Proof. ‘ $\Rightarrow $ ’: First, assume that $h \colon X \to Y$ is a homomorphism between the echeloned spaces $\mathbf {X}$ and $\mathbf {Y}$ . Define $\hat {h} \colon {X^2}/{\sim _{\mathbf {X}}} \to {Y^2}/{\sim _{\mathbf {Y}}},\, [(x_1, x_2)]_{\sim _{\mathbf {X}}} \mapsto [(h(x_1), h(x_2))]_{\sim _{\mathbf {Y}}}$ . First, we show that it is well defined. Let $x_1, x_2, x_1^{\prime }, x_2^{\prime } \in X$ and $(x_1, x_2) \sim _{\mathbf {X}} (x_1^{\prime }, x_2^{\prime })$ ; that is, $[(x_1, x_2)]_{\sim _{\mathbf {X}}} = [(x_1^{\prime }, x_2^{\prime })]_{\sim _{\mathbf {X}}}$ . Then, $(h(x_1), h(x_2)) \sim _{\mathbf {Y}} (h(x_1^{\prime }), h(x_2^{\prime }))$ since h preserves $\leqslant _{\mathbf {X}}$ . Consequently,

$$\begin{align*}\hat{h}([(x_1, x_2)]_{\sim_{\mathbf{X}}}) = [(h(x_1), h(x_2))]_{\sim_{\mathbf{Y}}} = [(h(x_1^{\prime}), h(x_2^{\prime}))]_{\sim_{\mathbf{Y}}} = \hat{h}([(x_1^{\prime}, x_2^{\prime})]_{\sim_{\mathbf{X}}}). \end{align*}$$

Next, we prove that $\hat {h}$ preserves $\leqslant _{E(\mathbf {X})}$ . Take any $x_1, x_2, x_1^{\prime }, x_2^{\prime } \in X$ such that $[(x_1, x_2)]_{\sim _{\mathbf {X}}} \leqslant _{E(\mathbf {X})} [(x_1^{\prime }, x_2^{\prime })]_{\sim _{\mathbf {X}}}$ . By definition, this means that $(x_1, x_2) \leqslant _{\mathbf {X}} (x_1^{\prime }, x_2^{\prime })$ . Consequently, given that h is a homomorphism, it holds that $(h(x_1), h(x_2)) \leqslant _{\mathbf {Y}} (h(x_1^{\prime }), h(x_2^{\prime }))$ . So,

$$\begin{align*}\hat{h}([(x_1, x_2)]_{\sim_{\mathbf{X}}}) = [(h(x_1), h(x_2))]_{\sim_{\mathbf{Y}}} \leqslant_{E(\mathbf{Y})} [(h(x_1^{\prime}), h(x_2^{\prime}))]_{\sim_{\mathbf{Y}}} = \hat{h}([(x_1^{\prime}, x_2^{\prime})]_{\sim_{\mathbf{X}}}).\end{align*}$$

Finally, we show that for our choice of $\hat {h}$ , the above diagram commutes. Take any $x_1, x_2 \in X$ . Then,

$$\begin{align*}\hat{h} \circ \eta_{\mathbf{X}} (x_1, x_2) = \hat{h} ([(x_1, x_2)]_{\sim_{\mathbf{X}}}) = [(h(x_1), h(x_2))]_{\sim_{\mathbf{Y}}} = \eta_{\mathbf{Y}}(h(x_1), h(x_2)) = \eta_{\mathbf{Y}} \circ h^2 (x_1, x_2), \end{align*}$$

which is what we needed to show.

‘ $\Leftarrow $ ’: Assume $\hat {h} \colon E(\mathbf {X}) \to E(\mathbf {Y})$ is a homomorphism and $h \colon X \to Y$ a map such that $\hat {h} \circ \eta _{\mathbf {X}} = \eta _{\mathbf {Y}} \circ h^2$ . We prove h is a homomorphism from $\mathbf {X}$ to $\mathbf {Y}$ . Take any $x_1, x_2,x_1^{\prime }, x_2^{\prime } \in X$ such that $(x_1, x_2) \leqslant _{\mathbf {X}} (x_1^{\prime }, x_2^{\prime })$ . This is equivalent to saying that

$$\begin{align*}\eta_{\mathbf{X}}(x_1, x_2) = [(x_1, x_2)]_{\sim_{\mathbf{X}}} \leqslant_{E(\mathbf{X})} [(x_1^{\prime}, x_2^{\prime})]_{\sim_{\mathbf{X}}} = \eta_{\mathbf{X}}(x_1^{\prime}, x_2^{\prime}), \end{align*}$$

by definition of $\leqslant _{E(\mathbf {X})}$ . It follows that

$$\begin{align*}\hat{h}\circ \eta_{\mathbf{X}} (x_1, x_2) = \hat{h} ([(x_1, x_2)]_{\sim_{\mathbf{X}}}) \leqslant_{E(\mathbf{Y})} \hat{h}([(x_1^{\prime}, x_2^{\prime})]_{\sim_{\mathbf{X}}}) = \hat{h}\circ \eta_{\mathbf{X}} (x_1^{\prime}, x_2^{\prime}).\end{align*}$$

Now, from the commutativity of the given diagram, that is equivalent to

$$\begin{align*}\eta_{\mathbf{Y}}(h(x_1), h(x_2)) = \eta_{\mathbf{Y}} \circ h^2 (x_1, x_2) \leqslant_{E(\mathbf{Y})} \eta_{\mathbf{Y}} \circ h^2 (x_1^{\prime}, x_2^{\prime}) = \eta_{\mathbf{Y}}(h(x_1^{\prime}), h(x_2^{\prime})).\end{align*}$$

In other words, $[(h(x_1), h(x_2))]_{\sim _{\mathbf {Y}}} \leqslant _{E(\mathbf {Y})} [(h(x_1^{\prime }), h(x_2^{\prime }))]_{\sim _{\mathbf {Y}}}$ or better yet, $(h(x_1), h(x_2)) \leqslant _{\mathbf {Y}} (h(x_1^{\prime }), h(x_2^{\prime }))$ , so h does preserve $\leqslant _{\mathbf {X}}$ and is thus a homomorphism.

Proof. ‘ $\Rightarrow $ ’: Assume h is an embedding from $\mathbf {X}$ to $\mathbf {Y}$ . Then by Lemma 1.3, there exists such a homomorphism $\hat {h} \colon E(\mathbf {X}) \to E(\mathbf {Y})$ for which $\hat {h} \circ \eta _{\mathbf {X}} = \eta _{\mathbf {Y}} \circ h^2$ . In order to prove that it does not only preserve, but also reflects $\leqslant _{E(\mathbf {X})}$ , take any $x_1, x_2, x_1^{\prime }, x_2^{\prime } \in X$ such that $\hat {h}([(x_1, x_2)]_{\sim _{\mathbf {X}}}) \leqslant _{E(\mathbf {Y})} \hat {h}([(x_1^{\prime }, x_2^{\prime })]_{\sim _{\mathbf {X}}})$ . Put differently,

$$\begin{align*}\eta_{\mathbf{Y}} \circ h^2 (x_1, x_2) = \hat{h} \circ \eta_{\mathbf{X}} (x_1, x_2) \leqslant_{E(\mathbf{Y})} \hat{h} \circ \eta_{\mathbf{X}} (x_1^{\prime}, x_2^{\prime}) = \eta_{\mathbf{Y}} \circ h^2 (x_1^{\prime}, x_2^{\prime}), \end{align*}$$

that is, $[(h(x_1), h(x_2))]_{\sim _{\mathbf {Y}}} \leqslant _{E(\mathbf {Y})} [(h(x_1^{\prime }), h(x_2^{\prime }))]_{\sim _{\mathbf {Y}}}$ . By definition of $\leqslant _{E(\mathbf {Y})}$ , we actually get $(h(x_1), h(x_2)) \leqslant _{\mathbf {Y}} (h(x_1^{\prime }), h(x_2^{\prime }))$ . Finally, as h reflects $\leqslant _{\mathbf {X}}$ , then $(x_1, x_2) \leqslant _{\mathbf {X}} (x_1^{\prime }, x_2^{\prime })$ , which is what we wanted. What remains is to show that $\hat {h}$ is injective. Take any $x_1, x_2, x_1^{\prime }, x_2^{\prime } \in X$ such that $\hat {h}([(x_1, x_2)]_{\sim _{\mathbf {X}}}) = \hat {h}([(x_1^{\prime }, x_2^{\prime })]_{\sim _{\mathbf {X}}})$ . Similarly as before, by the commutativity of the diagram, we get that $(h(x_1), h(x_2)) = (h(x_1^{\prime }), h(x_2^{\prime }))$ . As h itself is injective, it follows immediately that $(x_1, x_2) = (x_1^{\prime }, x_2^{\prime })$ .

