We know that under the lower limit topology, even though the addition map is still continuous, the negation and the subtraction map are not continuous any more.
However, what about the the continuity of reciprocation map and multiplication map under lower limit topology?
In other words, if we give $\mathbb{R}_{>0}$ the subspace topology coming from the lower limit topology on the real, is the multiplication map continuous? What about the reciprocation map?
I do not have an idea about this question.
Thank you so much for any hints or detailed proof!
Let $X = (0,+\infty)$ have the subspace topology w.r.t. the lower limit topology on $\mathbb{R}$.
Let $i: X \to X, i(x) = \frac1x$ be the inversion map.
Then $i^{-1}[[1,\infty)] = (0,1]$ where $[1,\infty)$ is open in $X$ and $(0,1]$ is not ($1$ is not an interior point). So $i$ is not continuous.
Let $m: X \times X \to X, m(x,y) = xy$ be the multiplication map.
It is well known that $m$ is continuous when we give the set $X$ the usual topology.
We can use this to see that our $m$ is also continuous: consider $(p,q) \in X \times X$ and let $[pq, pq+\varepsilon)$ be a basic open neighbourhood of $m(p,q)$.
By usual-topology continuity we can find $\delta>0$ such that
$$m[(p-\delta,p+\delta) \times (q-\delta, q+\delta)] \subseteq (pq-\varepsilon, pq+\varepsilon)$$
but as multiplication behaves nicely wrt ordering (and all is $>0$) we know that
$$ \forall x,y \in X: x \ge p, y \ge q \implies xy \ge pq$$ so that
$$m[[p,p+\delta) \times [q, q+\delta)] \subseteq [pq, pq+\varepsilon)$$
showing that $m$ is continuous at $(p,q)$, as required.