Proof that the set $S = \{(x, y)\in \mathbb{R}^2 \mid y=x^2\}$ is closed. Our professor defined that a closed set is a set whose complement is open.
Based on this definition, how do I prove that the set $S = \{(x, y)\in \mathbb{R}^2 \mid y=x^2\}$ is closed?
It makes intuitive sense to me, but I'm unable to pen down a proof.
Thanks in advance!
 A: As @peek-a-boo suggested, if you consider $f(x,y)=y-x^2$ which is a continuous function then the pre-image of a closed set is also a closed set. Because $\{0\}$ is a closed set in $\mathbb{R}$ then $f^{-1}(0)=\{(x,y)\in\mathbb{R}^2| f(x,y)=0\}$ is also a closed set.
An alternative way could be by considering $z=(x,y)\in S^{C}$ and so $D(z,r)\subset S^{C}$ where $r=\frac{\text{inf}\{d(z,w)| w\in S\}}{2}$. As $S^{C}$ is an open set then $S$ is a closed set.
A: This is the sketch of the proof, you must write down the details:
Take any point $(x_0,y_0)$ in the plane not residing on the curve $y=x^2$. Find the shortest distance from $(x_0,y_0)$ to the curve $y=x^2$; let $r$ denote this shortest distance. Now the ball centered at $(x_0,y_0)$ with radius $\frac{r}{2}$ does not intersect the curve; i.e. it is a subset of the complement of the curve. This proves that the complement of the curve is open.
A: I give a solution using sequence characterization:
$A\subseteq \mathbb{R}^n$ is closed if and only if for each sequence $(x_n)\subseteq A$ such that $x_n \to x$ then $x \in A$.
Let $z_n=(x_n,y_n)$ sequence in $S$ such that $z_n \to (x,y)$. So $y_n=x_n^2, \forall n\in \mathbb{N}$.
How $z_n \to (x,y)$ we have $x_n \to x$ and $y_n \to y$, then
$$ y = \lim y_n = \lim x_n^2 = x^2 $$
so $(x,y)\in A$.