‘ $\Leftarrow $ ’: Let $h \colon X \to Y$ be a map and assume the existence of such an $\hat {h}$ as described in the statement of the right-hand side of the corollary. First, by Lemma 1.3, we get that h is a homomorphism from $\mathbf {X}$ to $\mathbf {Y}$ . Then, take any $x_1, x_2, x_1^{\prime }, x_2^{\prime } \in X$ for which $(h(x_1), h(x_2)) \leqslant _{\mathbf {Y}} (h(x_1^{\prime }), h(x_2^{\prime }))$ . Notice how this leads to $[(h(x_1), h(x_2))]_{\sim _{\mathbf {Y}}} \leqslant _{E(\mathbf {Y})} [(h(x_1^{\prime }), h(x_2^{\prime }))]_{\sim _{\mathbf {Y}}}$ , which in turn translates to

$$\begin{align*}\hat{h} \circ \eta_{\mathbf{X}} (x_1, x_2) = \eta_{\mathbf{Y}} \circ h^2 (x_1, x_2) \leqslant_{E(\mathbf{Y})} \eta_{\mathbf{Y}} \circ h^2 (x_1^{\prime}, x_2^{\prime}) = \hat{h} \circ \eta_{\mathbf{X}} (x_1^{\prime}, x_2^{\prime}), \end{align*}$$

that is, $[(x_1, x_2)]_{\sim _{\mathbf {X}}} \leqslant _{E(\mathbf {X})} [(x_1^{\prime }, x_2^{\prime })]_{\sim _{\mathbf {X}}}$ . Thus, $(x_1, x_2) \leqslant _{\mathbf {X}} (x_1^{\prime }, x_2^{\prime })$ . Lastly, we show that h is injective. Take any $x_1, x_2 \in X$ , for which $h(x_1) = h(x_2)$ . Then $(h(x_1), h(x_2)) = (h(x_2), h(x_2))$ , and so

$$\begin{align*}\hat{h}([(x_1, x_2)]_{\sim_{\mathbf{X}}}) = \eta_{\mathbf{Y}} (h(x_1), h(x_2)) = \eta_{\mathbf{Y}} (h(x_2), h(x_2)) = \hat{h}([(x_2, x_2)]_{\sim_{\mathbf{X}}}). \end{align*}$$

As a result of $\hat {h}$ being injective, we get that $(x_1, x_2) \sim _{\mathbf {X}} (x_2, x_2)$ . However, by the axioms of an echeloned space, this leads to $x_1 = x_2$ . This concludes the proof.

Proof. We proceed to prove that $\leqslant _{\mathbf {M}}$ is an echelon on M. As $\leqslant $ is a linear order on $\mathbb {R}$ , $\leqslant _{\mathbf {M}}$ is trivially a prechain on $M^2$ by its definition. For any $x, y, z \in M$ ,

1. (i) $d_M(x,x) = 0 \leqslant d_M(y, z)$ ; thus, $(x,x) \leqslant _{\mathbf {M}} (y, z)$ .

2. (ii) if $(y, z) \leqslant _{\mathbf {M}} (x, x)$ , then $d_M(y, z) \leqslant d_M(x, x) = 0$ . So $d_M(y, z) = 0$ , and so $y = z$ .

3. (iii) $d_M(x, y) = d_M(y, x)$ , so $(x, y) \leqslant _{\mathbf {M}} (y, x)$ .

Proof. Let $(X, d_X)$ be a dull metric space. Define . Clearly, the diameter of $(X,d_X)$ is not larger than $2 \tau $ , due to dullness. Thus, any distance within $(X, d_X)$ is bounded by $2 \tau $ .

Without loss of generality, assume $|X| \not = 1$ . In other words, there exist such $x, y \in X$ that $d_X(x, y) \not = 0$ . As $\tau \leqslant d_X(x, y) \leqslant 2 \tau $ , it follows that $\tau> 0$ . Consequently, as $0 < \tau \leqslant d_X(x, y)$ for all distinct $x, y \in X$ , $(X, d_X)$ is uniformly discrete.

Proof. We show $(ii)\Rightarrow (i)\Rightarrow (iv)\Rightarrow (ii)$ , and $(iii)\iff (iv)$ . It is clear that $(ii)\Rightarrow (i)$ .

‘ $(i)\Rightarrow (iv)$ ’: Let X be a set and $\leqslant _{\mathbf {X}}$ an echelon on it induced by a metric space $(X, d_X)$ . Clearly, $\operatorname {\mathrm {im}} d_X$ is order-embeddable into the reals, given that $d_X$ is a metric. Observe that $E(\mathbf {X}) \cong (\operatorname {\mathrm {im}} d_X, \leqslant )$ .

‘ $(iv)\Rightarrow (ii)$ ’: Let f be an embedding of $E(\mathbf {X})$ into $(\mathbb {R},\le )$ . Without loss of generality, we may assume that the image of f is contained in $(\{0\} \cup (1, 2), \leqslant )$ , and that f maps the smallest element of $E(\mathbf {X})$ (containing all reflexive pairs) to $0$ .

Now we define a map $d_X$ on $X^2$ as follows: for any $(x, y) \in X^2$ , let . It remains to show that it is a well-defined metric.

To begin with, notice that for any $x \in X$ , the distance $d_X(x, x) = f ([(x,x)]_{\sim _{\mathbf {X}}}) = 0$ . For any $x, y \in X$ , $d_X (x, y) \geqslant 0$ . Also, since $(x, y) \sim _{\mathbf {X}} (y, x)$ , then $d_X (x, y) = d_X (y, x)$ . At last, take any $x, y, z \in X$ . If x, y, and z are pairwise distinct, then

$$\begin{align*}d_X(x, y) = f([(x, y)]_{\sim_{\mathbf{X}}}) < 2 = 1 + 1 < f([(x, z)]_{\sim_{\mathbf{X}}}) + f([(z, y)]_{\sim_{\mathbf{X}}}) = d_X(x, z) + d_X(z, y).\end{align*}$$

Otherwise, if $x=y$ , then $d_X(x, y) = 0 \leqslant d_X(x, z) + d_X(z, y)$ ; whereas if $x=z$ or $y=z$ , then the triangle inequality trivially holds.

Overall, $(X, d_X)$ is indeed a metric space (inducing the echeloned space $(X, \leqslant _{\mathbf {X}})$ ). Moreover, by definition, it is dull.

‘ $(iii)\Rightarrow (iv)$ ’: Let S be a countable order-dense subset of $X^2$ . Consider

Clearly, $S_{E(\mathbf {X})}$ is a countable order-dense subset of $E(\mathbf {X})$ . By Theorem 1.8, $E(\mathbf {X})$ embeds into $(\mathbb {R},\le )$ .

‘ $(iv)\Rightarrow (iii)$ ’: By Theorem 1.8, $E(\mathbf {X})$ contains a countable order-dense subset $S_{E(\mathbf {X})}$ . Let T be a transversal of $\sim _{\mathbf {X}}$ . Consider

Then it is easy to see that S is countable and order-dense in $(X^2,\le _{\mathbf {X}})$ .

Proof. The proof of Proposition 1.11 provides us with the existence of dull metrics $d_M$ and $d_N$ on M and N which induce the echelons $\leqslant _{\mathbf {M}}$ and $\leqslant _{\mathbf {N}}$ . By rescaling $d_M$ , we may assume that $\operatorname {\mathrm {im}} d_M \subseteq \{0\} \cup (2, 4)$ and that $\operatorname {\mathrm {im}} d_N \subseteq \{0\} \cup (1,2)$ . Now, take any $f \in \operatorname {\mathrm {Hom}}(\mathbf {M}, \mathbf {N})$ . As for any two distinct $x, y \in M$ ,

$$\begin{align*}d_N (f(x), f(y)) < 2 < d_M (x,y),\end{align*}$$

f is trivially a $1$ -Lipschitz map between the metric spaces $\mathcal {M}$ and $\mathcal {N}$ . Therefore, $f\in \operatorname {\mathrm {Lip}}^{\leqslant }_1(\mathcal {M}, \mathcal {N})$ . The reverse inclusion holds trivially.

Proof. Let f be an automorphism of $\mathbf {X}$ . As a consequence of Corollary 1.4, $\hat {f}$ is an automorphism of the echeloning $E(\mathbf {X})$ . Since $E(\mathbf {X})$ is well-ordered, a standard induction argument shows that $\hat {f}=\mathrm {id}_{E(\mathbf {X})}$ .

Proof. This follows immediately from Proposition 1.17 and Corollary 1.4.

Proof. Since every finite linear order is a well-order, this follows immediately from Corollary 1.18.

Proof. Let $\varepsilon> 0$ . Without loss of generality, assume $|X|> 1$ . As $(f[X], d_Y \mathord {\upharpoonright }_{f[X]})$ is not uniformly discrete, there exist $x_0, y_0 \in X$ such that $0 \not = d_Y(f(x_0), f(y_0)) \leqslant \varepsilon $ . Note that . We will show that for any $x, y \in X$ for which $d_X(x, y) < \delta $ , it follows that $d_Y(f(x), f(y)) < \varepsilon $ .

