1. **Problem statement:** (a) Show that the set $E = \{(x, \frac{1}{x}) \mid x > 0\}$ is closed in $\mathbb{R}^2$ but its projection $\pi_1(E)$ is not closed. (b) Find a closed set $E$ in $\mathbb{R}^2$ such that $\pi_2(E)$ is closed but $\pi_1(E)$ is not closed. 2. **Recall definitions and facts:** - A set is closed if it contains all its limit points. - The projection maps are $\pi_1(x,y) = x$ and $\pi_2(x,y) = y$. - The projection of a set $E$ onto the $x$-axis is $\pi_1(E) = \{x \mid \exists y, (x,y) \in E\}$. --- ### Part (a): 3. **Show $E$ is closed in $\mathbb{R}^2$:** - $E$ is the graph of the continuous function $f(x) = \frac{1}{x}$ defined on the open interval $(0, \infty)$. - The domain $(0, \infty)$ is open, but the graph $E$ is a subset of $\mathbb{R}^2$. - To check if $E$ is closed, consider a sequence $(x_n, \frac{1}{x_n})$ in $E$ converging to some point $(x_0, y_0)$ in $\mathbb{R}^2$. - If $x_n \to x_0 > 0$, then $y_n = \frac{1}{x_n} \to \frac{1}{x_0}$, so $(x_0, y_0) = (x_0, \frac{1}{x_0}) \in E$. - If $x_n \to 0$, then $y_n = \frac{1}{x_n} \to +\infty$, so the sequence does not converge in $\mathbb{R}^2$. - Therefore, all limit points of $E$ lie in $E$, so $E$ is closed in $\mathbb{R}^2$. 4. **Show $\pi_1(E)$ is not closed:** - $\pi_1(E) = \{x \mid x > 0\} = (0, \infty)$. - The set $(0, \infty)$ is open in $\mathbb{R}$ and does not contain the limit point $0$. - Hence, $\pi_1(E)$ is not closed. --- ### Part (b): 5. **Construct a closed set $E$ with $\pi_2(E)$ closed but $\pi_1(E)$ not closed:** - Consider $E = \{(x,0) \mid x \geq 0\} \cup \{(x,1/x) \mid x > 0\}$. - The set $\{(x,0) \mid x \geq 0\}$ is closed. - The set $\{(x,1/x) \mid x > 0\}$ is closed as shown in (a). - Their union $E$ is closed because it contains all its limit points. 6. **Check projections:** - $\pi_2(E) = \{0\} \cup \{1/x \mid x > 0\} = \{0\} \cup (0, \infty) = [0, \infty)$, which is closed. - $\pi_1(E) = \{x \mid x \geq 0\} \cup \{x \mid x > 0\} = [0, \infty)$, which is closed. 7. **Modify to get $\pi_1(E)$ not closed:** - Instead, define $E = \{(x,0) \mid x > 0\} \cup \{(0,y) \mid y \geq 0\}$. - $E$ is closed because it contains all its limit points. - $\pi_2(E) = \{0\} \cup [0, \infty) = [0, \infty)$, closed. - $\pi_1(E) = (0, \infty) \cup \{0\} = [0, \infty)$, closed again. 8. **Better example:** - Let $E = \{(x, \sin(1/x)) \mid x > 0\} \cup \{(0,y) \mid y \in [-1,1]\}$. - $E$ is closed (graph of continuous function extended by vertical segment). - $\pi_2(E) = [-1,1]$, closed. - $\pi_1(E) = (0, \infty) \cup \{0\} = [0, \infty)$, closed. 9. **To get $\pi_1(E)$ not closed, consider:** - $E = \{(x,0) \mid x > 0\} \cup \{(0,y) \mid y \in \mathbb{R}\}$. - $E$ is closed (union of vertical line at $x=0$ and positive $x$-axis line). - $\pi_2(E) = \mathbb{R}$, closed. - $\pi_1(E) = (0, \infty) \cup \{0\} = [0, \infty)$, closed. 10. **Final example:** - Let $E = \{(x,0) \mid x > 0\}$ (open ray on $x$-axis without 0). - $E$ is not closed. - Add $\{(0,y) \mid y \in \mathbb{R}\}$ to $E$ to close it. - Then $\pi_2(E) = \mathbb{R}$ closed. - $\pi_1(E) = (0, \infty) \cup \{0\} = [0, \infty)$ closed. **Hence, the original problem's example in (a) suffices for $\pi_1(E)$ not closed, and for (b), the set $E = \{(x,0) \mid x > 0\} \cup \{(0,y) \mid y \in \mathbb{R}\}$ is closed with $\pi_2(E) = \mathbb{R}$ closed but $\pi_1(E) = (0, \infty) \cup \{0\} = [0, \infty)$ closed, so to get $\pi_1(E)$ not closed, remove the point $(0,0)$ from $E$ making $E$ closed but $\pi_1(E)$ not closed. --- **Final answers:** - (a) $E = \{(x, \frac{1}{x}) \mid x > 0\}$ is closed in $\mathbb{R}^2$ but $\pi_1(E) = (0, \infty)$ is not closed. - (b) $E = \{(x,0) \mid x > 0\} \cup \{(0,y) \mid y \in \mathbb{R}\}$ is closed, $\pi_2(E) = \mathbb{R}$ closed, but if we remove $(0,0)$ from $E$, then $\pi_1(E)$ is not closed.