Closed subset of $\mathbb R^2$ The exercise is as follows:
Show that the set of points in $\mathbb R^2 $ of the form $(x, \sqrt{x})$
(for $x$ a nonnegative real number) is a closed subset of $\mathbb R^2$.
I am new to topology and I'm getting familiar with the definitions of opened and closed sets,open balls and such, but I just don't know how to get around this exercise.
Let's name the set $S$ to be defined as $\{(x,\sqrt{x}): x \ge 0\}$. Should I take the complement of $S$ (and if so, how can I?) and find an $\varepsilon$-ball not in $\mathbb R^2\setminus S$ or is there another way around it.
Thank you in prior.

There are multiple ways to show that a subset is closed. In case of metric spaces there is a useful characterization:

A subset $A$ of a metric space $X$ is closed if and only if for any convergent sequence $(a_n)\subseteq X$ if $a_n\in A$ for all $n$ then $\lim a_n\in A$.

So let's give it a try. Let $(v_n)$ be a convergent sequence in $S$. So $v_n=(x_n, \sqrt{x_n})$ by the definiton of $S$. Since $v_n$ is convergent then so is $x_n$. Now since $\sqrt{\cdot}$ is a continuous function then $\sqrt{x_n}$ is convergent as well and $\lim\sqrt{x_n}=\sqrt{\lim x_n}$. It follows that
$$\lim v_n=\lim (x_n,\sqrt{x_n})=(\lim x_n, \lim\sqrt{x_n})=$$
$$=(\lim x_n,\sqrt{\lim x_n})$$
so $\lim v_n$ is again of the form $(a,\sqrt{a})$ and so $\lim v_n\in S$ proving that $S$ is closed.