Thus, take any $x, y \in X$ , such that $d_X(x, y) < d_X(x_0, y_0)$ . The latter is equivalent to $(x, y) <_{\mathbf {X}} (x_0, y_0)$ . As f is a homomorphism between the induced echeloned spaces, $(f(x), f(y)) <_{\mathbf {Y}} (f(x_0), f(y_0))$ , which in turn is equivalent to

$$\begin{align*}d_Y((f(x), f(y))) < d_Y(f(x_0), f(y_0)) \leqslant \varepsilon.\\[-34pt] \end{align*}$$

Proof. If $(X, d_X)$ is not uniformly discrete, we obtain the claim by Proposition 2.1. Otherwise, the statement is trivial.

Observation 2.4. Let $(Y,\le _{\mathbf {Y}})$ be an echeloned space and let $(X, d_X)$ be a metric space inducing the echelon $\leqslant _{\mathbf {X}}$ . Let $f \colon (X, \leqslant _{\mathbf {X}}) \to (Y, \leqslant _{\mathbf {Y}})$ be a homomorphism. If $(X, d_X)$ is Cantor-connected, then f is constant or injective.

Proof. Assume f is not injective. That means that there exist distinct $x_0, y_0 \in X$ such that $f(x_0) = f(y_0)$ . Let . Take any $x, y \in X$ and let $x=m_0,m_1, m_2, \dots , m_n=y$ be an $\varepsilon $ -chain from x to y. Consequently, for all $i \in \{0, 1, \dots , n-1\}$ , we have $(m_i, m_{i+1}) \leqslant _{\mathbf {X}} (x_0, y_0)$ . Since f is a homomorphism, it follows that $(f(m_i), f(m_{i+1})) \leqslant _{\mathbf {Y}} (f(x_0), f(y_0)) = (f(x_0), f(x_0))$ . Thus, $f(m_i) = f(m_{i+1})$ for all $i \in \{0, 1, \dots , n-1\}$ , leading to $f(x) = f(y)$ . From the arbitrary choice of x and y, we conclude that f is constant.

Proof. Take any $x, y \in X$ and denote by m a midpoint of x and y. Then we have that $d_X(x, m) = d_X(m, y) = \frac {1}{2} d_X(x, y)$ . Thus, $(x, m) \sim _{\mathbf {X}} (m, y)$ , and so $(f(x), f(m)) \sim _{\mathbf {Y}} (f(m), f(y))$ ; that is, $d_Y(f(x), f(m)) = d_Y(f(m), f(y))$ . Now, from the triangle inequality, we know that $d_Y(f(x), f(y)) \leqslant d_Y(f(x), f(m)) + d_Y(f(m), f(y)) = 2 d_Y(f(x), f(m))$ . Hence, $\frac {1}{2} d_Y(f(x), f(y)) \leqslant d_Y(f(x), f(m))$ .

In what follows, we shall show that $f(m)$ is in fact a midpoint of $f(x)$ and $f(y)$ . As $(Y,d_Y)$ has midpoints and f is surjective, there exists $m'\in X$ such that $f(m')\in \operatorname {\mathrm {Mid}}(f(x),f(y))$ . Next, we show that

$$\begin{align*}d_Y(f(x),f(m)) \le d_Y(f(x),f(m'))=\tfrac{1}{2} d_Y(f(x),f(y)). \end{align*}$$

Without loss of generality, we may assume that $d_X(x,m')\le d_X(m',y)$ . Then $d_X(m,y)=d_X(x,m)\le d_X(m',y)$ since

$$\begin{align*}2d_X(x,m)= d_X(x,y)\le d_X(x,m')+d_X(m',y)\le 2 d_X(m',y). \end{align*}$$

Now, in case that $d_X(x,m)\le d_X(x,m')$ , we have that $d_Y(f(x),f(m))\le d_Y(f(x),f(m'))$ . If, on the other hand, $d_X(x,m')\le d_X(x,m)$ , then

$$ \begin{align*} d_Y(f(x),f(m')) \le d_Y(f(x),f(m))\le d_Y(f(m'),f(y)) = d_Y(f(x),f(m')). \end{align*} $$

In both cases, we arrive at the conclusion that $f(m)\in \operatorname {\mathrm {Mid}}(f(x),f(y))$ , as desired.

Proof. Let $\hat {f}$ be the action of f on the echeloning of $(X, \leqslant _{\mathbf {X}})$ (cf. Corollary 1.4). Define $\tilde {d}_X \colon E(X)\to \operatorname {\mathrm {im}} d_X,\, [(x,y)]_{\sim _{\mathbf {X}}}\mapsto d_X(x,y)$ . Since $\leqslant _{\mathbf {X}}$ is induced by $d_X$ , it is easy to see that $\tilde {d}_X$ is an order isomorphism (cf. Lemma 1.5). Let . In particular, for all $a,b\in X$ , we have $\tilde {f}(d_X (a,b))=d_X (f(a),f(b))$ , and the following diagram commutes:

We will show that for every $\delta \in \operatorname {\mathrm {im}} d_X\setminus \{0\}$ , the restriction of $\tilde {f}$ to the initial segment is linear. Since, for any $0 < \delta < \delta ' \in \operatorname {\mathrm {im}} d_X$ , we have $\{0\} \neq I(\delta ) \subseteq I(\delta ')$ , it will follow that $\tilde {f}$ as a whole is linear; that is, f is a dilation.

Let . Let $a, c \in X$ with $ d_X(a, c) = \delta $ , and therefore with $t = \tfrac { d_X (f(a), f(c))}{ d_X (a, c)}$ . Let b be a midpoint of a and c.

Let $s\in I(\delta )$ . Since $\tilde {f}(0)=0=t\cdot 0$ , we may and will assume that $s\neq 0$ . We choose recursively a sequence $(b_n,c_n)_{n\in \mathbb {N}}$ in $X^2$ with

$$\begin{align*}\forall n\in\mathbb{N}\,:\, s\le d_X(a,c_n), \text{ and } (a,b_n,c_n) \text{ is collinear.} \end{align*}$$

We proceed as follows: . If $(b_n,c_n)$ has been chosen, then $(b_{n+1},c_{n+1})$ is chosen according to the following cases:

Case 1: if $s=d_X(a,c_n)$ , then put ,

Case 2: if $d_X(a,b_n)<s<d_X(a,c_n)$ , then choose $m\in \operatorname {\mathrm {Mid}}(b_n,c_n)$ and put

Case 3: if $d_X(a,b_n)=s$ , then put ,

Case 4: if $0< s < d_X(a,b_n),$ then choose $m\in \operatorname {\mathrm {Mid}}(a,b_n)$ and put .

Let us call a pair $(x,y)\in X^2$ homothetic if

1. 1) $(a,x,y)$ is collinear,

2. 2) $(f(a),f(x),f(y))$ is collinear,

3. 3) $d_X(f(a),f(x))=t\cdot d_X(a,x),\quad d_X(f(x),f(y)) = t\cdot d_X(x,y)$ .

A straightforward induction using Remark 2.7 and Lemma 2.6 shows that for every $n\in \mathbb {N}$ , the pair $(b_n,c_n)$ is homothetic. Furthermore,

$$\begin{align*}s = \lim_{n\to\infty} d_X(a,b_n) = \lim_{n\to\infty} d_X(a,c_n). \end{align*}$$

Note that since $s>0$ , for all but finitely many n, we have $d_X(a,b_n)\le s$ .

Since $\tilde {f}(d_X(a,b_n)) = t\cdot d_X(a,b_n)$ and $\tilde {f}(d_X(a,c_n)) = t\cdot d_X(a,c_n)$ for all $n\in \mathbb {N}$ , monotonicity of $\tilde {f}$ implies that $\tilde {f}(s)=t\cdot s$ , as desired.

