Let
- $(E,\tau)$ be a topological space and $\Delta\not\in E$;
- $E^\ast:=E\cup\{\Delta\}$ and $$\tau^\ast:=\tau\cup\underbrace{\left\{E^\ast\setminus B:B\subseteq E\text{ is }\tau\text{-closed and }\tau\text{-compact}\right\}}_{=:\:\mathcal C}.$$
- $x:[0,\infty)\to E^\ast$ be right-continuous and $$\zeta:=\inf\underbrace{\left\{t\ge0:x(t)=\Delta\right\}}_{=:\:I}.$$
Are we able to show that either
- $I=\emptyset$ and hence $\zeta=\infty$;
- $\zeta\in I$ and hence $x(\zeta)=\Delta$.
I know that if $(M,d)$ is a metric space, $f:[0,\infty)\to M$ is right-continuous, $B\subseteq M$ is $d$-closed and $$\rho:=\inf\underbrace{\left\{t\ge0:f(t)\in B\right\}}_{=:\:J},$$ then either
- $J=\emptyset$ and hence $\rho=\infty$;
- $\rho\in J$ and hence $f(\rho)\in B$.
However, I've got trouble to show the same in the setting described in my question. I'm willing to assume that $\tau$ is metrizable, but I'm not sure whether this helpful at all. (I don't even know if this implies that $\tau^\ast$ is metrizable as well; I guess not?!)
Yes.
Noting that $E$ is open in $E^*$ (which is equivalent to $\{\Delta\}$ being closed in $E^*$), this follows from the more general
Theorem. Let $Z$ be any topological space and $z_0 \in Z$ be a point such that $\{z_0\}$ is closed in $Z$. Moreover, let $x : [0,\infty) \to Z$ be right-continuous, $I = \{ t \ge 0 \mid x(t) = z_0\} = x^{-1}(z_0)$ and $\zeta = \inf I$. Then either $I = \emptyset$ (and hence $\zeta = \infty$) or $\zeta \in I$ (and hence $x(\zeta) = z_0$).
Proof. Clearly either $I = \emptyset$ (which implies $\zeta = \inf \emptyset = \infty$) or $I \ne \emptyset$. Let us show that in the latter case we have $\zeta \in I$. Assume that $\zeta \notin I$. Then $x(\zeta) \in U = Z \setminus \{z_0\}$. Since $U$ is open and $x$ is right-continuous at $\zeta$, there exists $\delta > 0$ such that $x((\zeta,\zeta+\delta)) \subset U$. Thus $(\zeta,\zeta+\delta) \cap I = \emptyset$, hence $\zeta = \inf I \ge\zeta +\delta$ (recall the assumption $\zeta \notin I$) which is a contradiction.