Why is a line a closed subset of $\mathbb R^2$? I'm studying topology and I have a doubt in the following exercise. I'd appreciate some help. Let $m\neq0$ and $c$ be real numbers. Prove that the line $L=\{\langle x,y\rangle:y=mx+c\}$ is a closed subset of $\mathbb R^2$. I found similar questions here, but with answers involving continuity of functions. This exercise is in chapter $2$ of Topology without tears, in which continuity is presented in the fifth. Then, my attempt was to prove that $\mathbb R^2 \backslash L$ is open by setting, at each point $p\in L$, two open rectangles with a vertex at $p$: one above $L$ and the other below. As $\mathbb R^2 \backslash L$ is the union of these rectangles $(?)$, and every one of them is a open subset of $\mathbb R^2$, we have done. My guess is right? Thank you!
 A: Using the idea of complements (the complement of a closed set in $\mathbb{R}^2$ is an open set in $\mathbb{R}^2$), you could express $\mathbb{R}^2-L$ as a union of two open sets. I would do this as follows:
Define the complement set $L^c$ as the disjoint union $L^c = G_1 \cup G_2$, where $G_1 := \{(x,y) : x < y \}$ and $G_2 := \{(x,y) : x > y \}$. If we select a point $(x_0,y_0)$ that is not in $L$, then it must belong to either $G_1$ or $G_2$. Suppose WLOG that it belongs to $G_1$. Then you must find a number $r>0$ such that the open ball $B_{(x_0,y_0)}(r) \subset G_1$. For this purpose, you can just define $r$ to be the shortest distance (infimum) between the point and the line. I hope you can continue...
A: Hint:
Definition 1:  a set $S \subseteq \mathbb{R}^n$ is closed if it contains all its acumulation points. 
Definition 2: if every $n$-ball $B(x)$ contains at least one point  of $S$ distinct from $x$, then $x$ is an accumulation point of $S$.
A: First, note that any translate of an open set is open, and likewise for closed sets, therefore we may wlog assume that $ c = 0 $. In that case, the given set is the kernel of the continuous map $ \mathbb R^2 \to \mathbb R $ given by $ (x, y) \to y - mx $, and thus it is closed, being the preimage of a point in $ \mathbb R $ under a continuous map.
A: With sequences: let $ ((x_n,y_n))$ be a convergent sequence in $L$ with limit $(x_0,y_0)$. Then $x_n \to x_0$ and $y_n \to y_0$.
From
$$y_n=mx_n+c$$
for all $n$ we deduce that $y_0=mx_0+c$. This gives: $(x_0,y_0) \in L$
A: If you know that the real line is a closed subspace of $\Bbb{R}^2$ (indeed, it's a complete metric subspace, which is even stronger) the result follows easily because the real line can be mapped to any other line in $\Bbb{R}^2$ by a composition of a rotation and a translation. Since rotations and translations are isometries (i.e. homeomorphisms) of $\Bbb{R}^2$, they preserve closed sets.
