Proving that $L = \{(x,y)\in\mathbf{R}^2:y=mx+c\}$ is closed in $\mathbf{R}^2$. I am required to prove that the set $L = \{(x,y)\in\mathbf{R}^2:y=mx+c\}$ where $m,c\in\mathbf{R}$ is closed in $\mathbf{R}^2$.
I have tried to show that $\mathbf{R}\backslash L = \bigcup_{(a,b)\in\mathbf{R}\backslash L}D_{(a,b)}$ where $D_{(a,b)}$ is a disk having a radius dependent on the proximity of $(a,b)$ to $L$, but as you can see here https://www.desmos.com/calculator/dldadrhrdl the task proving a bit cumbersome. What is the best way to go about proving this?
 A: Perhaps that the shortest way consists in noticing that $L=f^{-1}\bigl(\{c\}\bigr)$, where $f(x,y)=y-mx$. Since $f$ is continuous and $L$ is closed, $L$ is then closed.

You can also easily prove that that set $M=\{(x,0)\,|\,x\in\mathbb{R}\}$ is closed. Now, consider the map$$\begin{array}{rccc}\varphi\colon&\mathbb{R}^2&\longrightarrow&\mathbb{R}^2\\&(x,y)&\mapsto&(x,y+mx+c).\end{array}$$It is clearly a homeomorphism (unless $m=0$) and $L=\varphi(M)$. Since $M$ is closed, $L$ is closed too.

Still another possibility consists in proving that $\mathbb{R}^2\setminus L$ is open. If $(x,y)\in\mathbb{R}^2\setminus L$, the distance from $(x,y)$ to $L$ is equal to $\frac{\lvert y-mx-c\rvert }{\sqrt{m^2+1}}$. Therefore, the open ball centered at $(x,y)$ with radius $\frac{\lvert y-mx-c\rvert }{\sqrt{m^2+1}}$ is contained in $\mathbb{R}^2\setminus L$.
A: hint
You can prove that each convergent sequence of elements in $L$, has its limit in $L$.
A: Consider $f(x,y) = y - mx$,and observe that $L = f^{-1}(\{c\})$ is closed since $f$ is continuous and $\{c\}$ is closed in $\mathbb{R}$.
A: The function $\Bbb{R}^{2}\to \Bbb{R}$ defined by $(x,y)\mapsto y-mx-c$ is continuous. Thus $L=f^{-1}(0)$ is closed.
A: The sets $L^1={0}\times\mathbb{R}$ and  $L^2=\mathbb{R}\times{0}$ are closed in $\mathbb{R}^2$ but there exists some omeomorphism to $\mathbb{R}^2$ from $\mathbb{R}^2$ that maps each $L$ to $L^1$ or $L^2$