Proof. Consider any three finite echeloned spaces $\mathbf {A}, \mathbf {B}_1, \mathbf {B}_2$ together with embeddings $f_1 \colon \mathbf {A} \hookrightarrow \mathbf {B}_1$ and $f_2 \colon \mathbf {A} \hookrightarrow \mathbf {B}_2$ . By Corollary 1.4, we obtain embeddings $\hat {f}_1 \colon E(\mathbf {A}) \hookrightarrow E(\mathbf {B}_1)$ and $\hat {f}_2 \colon E(\mathbf {A}) \hookrightarrow E(\mathbf {B}_2)$ , for which the following equations hold:

$$\begin{align*}\hat{f_1} \circ \eta_{\mathbf{A}} = \eta_{\mathbf{B}_1} \circ {f_1}^2 \quad \text{and} \quad \hat{f_2} \circ \eta_{\mathbf{A}} = \eta_{\mathbf{B}_2} \circ {f_2}^2.\end{align*}$$

Notice that $E(\mathbf {A})$ , $E(\mathbf {B}_1)$ and $E(\mathbf {B}_2)$ are all (finite) linear orders. Given that the class of finite linear orders has the amalgamation property, there exist a linear order $(D, \sqsubseteq )$ and embeddings

$$\begin{align*}{g}_1 \colon E(\mathbf{B}_1) \hookrightarrow (D, \sqsubseteq) \quad \text{and} \quad {g}_2 \colon E(\mathbf{B}_2) \hookrightarrow (D, \sqsubseteq)\end{align*}$$

for which ${g}_1 \circ \hat {f}_1 = {g}_2 \circ \hat {f}_2$ , that is, for which this diagram commutes:

For each $i\in \{1,2\}$ , observe that $\min E(\mathbf {B}_i)=\hat {f}_i (\min E(\mathbf {A}))$ , and therefore, without loss of generality, we can assign $g_i(\min E(\mathbf {B}_i))=\min (D, \sqsubseteq )$ . Moreover, we may assume that $\min (D, \sqsubseteq )\neq \max (D, \sqsubseteq )$ . Now, define and $\eta \colon C^2 \to (D, \sqsubseteq )$ as

Observe that $\eta $ is well defined. Now, we define a binary relation $\leqslant _{\mathbf {C}}$ on $C^2$ as follows:

$$\begin{align*}(c_1, c_2) \leqslant_{\mathbf{C}} (c_1^{\prime}, c_2^{\prime}) \quad :\Longleftrightarrow \quad \eta(c_1, c_2) \sqsubseteq \eta(c_1^{\prime}, c_2^{\prime}).\end{align*}$$

What we need to show now is that is indeed an echeloned space. The only nontrivial point to show is axiom (ii) from Definition 1.1. Take any $c_0, c_1, c_2 \in C$ . Let us assume that $(c_1, c_2) \leqslant _{\mathbf {C}} (c_0, c_0)$ . Then $\eta (c_1, c_2) \sqsubseteq \eta (c_0, c_0) = \min (D, \sqsubseteq )$ , and so $\eta (c_1, c_2) = \min (D, \sqsubseteq )$ . In other words, $(c_1, c_2) \in \mathbf {B}_i^2$ for some $i \in \{1, 2\}$ . Therefore, axiom (ii) for $\mathbf {B}_1$ or $\mathbf {B}_2$ implies $c_1=c_2$ . Thus, $\mathbf {C}$ is a well-defined finite echeloned space.

It remains to show that the dashed arrows in the following commuting diagram are embeddings:

Indeed, for any choice of $b_1, b_2, b_1^{\prime }, b_2^{\prime } \in \mathbf {B}_i$ , for $i \in \{1, 2\}$ , $(b_1, b_2) \leqslant _{\mathbf {B}_i} (b_1^{\prime }, b_2^{\prime })$ is equivalent to $\eta _{\mathbf {B}_i} (b_1, b_2) \leqslant _{E(\mathbf {B}_i)} \eta _{\mathbf {B}_i} (b_1^{\prime }, b_2^{\prime })$ , which in turn is equivalent to $\eta (b_1, b_2) \sqsubseteq \eta (b_1^{\prime }, b_2^{\prime })$ , that is, to $(b_1, b_2) \leqslant _{\mathbf {C}} (b_1^{\prime }, b_2^{\prime })$ . This concludes the proof.

Proof. What we need to show is that the class of finite echeloned spaces has the hereditary property ( $\operatorname {HP}$ ), the joint embedding property ( $\operatorname {JEP}$ ) and the amalgamation property ( $\operatorname {AP}$ ), and that up to isomorphism, there exist just countably many finite echeloned spaces.

The $\operatorname {AP}$ was already established in Proposition 3.1. With regards to establishing the $\operatorname {JEP}$ , observe that, by definition, an echeloned space is defined on a nonempty set. Note that the trivial one element echeloned space is embeddable into any echeloned space. Therefore, in this case, the $\operatorname {JEP}$ follows from the $\operatorname {AP}$ . Any subset of an echeloned space induces an echeloned space. Thus, this class has the $\operatorname {HP}$ . Having been defined over a finite relational signature, it clearly has only countably many isomorphism classes.

Proof. What we will show is that excluding $\bot _{\mathbf {F}}$ from the echeloning of $\mathbf {F}$ leaves us with a countable dense linear order without endpoints. In other words, the original structure is isomorphic to the set of nonnegative rational numbers equipped with the natural order.

As $\mathbf {F}$ is an echeloned space, $E(\mathbf {F})$ is a linear order. Take any $u_1, u_2, u_3, u_4 \in F$ such that $(u_1, u_2) <_{\mathbf {F}} (u_3, u_4)$ . Define and let $\mathbf {U}$ be the echeloned subspace of $\mathbf {F}$ induced by U. Further, let $v \not \in F$ be a new point and . We define an echeloned superspace of $\mathbf {U}$ on V, and call it $\mathbf {V}$ , for which $(u_1, u_2) <_{\mathbf {V}} (v, u_i)\sim _{\mathbf {V}} (v,u_j) <_{\mathbf {V}} (u_3, u_4)$ , for any $i,j \in \{1, 2, 3, 4\}$ .

Since $\mathbf {F}$ is homogeneous, it is also weakly homogeneous in the sense of [Reference Hodges12, p. 326]. Consequently, as $\mathbf {F}$ is also universal, there exists $\iota \colon \mathbf {V}\to \mathbf {F}$ , such that the following diagram commutes:

Let . Then, $(u_1, u_2) <_{\mathbf {F}} (u_1, u_5) <_{\mathbf {F}} (u_3, u_4)$ . Thus, $E(\mathbf {F})$ is indeed dense.

Similarly, one can prove the nonexistence of both the smallest and the greatest element of $E(\mathbf {F}) \setminus \{\bot _{\mathbf {F}}\}$ .

Proof. Let $(F, d)$ be some metric space which induces $\mathbf {F}$ . Further, let $\varphi \colon \operatorname {\mathrm {im}} d \to E(\mathbf {F})$ be the naturally induced order-isomorphism. Define . Recall that being dull is equivalent to $\operatorname {\mathrm {im}} d\subseteq [\tau ,2\tau ]\cup \{0\}$ (see the proof of Lemma 1.10).

Fix distinct $x_0, y_0 \in F$ . Recall that $\mathbf {F}$ is universal; thus, for any choice of distinct $x, y \in F$ , there have to exist $u, v, w \in F$ such that

$$\begin{align*}(u, v) \sim_{\mathbf{F}} (x, y), \quad (u, w) \sim_{\mathbf{F}} (v, w) \sim_{\mathbf{F}} (x_0, y_0).\end{align*}$$

Put differently, $ d(u, v) = c$ and $d(u, w) = d(v, w) = c_0$ , where and :

By the triangle inequality, we get $c\leqslant 2c_0$ .

Since $(x,y)$ (and hence c) was arbitrary, we have $\operatorname {\mathrm {im}} d\subseteq [0,2c_0]$ .

By letting $c_0$ converge to $\tau $ , we obtain that $\operatorname {\mathrm {im}} d\subseteq [0,2\tau ]$ . By the definition of $\tau $ , we actually have $\operatorname {\mathrm {im}} d\subseteq \{0\}\cup [\tau ,2\tau ]$ . Hence, $(F, d)$ is dull.

Proof. As the (bounded) rational Urysohn space is not dull, it follows by Proposition 3.5 that it cannot induce $\mathbf {F}$ .

Proof. By the uniqueness of Fraïssé limits, it is enough to prove that $\mathbf {M}$ is weakly homogeneous and universal for the class of all finite echeloned spaces.

• Universality Obviously, every finite substructure of $\mathbf {M}$ is a finite echeloned space. Conversely, let $\mathbf {A} = (A, \leqslant _{\mathbf {A}})$ be a finite echeloned space. By Proposition 1.11, we can realize $\leqslant _{\mathbf {A}}$ by a (dull) metric $ d_{\mathbf {A}}$ , whose image is in $\{0\} \cup (1, 2)$ . Since this image is finite, we can moreover assume that $ d_{\mathbf {A}}$ only takes rational values (up to applying an order-automorphism of $\{0\} \cup (1, 2)$ ). We have therefore realized $\mathbf {A}$ as a metric space with distances in S. By the universality of the Urysohn space $\mathbf {U}_S$ , there exists an isometric embedding f of $(A, d_{\mathbf {A}})$ into $(U, d_{\mathbf {U}_S})$ . Such an embedding is also an embedding of the respective induced echeloned spaces; that is, $\mathbf {A}$ indeed embeds into $\mathbf {M}$ .

