), just note that

$$ \begin{align*}R^{F_u}(\overline{v}^i)+\left(1-\displaystyle\frac{\epsilon_0}{\delta}\right)(R^{A_x}({\overline{v}^{i+1}}')-R^{A_x}({\overline{v}^i}'))<\left(1-\displaystyle\frac{\epsilon_0}{\delta}\right) R^{A_x}({\overline{v}^{i+1}}').\end{align*} $$

This finishes the definition of $R^{F_u}$ . It is easy to see that (a) and (b) hold.

We verify (UC_ℒ ) for R in $F_u$ . For this, let $\overline {v},\overline {w}\in F_u^n$ differ at exactly one coordinate, say the ith coordinate. If both $\overline {v},\overline {w}\in F^n$ then

$$ \begin{align*}|R^{F_u}(\overline{v})-R^{F_u}(\overline{w})|=|R^M(\overline{v})-R^M(\overline{w})|\leq K_{R,i}d^M(v_i,w_i)\end{align*} $$

by the (UC_ℒ ) for R in M. If both $\overline {v},\overline {w}\in F_u^n\setminus F^n$ , then by (b) we have

$$ \begin{align*} |R^{F_u}(\overline{v})-R^{F_u}(\overline{w})|&\leq \left( 1-\displaystyle\frac{\epsilon_0}{\delta}\right) |R^{A_x}(\overline{v}')-R^{A_x}(\overline{w}')| \\&\leq \left( 1-\displaystyle\frac{\epsilon_0}{\delta}\right)K_{R,i}d^{A_x}(v_i', w_i') \\&\leq K_{R,i}d^{A_x}(v_i',w_i') -K_{R,i}\epsilon_0. \end{align*} $$

Now if $v_i, w_i\in F$ then we have $v_i', w_i'\in A$ , $d^{A_x}(v_i', w_i')=d^M(v_i', w_i')$ ,

$$ \begin{align*}d^M(v_i', v_i), d^M(w_i', w_i)<\displaystyle\frac{\epsilon_0}{2NJ} \leq \frac{\epsilon_0}{2}\end{align*} $$

and

$$ \begin{align*} |R^{F_u}(\overline{v})-R^{F_u}(\overline{w})|&\leq K_{R,i}d^M(v_i',w_i')-K_{R,i}\epsilon_0 \\& \leq K_{R,i}d^M(v_i', w_i')-K_{R,i}(d^M(v_i',v_i)+d^M(w_i',w_i)) \\&\leq K_{R,i}d^M(v_i, w_i). \end{align*} $$

If $v_i\in F$ and $w_i=u$ , then $v_i'\in A$ and $w_i'=x$ , and

$$ \begin{align*}|R^{F_u}(\overline{v})-R^{F_u}(\overline{w})|\leq K_{R,i}d^{A_x}(v_i', x)-K_{R,i}\epsilon_0\leq K_{R,i}d^{F_u}(v_i, u).\end{align*} $$

This finishes the proof of (UC_ℒ ) for R for the case $\overline {v},\overline {w}\in F_u^n\setminus F^n$ . Finally, suppose $\overline {v}\in F_u^n\setminus F^n$ and $\overline {w}\in F^n$ . Since $\overline {v}$ and $\overline {w}$ differ at only coordinate i, we have $v_i=u$ and $w_i\in F$ . We have

$$ \begin{align*} & |R^{F_u}(\overline{v})-R^{F_u}(\overline{w})| \\& \quad \leq |R^{F_u}(\overline{v})-R^{A_x}(\overline{v}')|+|R^{A_x}(\overline{v}')-R^{M}(\overline{w}')| +|R^{M}(\overline{w}')-R^M(\overline{w})| \\& \quad \leq K\displaystyle\frac{\epsilon}{2}+d^{A_x}_R(\overline{v}', \overline{w}')+d^M_R(\overline{w}',\overline{w}) \\& \quad \leq K\displaystyle\frac{\epsilon}{2}+K_{R,i}d^{A_x}(x, w_i')+\sum_{j=1}^n K_{R,j}d^M(w_j', w_j) \\& \quad \leq K_{R,i}d^{A_x}(x, w_i')+K_{R,i}\displaystyle\frac{\epsilon}{2}+NJK_{R,i}\frac{\epsilon_0}{2NJ} \\& \quad < K_{R,i}d^{F_u}(u, w_i). \end{align*} $$

Now we have established (UC_ℒ ) for all $R\in \mathcal {L}$ in $F_u$ , we get that $\mathcal {F}_u$ is indeed a one-point extension of $\mathcal {F}$ as a continuous $\mathcal {L}$ -structure.

