In Munkres Chapter 2.22 Example 1, he provided the following example of a continuous map that is closed but not open:
Let $X$ be the subspace $[0, 1] \cup [2, 3]$ of $\mathbb{R}$, and let $Y$ be the subspace $[0, 2]$ of $\mathbb{R}$. The map $\rho: X \rightarrow Y$ defined by $$ \rho (x) = \begin{cases} x & x \in [0, 1]\\ x - 1 & x \in [2, 3] \end{cases} $$ is continuous, closed but not open.
I have the following adapted example of a continuous map that is open but not closed, and would like to check if my ideas are correct:
Consider the subspace $X = (0, 1) \cup (2, 3)$ of $\mathbb{R}$, and the subspace $Y = (0, 1)$ of $\mathbb{R}$. Let $\rho: X \rightarrow Y$ be the map defined by: $$ \rho(x) = \begin{cases} x & x \in (0, 1)\\ \frac{x - 2}{2} & x \in (2, 3) \end{cases} $$
Then, $\rho$ is continuous, open, but not closed.
- $\rho$ is a continuous map: Consider $\rho_1: (0, 1) \rightarrow (0, 1)$ defined by $x \mapsto x$. This is the identity map and is continuous. Similarly, consider $\rho: (2, 3) \rightarrow (0, 1)$ defined by $x \mapsto \frac{x - 2}{2}$. This is the composition of continuous maps (scaling and transformations), thus is continuous. By the pasting lemma, $\rho$ is continuous.
- $\rho$ is an open map: Take any open set $U \subseteq X$. Then, $U = U_1 \cup U_2$ where $U_1 = U \cap (0, 1)$ and $U_2 = U \cap (2, 3)$. $U_1$ and $U_2$ are open in $(0, 1)$ and $(2, 3)$ respectively with respect to the respective subspace topologies. Then:
$\rho \vert_{(0, 1)} (U_1) = U_1 \subseteq (0, 1)$ is open in $(0, 1)$
$\rho \vert_{(2, 3)} (U_2) = (\rho \vert^{-1}_{(2, 3)} (U_2))^{-1}$ is open in $(0, \frac{1}{2})$ (because $\rho \vert_{(2, 3)}$ is continuous with a continuous inverse). Moreover, $\rho \vert_{(2, 3)} (U_2)$ is open in $(0, 1)$ because $(0, \frac{1}{2})$ is open in $(0, 1)$ with respect to the subspace topology.
So, $\rho (U) = \rho \vert_{(0, 1)} (U_1) \cup \rho \vert_{(2, 3)} (U_2) \subseteq (0, 1)$ is open. Thus, $\rho$ is an open map.
- $\rho$ is not a closed map: The set $(2, 3)$ is closed in $X$ because $(2, 3) = [\frac{3}{2}, 3] \cap X$. But, $\rho((2, 3)) = (0, \frac{1}{2})$, which is not closed in $Y$.
Thank you.