Let $ \tau_S$ be the standard topology in $\mathbb{R}$, and let $ \tau_H$ be a topology generated by intervals $[a, b)$, with $a,b \in \mathbb{R}$. Let $f(x)= \begin{cases} 0 & x< 0 \\x & x\geq 0 \end{cases}$.
We know $f$ is continuous as $f:(\mathbb{R},\tau_S) \rightarrow (\mathbb{R},\tau_S)$, but is it continuous as $f:(\mathbb{R},\tau_S) \rightarrow (\mathbb{R},\tau_H)$?
My initial thought was that the preimage of an open set [𝑎,𝑏) in the second topology is not an open set in the standard topology, so the function is not continuous, is this true?
Best Answer
Yes, effectively. Take the set $[1,2) \in \tau_H$ as a counterexample. It's preimage is $[1,2)\subset\mathbb{R}$, which is not an open set in the standard topology. So the map cannot be continuous.
It seems that you had thought about all of this yourself. I suggest that next time you add all of your thoughts in the body of the question directly. You really didn't need an answer, but a confirmation of what you had already tried.