By Theorem 3.9, the canonical amalgam of $\mathcal {F}_u$ and $\mathcal {M}$ over $\mathcal {F}$ gives us a one-point extension $\mathcal {M}_u$ of $\mathcal {M}$ . Hence $u\in E(\mathcal {M},\omega )\cap E'(\mathcal {M})$ .

To complete the proof of the theorem, we claim that $d^E(x, u)<(NJ+1)\epsilon $ . For this, we note that for any $y\in M$ ,

$$ \begin{align*}d^{M_x}(x, y)=\inf\{d^{A_x}(x, a_i)+d^M(a_i, y)\,:\, 0\leq i\leq m\}\end{align*} $$

and

$$ \begin{align*}d^{M_u}(u, y)=\inf\{d^{F_u}(u, b_i)+d^M(b_i, y)\,:\, 0\leq i\leq m\}.\end{align*} $$

Fix an arbitrary $y\in M$ . Suppose first $d^{M_u}(u, y)< d^{M_x}(x,y)$ , and let $d^{M_u}(u,y)=d^{F_u}(u, b_i)+d^M(b_i,y)$ . Then

$$ \begin{align*} & d^{M_x}(x, y)-d^{M_u}(u,y) \\& \quad \leq d^{A_x}(x, a_i)+d^M(a_i, y)-(d^{F_u}(u, b_i)+d^M(b_i, y)) \\&\quad \leq d^{A_x}(x, a_i)-d^{F_u}(u, b_i)+d^M(a_i, b_i) \\&\quad \leq \displaystyle\frac{\epsilon_0}{2NJ}-\displaystyle\frac{\epsilon}{2}<0, \end{align*} $$

contradicting our assumption. Hence we must have $d^{M_u}(u, y)\geq d^{M_x}(x, y)$ . Let $d^{M_x}(x,y)=d^{A_x}(x, a_i)+d^M(a_i, y)$ . Then

$$ \begin{align*} & d^{M_u}(u, y)-d^{M_x}(x, y) \\& \quad \leq d^{F_u}(u, b_i)+d^M(b_i,y)-(d^{A_x}(x, a_i)+d^M(a_i, y)) \\& \quad \leq d^{F_u}(u,b_i)-d^{A_x}(x, a_i)+d^M(a_i, b_i) \\& \quad \leq (3m+2)\epsilon_0+\displaystyle\frac{\epsilon}{2}+\frac{\epsilon_0}{2NJ}<\frac{5}{6}\epsilon. \end{align*} $$

It follows that $\lambda (x, u)<\epsilon $ .

For any unary $R\in \mathcal {L}$ , we have

$$ \begin{align*}\mathfrak{u}^*_{R,1}(|R^{M_x}(x)-R^{M_u}(u)|)\leq \mathfrak{u}^*_{R,1}(\mathfrak{u}_{R,1}\left(\displaystyle\frac{\epsilon}{2}\right))=\frac{\epsilon}{2}.\end{align*} $$

Thus $\mu (x, u)<\epsilon $ .

To compute $\rho (x,u)$ , we fix an n-ary $R\in \mathcal {L}$ for $n\geq 2$ and $\overline {v}\in M_u^n\setminus M^n$ . Since $\mathcal {M}_u$ is the canonical amalgam of $\mathcal {F}_u$ and $\mathcal {M}$ over $\mathcal {F}$ , we have that

$$ \begin{align*}R^{M_u}(\overline{v})=\max\{0, \sup\{R^{M_u}(\overline{z})-d^{M_u}_R(\overline{v},\overline{z})\,:\, \overline{z}\in F_u^n\cup M^n\}\}.\end{align*} $$

Similarly,

$$ \begin{align*}R^{M_x}(\overline{v}(x|u))=\max\{0, \sup\{R^{M_x}(\overline{w})-d^{M_x}_R(\overline{v}(x|u), \overline{w})\,:\, \overline{w}\in A_x^n\cup M^n\}\}.\end{align*} $$