• Weak homogeneity Let $\mathbf {A} =(A, \leqslant _{\mathbf {A}})$ and $\mathbf {B} = (B, \leqslant _{\mathbf {B}})$ be two finite echeloned spaces such that $\mathbf {A}\le \mathbf {B}$ . Let f be an embedding of $\mathbf {A}$ into $\mathbf {M}$ . We can realize $\mathbf {A}$ as a metric space by using the metric inducing $\leqslant _{\mathbf {M}}$ , that is, by defining

  
  This trivially makes f an isometric embedding of $(A, d_{\mathbf {A}})$ into $(U, d_{\mathbf {U}_S})$ . Now choose any metric $ d_{\mathbf {B}}$ on B, extending $ d_{\mathbf {A}}$ , with image in S and inducing $\leqslant _{\mathbf {B}}$ (such a metric exists since $\mathbf {B}$ is finite and the order of $(1, 2) \cap \mathbb {Q}$ is dense and without minimum nor maximum). By the universality and the weak homogeneity of the Urysohn space $\mathbf {U}_S$ , there exists an isometric embedding g of $\mathbf {B}$ into $\mathbf {U}_S$ such that the diagram
  
  commutes. Obviously, g is also an embedding of the induced echeloned space.

Proof. ‘ $\Rightarrow $ ’: Let $\Gamma = (V, \chi )$ be a countable C-coloured graph for which the $(\ast _\infty )$ -property holds.

Let us start by showing that $\Gamma $ is universal. We proceed by induction on the size $\ell $ of the C-coloured graph to be embedded. The empty C-coloured graph embeds to $\Gamma $ trivially. Suppose that $\Delta =(W,\eta )$ is a C-coloured graph of size $\ell +1$ . Let $v\in W$ . Let $\Delta '$ be the C-coloured subgraph of $\Delta $ induced by $W\setminus \{v\}$ , and suppose that $\Delta '$ embeds into $\Gamma $ by an embedding $\iota $ . Let $c_1,\dots ,c_k$ be all the (pairwise distinct) colours appearing as a colour of an edge from v in $\Delta $ . For each $i\in \{1,\dots ,k\}$ , let . Note that $\iota (U_1),\dots ,\iota (U_k)$ are pairwise disjoint. By the $(*_{\infty })$ -property (and, therefore, $(*_{k})$ -property), there exists a vertex $z\in V$ such that for all $i\in \{1,\dots ,k\}$ and all $u\in \iota (U_i)\,:\, \chi (\{z,u\})=c_i$ . Hence, the map $\hat {\iota }\colon W\to V$ extending $\iota $ and sending v to z is an embedding of $\Delta $ into $\Gamma $ . This finishes the proof that $\Gamma $ is universal. By iterating the argument above, it becomes clear that $\Gamma $ is also weakly homogeneous in the sense of [Reference Hodges12], and hence homogeneous.

‘ $\Leftarrow $ ’: Consider any universal homogenous countable C-coloured graph $\Gamma = (V, \chi )$ . Fix any positive integer k and a tuple $(c_1, \dots , c_k) \in C^k$ . Then choose k finite, pairwise disjoint subsets $U_1, \dots , U_k$ of V. Define a finite C-coloured graph on the set of vertices , where $u \not \in V$ is a new vertex, with the edge-colouring function $\chi _2$ which maps $\{u, v\}$ to $c_i$ for any $v \in U_i$ and $\{v_1, v_2\}$ to $\chi (\{v_1, v_2\})$ for any two distinct $v_1, v_2 \in V_2 \setminus \{u\}$ . Let $\Gamma _1$ be the C-coloured subgraph of $\Gamma $ induced by $\bigcup \limits _{i = 1}^n U_i$ . Clearly, there exist identity embeddings $\iota _1 \colon \Gamma _1 \hookrightarrow \Gamma $ and $\iota _2 \colon \Gamma _1 \hookrightarrow (V_2, \chi _2)$ . Since $\Gamma $ is homogeneous, it is weakly homogeneous in the sense of [Reference Hodges12]. Hence, as $\Gamma $ is also universal, there exists an embedding $f \colon (V_2, \chi _2)\to \Gamma $ for which $\iota _1 = f \circ \iota _2$ . As a result, we get that $\chi (\{f(u), f(\iota _2(v))\}) = \chi (\{f(u), \iota _1(v)\})=\chi (\{f(u), v\})=c_i$ for all $i \in \{1, \dots , k\}$ . Thus, the $(\ast _k)$ -property of $\Gamma $ holds.

Observation 3.14. When a C-coloured graph $\Gamma = (V, \chi )$ , for $|C| = 2$ , has the $(\ast _2)$ -property, then the graph $(V, E)$ , with the set of edges E defined for a fixed $c \in C$ as follows:

$$\begin{align*}(u,v) \in E \quad :\Longleftrightarrow \quad \chi(\{u, v\}) = c, \end{align*}$$

is isomorphic to the Rado graph.

Proof. By Proposition 3.13, $\Gamma _C $ is isomorphic to $\mathcal {T}_C$ if it has the $(\ast _\infty )$ -property. As $|C| = k + 1$ , it suffices to show that $\Gamma _C$ has the $(\ast _{k+1})$ -property. Consider thus the $(k+1)$ –tuple $(c_1, \dots , c_k, c_\ast )$ . Without loss of generality, assume $c_1 <_{E(\mathbf {F})} c_2 <_{E(\mathbf {F})} \dots <_{E(\mathbf {F})} c_k$ .

Let $U_1, \dots , U_k, U_* \subseteq F$ be finite and pairwise disjoint. Set . Further, let U be a finite superset of $U'$ such that for all $i \in \{1 \dots , k\}$ , there exist $v_1, v_2 \in U$ for which $\chi (\{v_1, v_2\}) = c_i$ . Further, let $w \not \in F$ be a new point, , and let $\mathbf {U}$ be the echeloned subspace of $\mathbf {F}$ induced by U. We define an echelon $\leqslant _{\mathbf {V}}$ on V such that

• ○ $\leqslant _{\mathbf {V}} \cap ~U^2 =\ \leqslant _{\mathbf {U}} (=\ \leqslant _{\mathbf {F}} \cap ~U^2)$ ,

• ○ for all $i\in \{1,\dots ,k\}$ , for any $v_1,v_2 \in V$ such that $\chi (\{v_1,v_2\})=c_i$ , and for all $u\in U_i$ , $(w, u) \sim _{\mathbf {V}} (v_1, v_2)$ ,

• ○ for any $u_1,u_2 \in U_*$ and $(x, y) \in U^2$ , $(x, y) <_{\mathbf {V}} (w, u_1)\sim _{\mathbf {V}} (w,u_2)$ .

By the construction $\mathbf {U}$ is a subspace of $\mathbf {V}$ . Since $\mathbf {F}$ is universal and weakly homogeneous, there exists an embedding $h \colon \mathbf {V} \hookrightarrow \mathbf {F}$ such that the following diagram commutes:

(†)

Let . It remains to show that $\chi (\{z, u\}) = c_i$ , whenever $u \in U_i$ for any $i \in \{1, \dots , k\} \cup \{*\}$ .

First, fix $i \in \{1, \dots , k\}$ and $u \in U_i$ . Recall that there exist some $v_1, v_2 \in U$ such that $\chi (\{v_1, v_2\})=c_i$ . In particular, we have $\eta _{\mathbf {F}}(v_1, v_2) = \chi (\{v_1, v_2\}) =c_i$ . By construction, we have $(w,u) \sim _{\mathbf {V}} (v_1, v_2)$ . Thus, $(h(w), h(u)) \sim _{\mathbf {F}} (h(v_1), h(v_2))$ . By the commutativity of diagram (†), we obtain that $(z,u)\sim _{\mathbf {F}} (v_1,v_2)$ . As a result, $\chi (\{z, u\}) = \eta _{\mathbf {F}}(z, u) = \eta _{\mathbf {F}}(v_1, v_2) = c_i$ .

It remains to show that $\chi (\{z, u\}) = c_\ast $ for all $u \in U_*$ . Let $u \in U_*$ be arbitrary. Then by definition of $\leqslant _{\mathbf {V}}$ , we know that for $i \in \{1, \dots , k\}$ and any $(v_1, v_2) \in U^2$ with $\chi (\{v_1, v_2\}) = c_i$ , we have $(v_1, v_2) <_{\mathbf {V}} (w, u)$ . We get that $(h(v_1), h(v_2)) <_{\mathbf {F}} (h(w), h(u))$ , and hence, by the commutativity of diagram (†), $(v_1, v_2) <_{\mathbf {F}} (z, u)$ . In particular, $\eta _{\mathbf {F}}(z, u) \not = \eta _{\mathbf {F}}(v_1, v_2) = c_i$ for any $i \in \{1, \dots , k\}$ . Consequently, $\chi (\{z, u\}) = c_\ast $ .

