If $x:[0,\infty]\to E^\ast$ is a function into a topological space and $x^\ast([0,\zeta))⊆E$, how can we characterize left-regularity at $\zeta$?

Question

Let

• $(E^\ast,\tau^\ast)$ be a topological space and $$\mathcal N_{\tau^\ast}(x^\ast):=\left\{N^\ast\subseteq E^\ast:N^\ast\text{ is a neighborhood of }x^\ast\right\}\;\;\;\text{for }x^\ast\in E^\ast;$$
• $E\in\tau^\ast$ and $\tau:=\left.\tau^\ast\right|_E$;
• $x^\ast:[0,\infty]\to E^\ast$;
• $\zeta\in[0,\infty]$ with $x^\ast([0,\zeta))\subseteq E$.

Are we able to show that $x^\ast$ has a left-limit at $\zeta$ iff $$\exists x^\ast(\zeta-)\in E:\forall N\in\mathcal N_{\tau}(x^\ast(\zeta-)):\exists\delta>0:x^\ast((\zeta-\delta,\zeta))\subseteq N?\tag1$$

We may clearly exclude the trivial cases $\zeta=0$ and $\zeta=\infty$. So, assume $\zeta\in(0,\infty)$ and $x^\ast$ has a left-limit at $\zeta$. Then, by definition, there is a $x^\ast(\zeta-)\in E^\ast$ with $$\forall N^\ast\in\mathcal N_{\tau^\ast}(x^\ast(\zeta-)):\exists\delta>0:x^\ast((\zeta-\delta,\zeta))\subseteq N^\ast\tag2.$$ Now, $$\tau=\left\{N^\ast\cap E:N^ast\in\tau^\ast\right\}.\tag3.$$ But how can we conclude? Especially, how can we conclude that $x^\ast(\zeta-)\in E$?

And is it enough to assume that $(E,\tau)$ is Hausdorff in order to conclude that $x^\ast(\zeta-)$ is unique?

Answer 1

Let $E^\ast=\Bbb R$, $x^\ast(t)=t$ and $E=(-\infty,1)$, $\zeta=1$. Then left limit exists and is equal to $1$ but $1\notin E$.

Yes, it's enough to assume that the space is Hausdorff to conclude that the limit is unique if it exists.