Suppose first $R^{M_u}(\overline {v}){\kern-1pt}\geq{\kern-1pt} R^{M_x}(\overline {v}(x|u))$ . If $R^{M_u}(\overline {v}){\kern-1pt}={\kern-1pt}0$ we must have ${R^{M_x}(\overline {v}(x|u)){\kern-1pt}={\kern-1pt}0}$ and there is nothing to prove. Assume $R^{M_u}(\overline {v})>0$ . Let $\eta>0$ be arbitrary. Then for some $\overline {z}\in F_u^n\cup M^n$ ,

$$ \begin{align*}R^{M_u}(\overline{v})\leq R^{M_u}(\overline{z})-d^{M_u}_R(\overline{v},\overline{z})+\eta.\end{align*} $$

Consider first the case $\overline {z}\in M^n$ . Then

$$ \begin{align*} & R^{M_u}(\overline{v})-R^{M_x}(\overline{v}(x|u)) \\&\quad\leq R^{M_u}(\overline{z})-d^{M_u}_R(\overline{v},\overline{z})+\eta-(R^{M_x}(\overline{z})-d^{M_x}_R(\overline{v}(x|u), \overline{z})) \\&\quad\leq \|\overline{v}\|\lambda(u,x)+\eta. \end{align*} $$

Since $\eta $ is arbitrary, we have

$$ \begin{align*}\|\overline{v}\|^{-1}|R^{M_u}(\overline{v})-R^{M_x}(\overline{v}(x|u))|<\epsilon.\end{align*} $$

Next consider the case $\overline {z}\in F_u^n\setminus F^n$ . Then

$$ \begin{align*} & R^{M_u}(\overline{v})-R^{M_x}(\overline{v}(x|u)) \\&\quad\leq R^{M_u}(\overline{z})-d^{M_u}_R(\overline{v},\overline{z})+\eta-(R^{M_x}(\overline{z}')-d^{M_x}_R(\overline{v}(x|u), \overline{z}') \\&\quad= R^{M_u}(\overline{z})-R^{M_x}(\overline{z}')+(d^{M_x}_R(\overline{v}(x|u), \overline{z}')-d^{M_u}_R(\overline{v}, \overline{z}))+\eta \\& \quad\leq K\displaystyle\frac{\epsilon}{2}+NKJ \epsilon+\eta. \end{align*} $$

The last inequality follows from the observation that there are four cases for the values of $v_i, z_i\in M_u$ :

Since $\eta $ is arbitrary, we have

$$ \begin{align*}R^{M_u}(\overline{v})-R^{M_x}(\overline{v}(x|u))\leq K(NJ+1)\epsilon.\end{align*} $$

Since $\|\overline {v}\|\geq K$ , we have

$$ \begin{align*}\|\overline{v}\|^{-1} |R^{M_u}(\overline{v})-R^{M_x}(\overline{v}(x|u))|\leq (NJ+1)\epsilon.\end{align*} $$

For $R^{M_u}(\overline {v})\leq R^{M_x}(\overline {v}(x|u))$ we get a similar estimate by a symmetric argument. Thus we have $\rho (x, u)<(NJ+1)\epsilon $ and $d^E(x,u)<(NJ+1)\epsilon $ .

Proof Let $\mathcal {A}_x$ and $\epsilon>0$ be given. Let N, K, J be defined as in the proof of Theorem 4.5. Suppose $A=\{a_0,\dots , a_m\}\subseteq \bar {M}$ . Since M is dense in $\bar {M}$ , we get a finite subset $F=\{b_0,\dots , b_m\}\subseteq M$ such that

$$ \begin{align*}d^{\bar{M}}(a_i, b_i)<\displaystyle\frac{\epsilon}{2NKJ}\end{align*} $$

for all $0\leq i\leq m$ . By the proof of Theorem 4.5, we obtain an one-point extension $\mathcal {F}_u$ of $\mathcal {F}$ such that

$$ \begin{align*}|d^{A_x}(x, a)-d^{F_u}(u, b)|<\displaystyle\frac{\epsilon}{2}\end{align*} $$

for all $a\in A$ where $b=b_j$ if $a=a_j$ for $0\leq j\leq m$ , and for any n-ary $R\in \mathcal {L}$ and $\overline {y}\in A^n_x$ ,

$$ \begin{align*}|R^{A_x}(\overline{y})-R^{F_u}(\overline{y}')|<\displaystyle\frac{\epsilon}{2},\end{align*} $$

where $\overline {y}'$ is defined as in the proof of Theorem 4.5.