Proof. Fix a $c \in E(\mathbf {F}) \setminus \{\bot _{\mathbf {F}}\}$ and pick any $c_\ast \not \in E(\mathbf {F})$ . Define

$$\begin{align*}\chi \colon [F]^2 \to \{c, c_\ast\},\, \{u, v\} \mapsto \begin{cases} c& \text{if } \eta_{\mathbf{F}}(u, v) = c,\\ c_\ast& \text{otherwise}. \end{cases} \end{align*}$$

By Proposition 3.15, $(F, \chi ) \cong \mathcal {T}_C$ , with $C = \{c, c_\ast \}$ . Consequently, by Observation 3.14, $(F, E)$ is then isomorphic to the Rado graph.

Proof. Fix a positive integer $k \geqslant 2$ . Pick any k colours from C and enumerate them as $\{c_1, c_2, \dots , c_k\}$ . Also, choose any k pairwise disjoint finite sets of points $U_1, U_2, \dots , U_k$ from F. Let $c_{k+1}\in C\setminus \{c_1,\dots , c_k\}$ . Define and a colouring

$$\begin{align*}\chi_{C'} \colon [F]^2 \to C',\quad \{u,v\} \mapsto \begin{cases} c_i& \text{if } \eta_{\mathbf{F}}(u, v) = c_i, i \in \{1, 2, \dots, k\},\\ c_{k+1}& \text{otherwise}. \end{cases} \end{align*}$$

By Proposition 3.15, $(F, \chi _{C'})$ is isomorphic to $\mathcal {T}_{C'}$ . In particular, it enjoys the $(\ast _{k})$ -property. Therefore, there exists a vertex $z \in F$ such that for any $i \in \{1, 2, \dots , k\}$ , $\chi _{C'}(\{u, z\}) =\chi (\{u,z\})= c_i$ for all $u \in U_i$ . Owing to the arbitrary choice of $U_i$ ’s, we get that $(F, \chi )$ itself has the $(\ast _k)$ -property. Due to the arbitrary choice of k, $(F, \chi )$ has the $(\ast _\infty )$ -property and so by Proposition 3.13, it is indeed isomorphic to $\mathcal {T}_C$ .

Proof. Let . Let $\Gamma $ be the C-coloured graph $(H,\chi )$ with $\chi (\{u,v\})=d_i$ if $\eta _{\mathbf {H}}(u,v)=c_i$ . By the universality of $\mathcal {T}_{\mathbf {F}}$ established in Theorem 3.17, there exists an embedding $\kappa \colon \Gamma \hookrightarrow \mathcal {T}_{\mathbf {F}}$ of C-coloured graphs. It induces an embedding $\iota \colon \mathbf {H}\hookrightarrow \mathbf {F}$ with $\hat {\iota }(c_i)=d_i$ .

Proof. We show that $(V,\chi )$ has the $(\ast _\infty )$ -property with probability $1$ . Thus, we fix some integer $k \geqslant 2$ .

Let $U_1,\dots ,U_k$ be disjoint finite subsets of V. Let $c_1,\dots ,c_k\in C$ . Take arbitrary $z\in V$ . We compute the probability that z is not eligible for the $(\ast _\infty )$ -property; that is,

$$\begin{align*}P\left[\exists j\in \{1,\dots, k\}\; \exists u\in U_j\colon \chi(\{u,z\})\neq c_j\right] = 1 - \prod_{\ell = 1}^k \mu(c_\ell)^{|U_\ell|} < 1. \end{align*}$$

Since edges are coloured independently, we have

$$\begin{align*}P[\forall z\in V\;\exists j\in \{1,\dots, k\}\; \exists u\in U_j\colon \chi(\{u,z\})\neq c_j]=0. \end{align*}$$

There are only countably many choices for $U_j$ and $c_j$ , so

$$\begin{align*}P[(\ast_\infty)\text{-property fails}]=0. \end{align*}$$

Hence, our observed graph is isomorphic to $\mathcal {T}_C$ (with probability $1$ ).

Proof. The classes of finite echeloned spaces, finite C-coloured graphs and finite linear orders are all Fraïssé classes with the strong amalgamation property (for the case of finite echeloned spaces, see Remark 3.2). This immediately implies the result (see [Reference Cameron5, Section 3.9, p. 59]).

Proof. Let $(\mathbf {H},\preceq _{\mathbf {H}})$ be the ordered echeloned space whose echelon $\leqslant _{\mathbf {H}}$ is defined by

$$\begin{align*}(x,y)\leqslant_{\mathbf{H}} (u,v) \quad :\Longleftrightarrow \quad \chi_{\mathbf{H}}(\{x,y\})\leqslant \chi_{\mathbf{H}}(\{u,v\}), \end{align*}$$

from the edge colouring $\chi _{\mathbf {H}}\colon [H]^2\to C$ of a countable homogeneous ordered C-coloured graph $(H,\chi _{\mathbf {H}},\preceq _{\mathbf {H}})$ . We aim to show that $(\mathbf {H}, \preceq _{\mathbf {H}})$ is a universal homogeneous ordered echeloned space. To this end, it suffices to show that $(\mathbf {H},\preceq _{\mathbf {H}})$ is universal and weakly homogeneous. Let $(\mathbf {A},\preceq _{\mathbf {A}})$ be a finite ordered echeloned subspace of $(\mathbf {H},\preceq _{\mathbf {H}})$ , let $(\mathbf {B},\preceq _{\mathbf {B}})$ be a finite ordered echeloned space, and let $\iota \colon (\mathbf {A},\preceq _{\mathbf {A}})\hookrightarrow (\mathbf {B},\preceq _{\mathbf {B}})$ be an embedding. Note that $\kappa \colon E(\mathbf {A})\setminus \{\bot _{\mathbf {A}}\}\to C,\, [(x,y)]_{\sim _{\mathbf {A}}}\mapsto \chi _{\mathbf {H}}(\{x,y\})$ is an order embedding. The same holds for $\hat \iota \colon E(\mathbf {A})\hookrightarrow E(\mathbf {B})$ . Let $\tilde \iota \colon E(\mathbf {A})\setminus \{\bot _{\mathbf {A}}\}\hookrightarrow E(\mathbf {B})\setminus \{\bot _{\mathbf {B}}\}$ be the appropriate restriction of $\hat \iota $ . Recall that C is isomorphic to $\mathbb {Q}$ . In other words, it is a universal homogeneous chain. Thus, there exists an order embedding $\tilde \kappa \colon E(\mathbf {B})\setminus \{\bot _{\mathbf {B}}\}\hookrightarrow C$ that makes the following diagram commutative:

Next we define

$$ \begin{align*} \chi_{\mathbf{A}}&\colon [A]^2\to C, & \{x,y\} &\mapsto \kappa(\eta_{\mathbf{A}}(x,y)) = \chi_{\mathbf{H}}(\{x,y\}),\\ \chi_{\mathbf{B}}&\colon [B]^2\to C, & \{u,v\}&\mapsto \tilde\kappa(\eta_{\mathbf{B}}(u,v)). \end{align*} $$

Then $\iota \colon (A,\chi _{\mathbf {A}},\preceq _{\mathbf {A}})\to (B,\chi _{\mathbf {B}},\preceq _{\mathbf {B}})$ is an embedding of ordered C-coloured graphs. Indeed, $\iota $ preserves $\preceq $ , and

$$ \begin{align*} \forall \{x,y\}\in[A]^2\,:\, \chi_{\mathbf{B}}(\{\iota(x),\iota(y)\}) & = \tilde\kappa (\eta_{\mathbf{B}}(\iota(x),\iota(y)))\\ &= \tilde\kappa(\tilde\iota(\eta_{\mathbf{A}}(x,y))) = \kappa(\eta_{\mathbf{A}}(x,y)) = \chi_{\mathbf{A}}(\{x,y\}). \end{align*} $$

Next, note that $(A,\chi _{\mathbf {A}},\preceq _{\mathbf {A}})$ is an ordered C-coloured subgraph of $(H,\chi _{\mathbf {H}},\preceq _{\mathbf {H}})$ . Indeed, $A\subseteq H$ , ${\preceq _{\mathbf {A}}}\subseteq {\preceq _{\mathbf {H}}}$ , and

$$\begin{align*}\forall\{x,y\}\in[A]^2\,:\, \chi_{\mathbf{A}}(\{x,y\}) = \kappa(\eta_{\mathbf{H}}(x,y)) = \chi_{\mathbf{H}}(\{x,y\}), \end{align*}$$

by the definition of $\kappa $ .

Since $(H,\chi _{\mathbf {H}},\preceq _{\mathbf {H}})$ is universal and weakly homogeneous, there exists $\varepsilon \colon (B,\chi _{\mathbf {B}},\preceq _{\mathbf {B}})\hookrightarrow (H,\chi _{\mathbf {H}},\preceq _{\mathbf {H}})$ , such that the following diagram commutes:

(4.1)

We claim that $\varepsilon \colon (\mathbf {B},\preceq _{\mathbf {B}})\to (\mathbf {H},\preceq _{\mathbf {H}})$ is an embedding. Clearly, $\varepsilon \colon (B,\preceq _{\mathbf {B}})\hookrightarrow (H,\preceq _{\mathbf {H}})$ is an order embedding. So let $(x,y)\leqslant _{\mathbf {B}}(u,v)$ . If $x=y$ , then $\varepsilon (x)=\varepsilon (y)$ , and thus, $(\varepsilon (x),\varepsilon (y))\leqslant _{\mathbf {H}}(\varepsilon (u),\varepsilon (v))$ . If, however, $x\neq y$ , then we compute

From (4.1), we obtain that

commutes. This shows that $(\mathbf {H},\preceq _{\mathbf {H}})$ is universal and weakly homogeneous.

