1. **Problem statement:** We want to find a closed set $E \subseteq \mathbb{R}^2$ such that the projection onto the second coordinate $\pi_2(E)$ is closed, but the projection onto the first coordinate $\pi_1(E)$ is not closed. 2. **Recall definitions:** - The projection maps are $\pi_1(x,y) = x$ and $\pi_2(x,y) = y$. - A set is closed if it contains all its limit points. 3. **Key idea:** - The projection $\pi_2(E)$ is the set of all $y$ such that $(x,y) \in E$ for some $x$. - Similarly for $\pi_1(E)$. - We want $\pi_2(E)$ to be closed but $\pi_1(E)$ not closed. 4. **Constructing $E$:** - Consider the set $E = \{(x, \sin(1/x)) : x \in (0,1] \} \cup \{(0,0)\}$. - This set is the graph of $y = \sin(1/x)$ for $x$ in $(0,1]$ plus the point $(0,0)$. 5. **Check if $E$ is closed:** - The graph of $\sin(1/x)$ is continuous on $(0,1]$. - Adding the point $(0,0)$ closes the set because $\lim_{x \to 0} \sin(1/x)$ does not exist, but the point $(0,0)$ is included. - The closure of the graph includes $(0,y)$ for all $y$ in $[-1,1]$, but since we only add $(0,0)$, $E$ is closed in $\mathbb{R}^2$. 6. **Projection $\pi_2(E)$:** - $\pi_2(E) = \{\sin(1/x) : x \in (0,1]\} \cup \{0\}$. - The set $\{\sin(1/x) : x \in (0,1]\}$ is dense in $[-1,1]$. - However, since $E$ only contains $(0,0)$ at $x=0$, $\pi_2(E)$ is not the entire interval $[-1,1]$, but it contains $0$. - The closure of $\pi_2(E)$ is $[-1,1]$, but $\pi_2(E)$ itself is not closed. 7. **Projection $\pi_1(E)$:** - $\pi_1(E) = (0,1] \cup \{0\} = [0,1]$. - This is a closed interval. 8. **We want the opposite:** - So switch the roles: define $E = \{(x,0) : x \in (0,1)\} \cup \{(0,y) : y \in [0,1]\}$. 9. **Check $E$ closed:** - $E$ is the union of the vertical segment at $x=0$ from $y=0$ to $y=1$ and the horizontal segment along $y=0$ for $x$ in $(0,1)$. - The set $E$ is closed because it contains all its limit points. 10. **Projection $\pi_2(E)$:** - $\pi_2(E) = \{0\} \cup [0,1] = [0,1]$, which is closed. 11. **Projection $\pi_1(E)$:** - $\pi_1(E) = (0,1) \cup \{0\} = [0,1)$, which is not closed because it does not contain $1$. 12. **Final answer:** - The set $E = \{(x,0) : x \in (0,1)\} \cup \{(0,y) : y \in [0,1]\}$ is closed in $\mathbb{R}^2$. - $\pi_2(E) = [0,1]$ is closed. - $\pi_1(E) = [0,1)$ is not closed. Thus, this $E$ satisfies the problem conditions.