Since M has the Urysohn property, we may find such a $u\in M$ . Now for any $a\in A$ , letting $b\in F$ be defined as above, we have

$$ \begin{align*} & |d^{A_x}(x, a)-d^{\bar{M}}(u,a)| \\&\quad \leq |d^{A_x}(x, a)-d^{F_u}(u, b)|+|d^{F_u}(u, b)-d^{\bar{M}}(u, a)| \\&\quad < \displaystyle\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. \end{align*} $$

For n-ary $R\in \mathcal {L}$ and $\overline {y}\in A_x^n$ ,

$$ \begin{align*}& |R^{A_x}(\overline{y})-R^{\bar{M}}(\overline{y}(u|x))| \\& \quad \leq |R^{A_x}(\overline{y})-R^{F_u}(\overline{y}')|+|R^{F_u}(\overline{y}')-R^{\bar{M}}(\overline{y}(u|x))| \\& \quad \leq \displaystyle\frac{\epsilon}{2}+d_R^{\bar{M}}(\overline{y}',\overline{y}(u|x)) \\& \quad \leq \displaystyle\frac{\epsilon}{2}+NKJ\frac{\epsilon}{2NKJ}=\epsilon.\\[-37pt] \end{align*} $$

Proof Take $\mathcal {M}_{xy}$ to be the substructure of $\mathcal {E}(\mathcal {M})$ with domain $M_{xy}$ .

Proof Let $A\subseteq \bar {M}$ and $\mathcal {A}_x$ be a one-point extension of $\mathcal {A}$ . Let

$$ \begin{align*}\delta=\min\{ d^{A_x}(a, x)\,:\, a\in A\}>0.\end{align*} $$

We define a sequence $(y_m)$ in M by induction on m. By Lemma 4.6 there is $y_0\in M$ such that

$$ \begin{align*}|d^{A_x}(x,a)-d^{\bar{M}}(y_0, a)|<\delta \mbox{ for all}\ a\in A,\end{align*} $$

for any unary $R\in \mathcal {L}$ ,

$$ \begin{align*}|R^{A_x}(x)-R^{\bar{M}}(y_0)|<\mathfrak{u}_{R,1}(\delta),\end{align*} $$

and for any n-ary $R\in \mathcal {L}$ and $\overline {z}\in A^n_x$ ,

$$ \begin{align*}|R^{A_x}(\overline{z})-R^{\bar{M}}(\overline{z}(y_0|x))|<\delta K,\end{align*} $$

where K is defined as in the proof of Theorem 4.5. Let $B_0=A\cup \{y_0\}\subseteq \bar {M}$ . By Lemma 4.7 we obtain a one-point extension $(\mathcal {B}_0)_x=\mathcal {A}_{xy_0}$ such that

$$ \begin{align*}d^{(B_0)_x}(x, y_0)=\max\{\mu(x, y_0), \rho(x,y_0),\lambda(x,y_0)\}<\delta.\end{align*} $$

In general, assume $y_m$ has been defined to satisfy the following inductive hypothesis: for

$$ \begin{align*}B_m=A\cup \{y_0,\dots, y_m\}\subseteq \bar{M},\end{align*} $$

there is an one-point extension $(\mathcal {B}_m)_x$ of $\mathcal {B}_m$ (obtained from Lemma 4.7) such that

$$ \begin{align*}d^{(B_m)_x}(x, y_m)=\max\{\mu(x, y_m), \rho(x,y_m), \lambda(x,y_m)\}<\displaystyle\frac{\delta}{2^{m}}.\end{align*} $$

By Lemma 4.6 there is $y_{m+1}\in M$ such that

$$ \begin{align*}|d^{(B_m)_x}(x, b)-d^{\bar{M}}(y_{m+1},b)|<\displaystyle\frac{\delta}{2^{m+1}} \mbox{ for all}\ b\in B_m,\end{align*} $$

for any unary $R\in \mathcal {L}$ ,

$$ \begin{align*}|R^{(B_m)_x}(x)-R^{\bar{M}}(y_{m+1})|<\mathfrak{u}_{R,1}\left(\displaystyle\frac{\delta}{2^{m+1}}\right),\end{align*} $$

and for any n-ary $R\in \mathcal {L}$ and $\overline {z}\in (B_m)_x^n$ ,

$$ \begin{align*}|R^{(B_m)_x}(\overline{z})-R^{\bar{M}}(\overline{z}(y_{m+1}|x))|<\displaystyle\frac{\delta K}{2^{m+1}}.\end{align*} $$

