TITLE: Open map that is not continuous
QUESTION [5 upvotes]: I was trying to find an open map that is not continuous, here was my effort, is it ok?
Consider the map $f:\mathbb{R} \rightarrow \mathbb{Z}$ given by $f(x)=\lfloor x \rfloor$ where $\mathbb{R}$ has the standard topology and $\mathbb{Z}$ has the discrete topology. Then $\{1\}$ is open in $\mathbb{Z}$. However $f^{-1}(\{1\})=[1,2)$, which is not open in $\mathbb{R}$, so $f$ is not continuous. Since $\mathbb{Z}$ has the discrete topology, the image of any open set is necessarily open. Thus, $f$ is an open map.

REPLY [5 votes]: Yes, that example is fine. In fact the identity from $(X, \tau)$ ( a non-discrete space) to $X$ in the discrete topology is always open and non-continuous.