Rays in $\mathbb{R}^n$ are closed

You can prove directly that $R_{x,y}$ is closed by showing that the limit of all converging sequence $\{x_n\in R_{x,y}\}_{n=1}^\infty$ is in $R_{x,y}$ itself.

Now here's the proof. If $\{x_n\in R_{x,y}\}_{n=1}^\infty$ is a converging sequence, there exists $\{t_n\geq0\}_{n=1}^\infty$ such that $\{x + t_n(y-x)\}_{n=1}^\infty$ is a converging sequence, which implies that $\{t_n\geq0\}_{n=1}^\infty$ is a converging sequence. Since $\newcommand{\reals}{{\mathbf R}}\{t\in\reals\|t\geq0\}$ is closed, $\lim_{n\to \infty} t_n \geq0$. Thus \begin{equation} \lim_{n\to\infty} x_n = \lim_{n\to\infty} (x + t_n(y-x)) = x + \left(\lim_{n\to\infty} t_n\right)(y-x) \in R_{x,y}, \end{equation} hence $R_{x,y}$ is closed!