Now let

$$ \begin{align*}B_{m+1}=B_m\cup\{y_{m+1}\}=A\cup \{y_0, \dots, y_m, y_{m+1}\}\subseteq \bar{M}\end{align*} $$

and $(\mathcal {B}_{m+1})_x$ be given by Lemma 4.7 such that

$$ \begin{align*}d^{(B_{m+1})_x}(x, y_{m+1})=\max\{\mu(x, y_{m+1}), \rho(x, y_{m+1}), \lambda(x, y_{m+1})\}<\displaystyle\frac{\delta}{2^{m+1}}.\end{align*} $$

The inductive step of the definition is complete, and the inductive hypothesis is maintained. Note that

$$ \begin{align*} d^M(y_m, y_{m+1})&\leq d^{(B_m)_x}(x, y_{m})+d^{(B_{m+1})_x}(x, y_{m+1}) \\ &<\displaystyle\frac{\delta}{2^m}+\frac{\delta}{2^{m+1}}<\frac{\delta}{2^{m-1}}. \end{align*} $$

Thus $(y_m)$ is a $d^M$ -Cauchy sequence in M. Let $y=\lim _m y_m\in \bar {M}$ . Then for any $a\in A$ , since

$$ \begin{align*}|d^{A_x}(x, a)- d^{\bar{M}}(y_m, a)|<\displaystyle\frac{\delta}{2^m},\end{align*} $$

letting $m\to \infty $ , we get that

$$ \begin{align*}d^{A_x}(x,a)=d^{\bar{M}}(y, a).\end{align*} $$

For any unary $R\in \mathcal {L}$ , we have

$$ \begin{align*}|R^{A_x}(x)-R^{\bar{M}}(y_m)|<\mathfrak{u}_{R,1} \left(\displaystyle\frac{\delta}{2^{m}}\right).\end{align*} $$

Since $\mathfrak {u}$ is continuous at $0$ , by letting $m\to \infty $ we get

$$ \begin{align*}R^{A_x}(x)=R^{\bar{M}}(y).\end{align*} $$

For any n-ary $\mathcal {R}\in \mathcal {L}$ and $\overline {z}\in A_x^n$ ,

$$ \begin{align*}|R^{A_x}(\overline{z})-R^{\bar{M}}(\overline{z}(y_m|x))|<\displaystyle\frac{\delta K}{2^m}.\end{align*} $$

Letting $m\to \infty $ , we obtain

$$ \begin{align*}R^{A_x}(\overline{z})=R^{\bar{M}}(\overline{z}(y|x)).\end{align*} $$

This shows that $\mathcal {A}_x$ and $\mathcal {A}_y$ are isomorphic, and in particular there is an isomorphic embedding $\varphi : \mathcal {A}_x\to \mathcal {\bar {M}}$ with $\varphi (a)=a$ for all $a\in A$ .

Proof We first claim that for any continuous $\mathcal {L}$ -pre-structure $\mathcal {M}$ and finite subsets $F\subseteq G$ of M, if the one-point extension $\mathcal {M}_x$ has support F, then it also has support G. It is easy to verify that for any $y\in M$ ,

$$ \begin{align*}d^{M_x}(x, y){\kern-1pt}={\kern-1pt}\min\{d^{M_x}(x, z)+d^{M}(z, y)\,:\, z\in F\}=\min\{d^{M_x}(x, z)+d^M(z, y)\,:\,z\in G\}.\end{align*} $$

If $R\in \mathcal {L}$ is n-ary, where $n\geq 2$ , and $\bar {u}\in M^n_x$ , then

$$ \begin{align*} R^{M_x}(\bar{u})&=\max\{0, \sup\{R^{M_x}(\bar{v})-d^{M_x}_R(\bar{u}, \bar{v}): \bar{v}\in M^n\cup F^n_x\}\}\\ &\leq\max\{0, \sup\{R^{M_x}(\bar{v})-d^{M_x}_R(\bar{u}, \bar{v}): \bar{v}\in M^n\cup G^n_x\}\}. \end{align*} $$

