Union of lines that passes through some point of a closed set is closed in $R^n\setminus\{0\}$

Question

I have the following problem: Let $U,F\subset \mathbb{R}^n\setminus\{0\}$ open and closed set of $\mathbb{R}^{n}\setminus\{0\}$, respectively. I already proved that $\cup_{\lambda\in\mathbb{R}\setminus\{0\}}\lambda U$ is open, where $\lambda U=\{\lambda\cdot x:x\in U\}$, but I have not been able to prove that $\cup_{\lambda\in\mathbb{R}\setminus\{0\}}\lambda F$ is closed, may it is not, but I don't find a counterexample.

Answer 1

Let $\widehat F=\bigcup_{\lambda\in\Bbb R\setminus\{0\}}\lambda F$. It’s not hard to show that $\widehat F$ is closed if $F$ is bounded. (If $p\in\operatorname{cl}\widehat F$, there are sequences $\langle\lambda_k:k\in\Bbb N\rangle$ in $\Bbb R\setminus\{0\}$ and $\langle x_k:k\in\Bbb N\rangle$ in $F$ such that $\langle\lambda_kx_k:k\in\Bbb N\rangle$ converges to $p$. By passing to a subsequence if necessary we may assume that $\langle x_k:k\in\Bbb N\rangle$ converges to some $x\in F$, and from there it’s straightforward to check that $p$ is on the line through $x$.) Thus, we should look at unbounded closed sets if we’re looking for a counterexample, and we should make use of their unboundedness.

Try $F=\left\{\left\langle n,\frac1n\right\rangle:n\in\Bbb Z^+\right\}$; it doesn’t matter whether you interpret the coordinates as rectangular or polar coordinates, though it’s especially easy to see what’s going on if you think in polar coordinates.

Answer 2

Let $F=\{(x,\frac1x)\in\mathbb{R}^2\backslash\{0\}|\,\,x>0\}$. $F$ is closed. However $\cup_{\lambda\in\mathbb{R}\setminus\{0\}}\lambda F$ contains the positive quadrant, but not the axes, so is not closed.