By the uniqueness of Fraïssé limits, there exists an isomorphism $\varphi \colon (\mathbf {F},\preceq _{\mathbf {F}})\to (\mathbf {H},\preceq _{\mathbf {H}})$ . Since

commutes, by the definition of $\chi _{\mathbf {F}}$ and $\chi _{\mathbf {H}}$ , also the following diagram commutes:

where $[\varphi ]^2\colon \{x,y\}\mapsto \{\varphi (x),\varphi (y)\}$ , and where $\tilde \varphi $ is the appropriate restriction of $\hat \varphi $ to C.

However, this means that $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ and $(H,\chi _{\mathbf {H}},\preceq _{\mathbf {H}})$ are practically equal, up to names of vertices and names of colours. In other words, $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ is a universal homogeneous C-coloured graph, too.

Proof. By Theorem 4.1, it suffices to prove that the class of finite ordered C-coloured graphs is Ramsey. In order to show this, we consider the class of all finite ordered C-coloured graphs, where . Let $c\in C$ .

Note that there exists a canonical bijection $\Phi $ between the class of all ordered C-coloured graphs and all ordered $C\setminus \{c\}$ -edge-coloured (simple) graphs such that, for all ordered C-coloured graphs $\mathbf {A}$ and $\mathbf {B}$ ,

1. (i) $\Phi (\mathbf {A})$ has the same vertex set as $\mathbf {A}$ , and

2. (ii) the set of embeddings from $\mathbf {A}$ to $\mathbf {B}$ coincides with the set of embeddings from $\Phi (\mathbf {A})$ to $\Phi (\mathbf {B})$ .

In particular, $\Phi (\mathbf {A})$ is obtained from $\mathbf {A}$ through replacing all c-coloured edges by non-edges.

Clearly, the class of finite $C\setminus \{c\}$ -edge-coloured graphs has the free amalgamation property (in the sense of [Reference Macpherson18, page 1602]); thus, the class of finite ordered $C\setminus \{c\}$ -edge-coloured graphs is a Ramsey class, by a result of Hubička and Nešetril [Reference Hubička and Nešetřil14, Corollary 4.2, p. 51].

Thanks to the properties of $\Phi $ ( $\Phi $ is an isomorphism between the category of C-coloured graphs with embeddings and the category of $C\setminus \{c\}$ -edge-coloured graphs with embeddings), the class of finite ordered C-coloured graphs is Ramsey, too. So $\operatorname {\mathrm {Aut}} (\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ is extremely amenable due to Theorem 4.1.

Proof. Recall from Lemma 4.3 that $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}}) = (F,\chi ,\preceq _{\mathbf {F}})$ is a universal homogeneous ordered C-coloured graph, where . Denote the elements of $\operatorname {\mathrm {dom}}\beta $ by $c_1,\dots ,c_k$ and let for each $i\in \{1,\dots ,k\}$ . Let $c_\ast \in C$ be strictly greater than any element of $\{c_1,\dots ,c_k,c_{k+1},\dots ,c_{2k}\}$ . Consider the ordered C-coloured graphs $(\Delta ,\preceq _\Delta )=(D,\chi _{\Delta },\preceq _\Delta )$ and $(\tilde \Delta ,\preceq _{\tilde \Delta })=(\tilde {D},\chi _{\tilde \Delta },\preceq _{\tilde \Delta })$ given by

• ○ $D=\{v_1,\dots ,v_{k}\}\mathrel {\dot \cup } \{w_1,\dots ,w_{k}\}$ , $v_1\preceq _\Delta \dots \preceq _\Delta v_k\preceq _\Delta w_1\preceq _\Delta \dots \preceq _\Delta w_k$ , and

• ○ $\chi _{\Delta }(\{x,y\})=\begin {cases} c_i & \text {if } \{x,y\}=\{v_i,w_i\} \text { for some } i\in \{1,\dots ,k\},\\ c_\ast & \text {else}. \end {cases}$

• ○ $\tilde {D}=\{\tilde {v}_1,\dots ,\tilde {v}_{k}\}\mathrel {\dot \cup } \{\tilde {w}_1,\dots ,\tilde {w}_{k}\}$ , $\tilde {v}_1\preceq _\Delta \dots \preceq _\Delta \tilde {v}_k\preceq _\Delta \tilde {w}_1\preceq _\Delta \dots \preceq _\Delta \tilde {w}_k$ , and

• ○ $\chi _{\tilde \Delta }(\{x,y\})=\begin {cases} c_{k+i} & \text {if } \{x,y\}=\{\tilde {v}_i,\tilde {w}_i\} \text { for some } i\in \{1,\dots ,k\},\\ c_\ast & \text {else}. \end {cases}$

Since $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ is universal, both $(\Delta ,\preceq _\Delta )$ and $(\tilde \Delta ,\preceq _{\tilde \Delta })$ embed into it. Without loss of generality, we may assume that $(\Delta ,\preceq _\Delta )$ and $(\tilde \Delta ,\preceq _{\tilde \Delta })$ are substructures of $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ . Consider

$$\begin{align*}\varphi\colon D\to F \text{ given by } v_i\mapsto\tilde{v}_i,\, w_i\mapsto\tilde{w}_i,\, i=1,\dots, k. \end{align*}$$

We claim that $\varphi $ is an embedding of the ordered echeloned space into $(\mathbf {F},\preceq _{\mathbf {F}})$ . Clearly, $\varphi $ is an order-embedding from $(D,\preceq _\Delta )$ into $(F,\preceq _{\mathbf {F}})$ . Let $(x,y),(u,v)\in D^2$ . Then

By the homogeneity of $(\mathbf {F},\preceq _{\mathbf {F}})$ , there exists an automorphism $\alpha $ of $(\mathbf {F},\preceq _{\mathbf {F}})$ that makes the following diagram commutative:

From the previous calculations, it is clear that $\hat \alpha \mathord {\upharpoonright }_T=\beta $ .

Proof. Due to Theorem 4.1, $(\mathbf {F}, \preceq _{\mathbf {F}})$ has the Ramsey property if and only if $\operatorname {\mathrm {Aut}} (\mathbf {F}, \preceq _{\mathbf {F}})$ is extremely amenable.

Let . Let $\mathcal {T}_{\mathbf {F}}=(F,\chi )$ be the C-coloured graph constructed in Proposition 3.17. Recall that $(\mathbf {F},\preceq _{\mathbf {F}})$ is the Fraïssé limit of the class of finite ordered echeloned spaces. Therefore, $\operatorname {\mathrm {Aut}}(\mathcal {T}_{\mathbf {F}}, \preceq _{\mathbf {F}})$ is a subgroup of $\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})$ – namely, the one of automorphisms that setwise preserve each equivalence class of $\sim _{\mathbf {F}}$ . Let $\pi \colon \operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}}) \to \operatorname {\mathrm {Aut}}(E(\mathbf {F})) ,\, \alpha \mapsto \hat {\alpha }$ . Clearly, $\ker (\pi ) = \operatorname {\mathrm {Aut}}(\mathcal {T}_{\mathbf {F}}, \preceq _{\mathbf {F}})$ , which is extremely amenable by Lemma 4.4.

Proof of the claim

As $\pi $ is a homomorphism, it is enough to show that, for every finite subset E of F, there exists a finite subset $\tilde {E}$ of $E(\mathbf {F})$ such that

$$\begin{align*}\pi[\operatorname{\mathrm{Aut}}(\mathbf{F}, \preceq_{\mathbf{F}})] \cap V_{\operatorname{\mathrm{Aut}}(E(\mathbf{F}))}(\tilde{E}) \subseteq \pi[V_{\operatorname{\mathrm{Aut}}(\mathbf{F}, \preceq_{\mathbf{F}})}(E)].\end{align*}$$

Let E be a finite subset of F. Define . Let $\gamma \in \pi [\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})] \cap V_{\operatorname {\mathrm {Aut}}(E(\mathbf {F}))}(\tilde {E})$ . There then exists some $\alpha _{0} \in \operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})$ with $\pi (\alpha _{0}) = \gamma $ . From $\pi (\alpha _{0}) = \gamma \in V_{\operatorname {\mathrm {Aut}}(E(\mathbf {F}))}(\tilde {E})$ , we infer that $E \to \alpha _{0}[E], \, x \mapsto \alpha _{0}(x)$ is a local isomorphism in $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ . Since $(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ is homogeneous by Lemma 4.3, there exists $\alpha _{1} \in \operatorname {\mathrm {Aut}}(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ such that $\alpha _{1}\mathord {\upharpoonright }_{E} = \alpha _{0}\mathord {\upharpoonright }_{E}$ . We see that . Moreover, as $\alpha _{1} \in \operatorname {\mathrm {Aut}}(\mathcal {T}_{\mathbf {F}},\preceq _{\mathbf {F}})$ , one has $\pi (\alpha _{1}) = \mathrm {id}_{E(\mathbf {F})}$ , so that $\pi (\beta ) = \pi (\alpha _{1})^{-1}\pi (\alpha _{0}) = \pi (\alpha _{0}) = \gamma $ . Hence, $\gamma = \pi (\beta ) \in \pi [V_{\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})}(E)]$ , as desired.