On the other hand, if $\bar {v}\in M^n\cup G^n_x$ ,

$$ \begin{align*} R^{M_x}(\bar{v})-d^{M_x}_R(\bar{u}, \bar{v})&\leq\sup\{R^{M_x}(\bar{w})-d^{M_x}_R(\bar{v}, \bar{w})-d^{M_x}_R(\bar{u}, \bar{v}): \bar{w}\in M^n\cup F^n_x\}\\ &\leq\sup\{R^{M_x}(\bar{w})-d^{M_x}_R(\bar{u}, \bar{w}): \bar{w}\in M^n\cup F^n_x\}. \end{align*} $$

So we have

$$ \begin{align*}R^{M_x}(\bar{u})=\max\{0, \sup\{R^{M_x}(\bar{v})-d^{M_x}_R(\bar{u}, \bar{v}): \bar{v}\in M^n\cup G^n_x\}\}\end{align*} $$

as required.

To complete the proof, it suffices to show that $\mathcal {E}(\mathcal {M}, \omega )$ is a continuous $\mathcal {L}$ -pre-structure, since $\mathcal {M}_\omega $ would be easily defined from iterating the definition of $\mathcal {E}(\mathcal {M},\omega )$ , and $\mathcal {M}_\omega $ would have the Urysohn property by the construction. For this, we only observe that Lemmas 4.3 and 4.4 can be proved for $\mathcal {E}(\mathcal {M},\omega )$ without using upper semicontinuity. For Lemma 4.3, only note that the use of Lemma 2.3 to obtain subadditivity of $\mathfrak {u}^*_{R,1}$ does not require upper semicontinuity of $\mathfrak {u}_{R,1}$ . For Lemma 4.4, the use of Lemma 2.3(3) in its proof does require upper semicontinuity. Below we provide an alternative proof of the relevant part for $\mathcal {E}(\mathcal {M},\omega )$ without using upper semicontinuity.

Let $R\in \mathcal {L}$ be unary and $x,y\in E'(\mathcal {M},\omega )$ , we need to show

$$ \begin{align*}|R^{M_x}(x)-R^{M_y}(y)|\leq \mathfrak{u}_{R,1}(\inf\{d^{M_x}(x,z)+d^{M_y}(z,y)\,:\, z\in M\}).\end{align*} $$

Suppose $\mathcal {M}_x$ has finite support $F\subseteq M$ and $\mathcal {M}_y$ has finite support $G\subseteq M$ . Then by the above claim both $\mathcal {M}_x$ and $\mathcal {M}_y$ have finite support $H=F\cup G$ . It follows that

$$ \begin{align*}\inf\{d^{M_x}(x, z)+d^{M_y}(z, y)\,:\, z\in M\}=\min\{d^{H_x}(x, z)+d^{H_y}(z, y)\,:\, z\in H\}.\end{align*} $$

By the superadditivity of $\mathfrak {u}_{R,1}$ , we have that for any $z\in H$ ,

$$ \begin{align*} |R^{M_x}(x)-R^{M_y}(y)|&= |R^{H_x}(x)-R^{H_y}(y)| \\ &\leq |R^{H_x}(x)-R^H(z)|+|R^H(z)-R^{H_y}(y)| \\ &\leq \mathfrak{u}_{R,1}(d^{H_x}(x,z))+\mathfrak{u}_{R,1}(d^{H_y}(z,y)) \\ &\leq \mathfrak{u}_{R,1}(d^{H_x}(x,z)+d^{H_y}(z,y)).\end{align*} $$

The required inequality follows from the monotonicity of $\mathfrak {u}_{R,1}$ because H is finite.

Proof (1) By Theorem 3.8 it suffices to show that the AP holds for ${\mathcal {K}^{\mathcal {L}}_{\mathrm {\scriptsize fin}}}$ . Let $\mathcal {M}$ , $\mathcal {P}$ , $\mathcal {Q}$ be finite continuous $\mathcal {L}$ -structures with isomorphic embeddings $\varphi : M\to P$ and $\psi : M\to Q$ . By the Urysohn property of $\mathcal {U}$ , we can find a substructure $\mathcal {P}'$ of $\mathcal {U}$ that is isomorphic to $\mathcal {P}$ . Let $i:P\to P'$ be an isomorphism. Let $M'=i\circ \varphi (M)\subseteq U$ . Then $i\circ \varphi $ is an isomorphism between $\mathcal {M}$ and $\mathcal {M}'$ as a substructure of $\mathcal {U}$ (or $\mathcal {P}'$ ). Let $e:\psi (M)\to M'$ be $i\circ \varphi \circ \psi ^{-1}$ . Then e is an isomorphism, and we have $i\circ \varphi =e\circ \psi $ .

