{\mathfrak{L}}_2) = \inf\{\varepsilon \, : \,  {\mathfrak{L}}_1 \subseteq_{\varepsilon} {\mathfrak{L}}_2\},$$ that is, $$c({\mathfrak{L}}_1, {\mathfrak{L}}_2) = \sup_{\tau_1 \in {\mathfrak{L}}_1} \inf_{\tau_2 \in {\mathfrak{L}}_2} d(\tau_1, \tau_2) = \sup_{\tau_1 \in {\mathfrak{L}}_1} d(\tau_1, {\mathfrak{L}}_2).$$ The distance $d({\mathfrak{L}}_1,{\mathfrak{L}}_2)$ between ${\mathfrak{L}}_1$ and ${\mathfrak{L}}_2$ is $$\label{qe:lang_dist}
    d({\mathfrak{L}}_1,{\mathfrak{L}}_2) = \max\{c({\mathfrak{L}}_1, {\mathfrak{L}}_2), c({\mathfrak{L}}_2, {\mathfrak{L}}_1)\}.$$

In case of a finite conformance distance $n$ that is reached as a limit of a sequence of distances, we can denote it as $n^{+}$ (for a limit from above) or as $n^{-}$ (for a limit from below). Thus, ${\mathfrak{L}}_1 \subseteq {\mathfrak{L}}_2$ if and only if $c({\mathfrak{L}}_1,{\mathfrak{L}}_2) = 0$. But when $c({\mathfrak{L}}_1,{\mathfrak{L}}_2) = 0^{+}$ then ${\mathfrak{L}}_1 \nsubseteq {\mathfrak{L}}_2$ but ${\mathfrak{L}}_1 \subseteq \overline{{\mathfrak{L}}_2}$, where $\overline{{\mathfrak{L}}_2}$ is the closure of ${\mathfrak{L}}_2$ in the Euclidean topology, defined as follows. Fixing an untimed word $w \in \Sigma^*$ of length $n$, let ${\mathfrak{L}}_2(w)$ be the timed traces in ${\mathfrak{L}}_2$ whose untimed word is $w$ and let ${\mathbb{R}}^n_w$ be a copy of ${\mathbb{R}}^n$ indexed by $w$. There is a natural embedding $\iota: {\mathfrak{L}}_2(w) \to {\mathbb{R}}^n_w$. Then, $c({\mathfrak{L}}_1, {\mathfrak{L}}_2)=0^{+}$ implies that $\iota({\mathfrak{L}}_1) \subseteq \overline{\iota({\mathfrak{L}}_2)}$, where $\iota({\mathfrak{L}}_j) = \bigcup_{w \in \Sigma^*}\iota({\mathfrak{L}}_j(w))$, $j=1,2$, and $\overline{S}$ is the closure of $S$ in the Euclidean topology.

Subadditivity (triangle inequality) holds for the conformance distance: $$c({\mathfrak{L}}_1, {\mathfrak{L}}_3) \leq c({\mathfrak{L}}_1, 