TITLE: Show that any line in $\Bbb R^2$ is closed
QUESTION [5 upvotes]: Show that any line in $\Bbb R^2$ is closed under the usual topology
I think showing the complement is open is best way. 
Let $ax+by+c=0$ be a line in $\Bbb R^2$ as a line it is the set of $(x,y) \in \Bbb R^2 $ such that it satisfies the above equation for constants $a,b,c$ for simplicity call the set $L$ 
Take $(x,y) \in L^c$, the distance from $(x,y)$ to $L$ is $\displaystyle \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$
Let $\epsilon = \frac{\mid ax+by+c \mid}{\sqrt{a^2+b^2}}$ then take as $U$ the ball $((x,y), \epsilon)$ then $U \cap L = \emptyset$ so $L^c$ is open.

REPLY [4 votes]: More generally
Any graph of a continuous function is closed. Indeed, let $$\Gamma=\{(x,f(x))\mid x\in \mathbb R\},$$
the graph of a continuous function $f:\mathbb R\longrightarrow \mathbb R$. Let $\{(x_n,f(x_n))\}_n$ a sequence of $\Gamma$ that converge. Let denote $(a,b)$ it's limit. Then, by continuity of $f$, $$b=\lim_{n\to \infty }f(x_n)=f(a)$$
and thus, $(a,b)\in\Gamma $. Therefore, $\Gamma$ is closed.