It follows by these three claims that the group $\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})$ has an extremely amenable closed normal subgroup $\ker (\pi ) = \operatorname {\mathrm {Aut}}(\mathcal {T}_{\mathbf {F}}, \preceq _{\mathbf {F}})$ whose corresponding quotient ${\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})}/{\ker \pi }$ , being isomorphic to $\operatorname {\mathrm {Aut}}(\mathbb {Q}, <)$ by Lemma 3.4, is also extremely amenable [Reference Pestov21]. Hence, $\operatorname {\mathrm {Aut}}(\mathbf {F}, \preceq _{\mathbf {F}})$ itself is extremely amenable (see, for example, [Reference Pestov22, Corollary 6.2.10]) and therefore $(\mathbf {F}, \preceq _{\mathbf {F}})$ has the Ramsey property by Theorem 4.1.

Proof. Since all the morphisms of $\mathscr {A}$ and $\mathscr {C}$ are embeddings, the condition that K preserves embeddings is trivially fulfilled.

For every finite echeloned space $\mathbf {X}$ , let $\lambda _{\mathbf {X}}\colon \mathbf {X}\hookrightarrow K(\mathbf {X})$ be the identical embedding. It is not hard to check that this defines a natural transformation $\lambda \colon \operatorname {Id}\hookrightarrow K$ .

Let now $\mathbf {X}$ be a finite echeloned space and let $\mathbf {Y}$ be a one-point extension of $\mathbf {X}$ (i.e., $Y=X\mathrel {\dot \cup }\{y\}$ and the identical embedding of X into Y is an embedding of echeloned spaces). Denote the identical embedding of $\mathbf {X}$ into $\mathbf {Y}$ by e. Suppose $E(\mathbf {X})=\{\bot _{\mathbf {X}},c_1,\dots ,c_n\}$ , where

$$\begin{align*}\bot_{\mathbf{X}}<_{E(\mathbf{X})} c_1 <_{E(\mathbf{X})} c_2 <_{E(\mathbf{X})}\dots <_{E(\mathbf{X})} c_n. \end{align*}$$

Then $E(\mathbf {Y})$ is of the shape

$$ \begin{align*} &\bot_{\mathbf{Y}} <_{E(\mathbf{Y})} d_{1,0} <_{E(\mathbf{Y})} \dots d_{i_0,0} <_{E(\mathbf{Y})} \hat{e}(c_1) <_{E(\mathbf{Y})} d_{1,1} <_{E(\mathbf{Y})} \dots \\ &\quad \dots<_{E(\mathbf{Y})} d_{i_1,1} <_{E(\mathbf{Y})} \hat{e}(c_2)<_{E(\mathbf{Y})}\dots <_{E(\mathbf{Y})} \hat{e}(c_n) <_{E(\mathbf{Y})} d_{1,n} <_{E(\mathbf{Y})} \dots <_{E(\mathbf{Y})} d_{i_n,n}, \end{align*} $$

where $i_0+i_1+\dots +i_n\leqslant |X|$ . Define

$$\begin{align*}h\colon X\to C_{\mathbf{X}}\setminus\{\bot_{\mathbf{X}}\},\quad x\mapsto\begin{cases} (k,j) & \text{if } \eta_{\mathbf{Y}}(y,e(x)) = d_{k,j}, \text{ where }j=0,\dots,n,\, k=1,\dots, i_j,\\ c_j & \text{if } \eta_{\mathbf{Y}}(y,e(x)) = \hat{e}(c_j),\text{ where } j=1,\dots,n. \end{cases} \end{align*}$$

Finally, define

$$\begin{align*}g\colon Y\to X\mathrel{\dot\cup} C_{\mathbf{X}}^{(X)},\quad x\mapsto\begin{cases} x & \text{if } x\in X,\\ h & \text{if } x=y. \end{cases} \end{align*}$$

By construction, g is injective. Next, we show that $g\colon \mathbf {Y}\to K(\mathbf {X})$ is an embedding. Let $\gamma \colon E(\mathbf {Y})\to C_{\mathbf {X}}$ be given by

$$\begin{align*}\gamma\colon \hat{e}(c_j)\mapsto c_j,\quad \bot_{\mathbf{Y}}\mapsto\bot_{\mathbf{X}},\quad d_{k,j}\mapsto (k,j)\quad \text{ for all } j\in\{0,\dots,n\},\, k\in \{1,\dots,i_j\}. \end{align*}$$

Clearly, $\gamma $ is an order embedding and the following diagram commutes:

Next, we show that the following diagram commutes, too:

(5.6)

To this end, let $(x,z)\in Y^2$ . We distinguish six cases:

Case 1: Suppose that $x=z\neq y$ . Then

Case 2: Suppose that $x=z=y$ . Then

Case 3: Suppose that $x\neq z$ and that $x,z\in X$ . Then $\eta _{\mathbf {X}}(x,z)=c_i$ for some i and

Case 4: Suppose that $x\neq z$ and $x=y$ , and that $\eta _{\mathbf {Y}}(y,e(z))= d_{k,j}$ for some $j\in \{0,\dots ,n\}$ , and $k\in \{1,\dots ,i_j\}$ . Then

Case 5: Suppose that $x\neq z$ and $x=y$ , and that $\eta _{\mathbf {Y}}(x,e(z))= \hat {e}(c_j)$ for some $j\in \{0,\dots ,n\}$ . Then

Case 6: If $x\neq z$ and $z=y$ , then we proceed analogously as in cases 4 and 5.

At this point, we may conclude that (5.6) commutes. Combining this with (5.1), we obtain that the following diagram commutes, too:

Hence, by Corollary 1.4, g is an embedding with $\hat {g}=\zeta _{\mathbf {X}}\circ \gamma $ .

Finally, by the definition of g, the following diagram commutes:

Thus, indeed, K is a Katětov functor.

Observation 6.5. We describe the connection between echeloned spaces and Definition 6.1 in some detail.

Let $\mathscr {A}$ be the category of echeloned spaces with homomorphisms, and let $\mathscr {B}$ be the category of echeloned maps with homomorphisms. Clearly, if $\mathbf {X} = (X, \leqslant _{\mathbf {X}})$ is an echeloned space, then $\eta _{\mathbf {X}}$ is an echeloned map. Moreover, the homomorphisms between two echeloned spaces are the same as those between their corresponding echeloned maps (see Lemma 1.3). This means that $F \colon \mathscr {A} \to \mathscr {B},\, \mathbf {X} \mapsto \eta _{\mathbf {X}},\, h \mapsto h$ is a well-defined functor.

Now, define a functor $G \colon \mathscr {B} \to \mathscr {A}$ . To every echeloned map $f \colon X^2 \twoheadrightarrow (C, \leqslant )$ , we associate an echeloned space $\mathbf {X}_f = (X, \leqslant _{\mathbf {X}})$ according to the following rule:

$$\begin{align*}(x,y) \leqslant_{\mathbf{X}} (u,v) \quad :\Longleftrightarrow \quad f(x,y) \leqslant f(u, v).\end{align*}$$

Again, by Lemma 1.3, $h \colon f \to g$ is a homomorphism of echeloned maps. Then $h \colon \mathbf {X}_f \to \mathbf {X}_g$ is a homomorphism, too. Hence, the assignment $G\colon \mathscr {B}\to \mathscr {A},\, f \mapsto \mathbf {X}_f,\, h \mapsto h$ is a well-defined functor. Note that $G\circ F = \operatorname {Id}_{\mathscr {A}}$ . Conversely, for each echeloned map $f \colon X^2 \twoheadrightarrow (C, \leqslant )$ , note that the identity function $\mathrm {id}_X$ is an isomorphism from $F(G(f))$ to f. Moreover, $(\varphi _f)_{f\in \operatorname {\mathrm {ob}}(\mathscr {B})}\colon F\circ G\to \operatorname {Id}_{\mathscr {B}}$ with is a natural isomorphism. This shows that $\mathscr {A}$ and $\mathscr {B}$ are equivalent categories. Note that F and G, both, preserve finiteness and embeddings.

All the proofs from the previous sections can be rewritten easily in this alternative approach.