A general question about continuity is that why it can not be defined as "Mapping open sets to open sets". The discussion can be seen in Why aren’t continuous functions defined the other way around? or Continuity vs. Mapping open sets to open sets?
A typical counterexample is to show that: a continuous function can map an open set to non-open set. E.g., the continuous mapping $f(x)=x^2$ on $\mathbb R \rightarrow \mathbb R$, which send the open set $(-1,1)$ to the non-open set $[0,1)$.
In Charles Nash's book, Topology and Geometry for Physicists (page 11), there is another type of counterexamples – a discontinuous function can always map an open set to open set. However, the example itself confuse me.
The example is the discontinuous function:
$$f\left(x\right)=\begin{cases}
x, & x\leq0\\
1+x, & x>0
\end{cases}$$
In the book, it is said "$f$ always maps open sets into open sets" and gives no enough details. But if we choose $x\in\left(-\epsilon,\epsilon\right)$, then $f\left(x\right)\in\left(-\epsilon,0\right]\cup\left(1,1+\epsilon\right)$, which is not an open set by the usual topology.
To make sense, should we use other topology here?
Best Answer
You are correct in that if we take $f: \mathbb{R} \to \mathbb{R}$ with $\mathbb{R}$ in the usual topology, then $f$ is not an open map.
However, an open map is obtained if the range is restricted to the image of the domain, $Y = (-\infty, 0] \cup (1, \infty)$, and take $g: \mathbb{R} \mapsto Y$ and $Y$ in the subspace topology. Since then, $g ( (-\epsilon, \epsilon) ) = (-\epsilon, 0] \cup (1, 1 + \epsilon) = Y \cap (-\epsilon, 1 + \epsilon)$. And, $g$ is not continuous because the inverse image of the set $(-1, 0] = Y \cap (-1, 1)$, open in $Y$, is $(-1, 0]$, which is not open in $\mathbb{R}$.