PomeLemma 19.7.28.323C6B6B61D0 Let $S,S'\in \mathcal{S}'(C)$. Assume that $d_{\mathrm{sg}}(S,S')=0$. Then $\Delta (S)=\Delta (S')$.
Lemma 19.7.28323C6B6B61D0
Proof. Observe that $\operatorname{vol}(S\cap S')>0$, as otherwise
\[ d_{\mathrm{sg}}(S,S')\geq \operatorname{vol}S+\operatorname{vol}S'>0, \]
which is a contradiction.
It follows from 19.7.25 that $S\cap S'\in \mathcal{S}'(C)$. It suffices to show that $\Delta (S)=\Delta (S\cap S')$. In fact, suppose that this holds, since $\operatorname{vol}\Delta (S')=\operatorname{vol}S'=\operatorname{vol}S=\operatorname{vol}\Delta (S)$, the inclusion $\Delta (S')\supseteq \Delta (S\cap S')=\Delta (S)$ is an equality.
After replacing $S'$ by $S\cap S'$, we still have $d_{\mathrm{sg}}(S,S')=0$ by 19.7.19, and now $S'\subseteq S$. Hence corollary 19.2.8 gives
\[ 0=d_{\mathrm{sg}}(S,S')=\operatorname{vol}S-\operatorname{vol}S', \]
so $\operatorname{vol}S=\operatorname{vol}S'$. Therefore
\[ \operatorname{vol}\Delta (S)=\operatorname{vol}\Delta (S')>0 \]
by (19.2.6). Therefore, they are equal. $\square$
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