$|\alpha \subset z| = \inf_{\beta \in dom(\alpha)}(\alpha(\beta) \implies |\beta \in z|) = \inf_{\beta \in dom(\alpha)}( |\beta \in z| \implies |\beta \in z|) = 1 $

• Vamos tomar agora uma nome $\beta$ qualquer:

$$ \begin{array}{ll} |\beta \in \dot{x}(t) \wedge \beta \in z| &= \sup_t \dot{x}(t)|t=\beta||\beta \in z| \\ \\ &= \sup_t \dot{x}(t)|t=\beta||\beta \in z||t=\beta| \\ \\ &\leq \sup_t \dot{x}(t)|\beta \in z||t=\beta| \\ \\ &\leq \sup_t |\beta \in z||t=\beta| \\ \\ &\leq \sup_t \alpha(t)|t=\beta| \\ \\ &= \alpha(\beta) = |\beta \in \alpha| \end{array} $$

• Assim $|(\dot{x}\cap z)\subset \alpha| = 1$, pois tomamos qualquer $\beta \in \dot{x},z$ e $\beta \in \alpha$.
• Dessa forma:
  • $|z \subset \dot{x}| = |z \subset \dot{x}||(\dot{x}\cap z) \subset \alpha| \leq |z \subset \alpha|$
  • Com isso temos que $|z\subset \dot{x} \Longrightarrow z = \alpha|=1$ $\square$