To show $(0,1)$ is open in LL topology note the following
$$
x \in [x,1) \subset (0,1) ~~\text{for all $x \in (0,1)$}.
$$
Hence $(0,1)$ is open in LL topology.
To show $(0,1)$ not closed in LL topology, we shall show that closure of $(0,1)$ in LL topology is not $(0,1)$.
Take any neighborhood $N$ of $0$. There exists $a>0$ such that $0 \in [0,a) \subset N$
Hence $N \cap (0,1)$ is not empty.
Hence $0 \in \overline{(0,1)}$. But $0 \notin (0,1)$.
Hence $(0,1)$ is not closed in LL topology.
Hope this helps.
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Sets of the form $[a,x)$ are open in the lower-limit topology by definition (at least in mine). You can show that they are not open in the metric topology as there is no neighborhood about $a$ such that $(a-\epsilon,a+\epsilon)\subseteq [a,x)$ for any $\epsilon>0$.
To show that the topology is not discrete, you can show that singletons are not open. And this is true because there is no $\epsilon>0$ such that $[a,a+\epsilon)\subseteq\{a\}$.
Best Answer
Suppose $f$ is right continuous at $p$. This means (by my definition) that for each $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $x$ with $p < x < p + \delta$ we have that $|f(x) - f(p)| < \varepsilon$.
This implies, in set theory terms, that $f[[p,p+\delta)] \subseteq B_{\varepsilon}(f(p))$, or: for every open basic neighbourhood $U$ of $f(p)$ we have a basic open neighbourhood $V$ of $p$ (in the lower limit topology) such that $f[V] \subseteq U$. So $f: (\mathbb{R},\mathcal{T}_l) \rightarrow (\mathbb{R}, \mathcal{T}_e)$ from the lower limit topology to the Euclidean topology is continuous at $p$.