5

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.

oliverjones
  • 4,199
  • 3
    sounds correct. – Mustafa Apr 20 '16 at 17:12
  • 4
    Of course if you define $f(x,y) = ax + by + c$, which is continuous, then $L = f^{-1}[{0}]$ which is the inverse image of a closed set under a continuous function (and a singleton is a closed set). – Henno Brandsma Apr 20 '16 at 17:14
  • 1
    I think you are confusing points $(x,y)$ and $(w,z)$. The fraction you mention is the distance from $(x,y)$ to $L$, which is $0$. – Wojowu Apr 20 '16 at 17:14

2 Answers2

8

The line defined by the equation $ax+by+c=0$ is the inverse image of the closed subset $0$ in $\mathbf R$, by the (continuous) map $\begin{aligned}[t]\mathbf R^2&\longrightarrow \mathbf R,\\ (x,y)&\longmapsto ax+by+c.\end{aligned}$

Bernard
  • 175,478
4

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.

Surb
  • 55,662