By the Urysohn property of $\mathcal {U}$ we can find an extension $\mathcal {Q}'$ of $\mathcal {M}'$ with ${Q'\subseteq U}$ such that $\mathcal {Q}'$ is isomorphic to $\mathcal {Q}$ , and letting $j: Q\to Q'$ be the isomorphism, we have $j\,{\!\upharpoonright \!}\,\, \psi (M)=e$ . Now let $N=P'\cup Q'\subseteq U$ . Then $\mathcal {N}$ is a substructure of U, $i: P\to N$ is an isomorphic embedding from $\mathcal {P}$ into $\mathcal {N}$ , and $j: P\to N$ is an isomorphic embedding from $\mathcal {Q}$ into $\mathcal {N}$ . By our construction, $i\circ \varphi =j\circ \psi $ . Thus $\mathcal {N}$ is an amalgam of $\mathcal {P}$ and $\mathcal {Q}$ over $\mathcal {M}$ .

(2) Assume that $\mathcal {U}$ is a continuous $\mathcal {L}$ -structure with the Urysohn property. By (1), $\mathcal {L}$ is semiproper. It remains to show that given any unary $R\in \mathcal {L}$ , $\mathfrak {u}_{R, 1}$ is upper semicontinuous on $I_{R, 1}$ . Let $I^\circ _{R, 1}$ be the interior of $I_{R, 1}$ and assume $a\in I^\circ _{R, 1}$ . Since $\mathcal {U}$ has the Urysohn property, there is a countable substructure $\mathcal {M}$ of $\mathcal {U}$ where:

• • $M=\{x_n\}_{n\geq 1}\cup \{y\}$ ,

• • for all $m, n\geq 1$ , $d^M(x_n,y)=a+2^{-n}$ , $d^M(x_m, x_n)=|2^{-m}-2^{-n}|$ ,

• • $R^M(y)=\lim _{\delta \rightarrow 0^+}\mathfrak {u}_{R, 1}(a+\delta )<1$ , $R^M(x_n)=0$ for all $n\geq 1$ , and

• • for all other $R'\in \mathcal {L}$ , ${R'}^M$ is identically $0$ .

Then by the completeness of $\mathcal {U}$ , there is $z\in \mathcal {U}$ such that $d^U(y, z)=a$ and ${R^U(z)=0}$ , which implies that

$$ \begin{align*}\lim_{\delta\rightarrow 0^+}\mathfrak{u}_{R, 1}(a+\delta)=|R(y)-R(z)|\leq\mathfrak{u}_{R, 1}(a).\\[-37pt] \end{align*} $$

Proof For (i) $\Rightarrow $ (ii) let $A\subseteq \mathbb {U}_{\mathcal {L}}$ be finite and $A_x$ be a metric space which is a one-point extension of A as a metric space. It suffices to show that x can be realized as a point in $\mathbb {U}_{\mathcal {L}}$ . For this we only need to make a continuous $\mathcal {L}$ -pre-structure $\mathcal {A}_x$ as a one-point extension of the substructure $\mathcal {A}$ . Observe that for any n-ary $R\in \mathcal {L}$ , since $\mathcal {L}$ is Lipschitz, we have that for any metric space $(X, d^X)$ and $\overline {u}=(u_1,\dots , u_n), \overline {v}=(v_1,\dots , v_n)\in X^n$ ,

$$ \begin{align*}d^X_R(\overline{u},\overline{v})=\displaystyle\sum_{i=1}^n K_id^X(u_i, v_i).\end{align*} $$

Now for any n-ary $R\in \mathcal {L}$ , $R^A$ is defined on $A^n$ and is $1$ -Lipschitz with respect to $d^A_R$ by Lemma 3.5. It follows that $R^{A_x}$ is partially defined on ${\mathrm {dom}}(R^{A_x})=A^n$ and is $1$ -Lipschitz with respect to $d^{A_x}_R$ . Thus by Lemma 3.7 we can define $R^{A_x}$ on $A_x^n$ so that $\mathcal {A}_x$ is an extension of $\mathcal {A}$ . By the Urysohn property of $\mathbb {U}_{\mathcal {L}}$ , x can now be realized as a point in $\mathbb {U}_{\mathcal {L}}$ as desired.