Rays in $\mathbb{R}^n$ are closed A ray in $\mathbb{R}^n$ is a subset of the form $\{ x + t(y - x) \mid t \geq 0 \}$ for some distinct $x,y \in \mathbb{R}^n$. 
Bourbaki's "General Topology" claim that such a ray is closed, but provide no proof.
I can't find a proof myself. Let $R_{x,y} = \{ x + t(y - x) \mid t \geq 0 \}$. To prove that $R_{x,y}$ is closed, we need to prove that $\mathbb{R}^n\setminus R_{x,y}$ open, that is, that for every $z \in \mathbb{R}^n\setminus R_{x,y}$ there is $\epsilon > 0$ such that $B_{\mathbb{R}^n}(z,\epsilon) \subseteq \mathbb{R}^n\setminus R_{x,y}$. 
$\mathbb{R}^n\setminus R_{x,y}$ seems to consist of points of the form $x + t(y - x)$ where $t < 0$ and those who can't be written in the form $x + t(y - x)$ at all.
 A: 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!
A: For any fixed vectors $x,z \neq 0$ we have that $L(x,z):=\{x + tz: t \in \Bbb R\}$ is homeomorphic to $\Bbb R$ via the map $h:t \to x + tz$. It's clearly a bijection and continuous as vector operations are continuous. $L(x,z)$ is itself closed in $\Bbb R^n$ too (as all translations are homeomorphisms of $\Bbb R^n$ and linear subspaces are closed etc.).
The ray is just $h[[0,+\infty)]$ which is closed as $h$ is closed onto $L(x,z)$ and the latter set is closed in $\Bbb R^n$.
A: Same idea as Sunghee Yun's answer.
Let $z_n \in S $ (ray) $\rightarrow z$   be a convergent sequence.
Need to show that  $z \in S$.
$z_n=x+t_n(y-x)$ , $x,y \in \mathbb{R^n}$ fixed, $t_n \ge 0$.
Since $z_n$ is Cauchy:
$\epsilon >0$ given; There is a $n_0$ s.t. for 
$m \ge n \ge n_0$
$||z_m-z_n||=||(t_m-t_n)(x-y)|| =$
$|t_m-t_n|$ $ ||(x-y)|| < \epsilon$.
$||(x-y)|| =c >0$ ( assuming $x\not =y$, else no ray).
This implies  the sequence  $t_n \ge 0$ is Cauchy in $\mathbb{R_0^+}$, and has a limit $t\ge 0$.
Hence
$\lim_{n \rightarrow \infty}z_n= z$, with
$z=(x +t(y-x)) \in S$.
Remark: 
$||z-z_n|| =|t-t_n|c \rightarrow 0$.
