Is this set closed? (Finding the limit points of a set) 
Prove that this set is closed:
$$ \left\{ \left( (x, y) \right) : \Re^2 : \sin(x^2 + 4xy) = x + \cos y \right\} \in (\Re^2, d_{\Re^2}) $$

I've missed a few days in class, and have apparently missed some very important definitions if they existed.  I know that a closed set is a set which contains its limit points (or, equivalently, contains its boundary), but I have no idea how to calculate the limit points of an arbitrary set like this.  The only intuition I have to that end is to fix either $x$ or $y$ and do calculations from there, since doing it in parallel often ends up in disaster.  Besides this, I don't know how to approach this problem.
If there is any other extant definition of closed-ness in a metric space, I welcome them wholeheartedly.
 A: *

*Prove that the function $f(x,y) = \sin(x^2 + 4xy) - x\cos y $ is continuous on all of $\Bbb{R}^2$.

*The point $0 \in \Bbb{R}$ is closed. (Why?)

*Continuous functions take open sets to open sets and so take closed sets to closed sets, because taking the complement of a set commutes with taking the 
inverse image.

*Conclude.
A: Your set is the preimage of the closed set $\{0\}$ by the continuous function $\Bbb R^2\ni(x,y)\mapsto\sin (x^2+4xy)- x\cos y$, hence it's closed. 
Let $f:X\to Y$ a continuous function (between two metric spaces for example) and $F\subset Y$ a closed set, $G=f^{-1}(F)$. Here are two ways of proving that $G$ is a closed set of $X$.


*

*The easiest, evident way : $f$ is continuous, so preimage of open sets are open. Since $F$ is a closed subset of $Y$, $Y\setminus F$ is an open subset of $Y$, and $X\setminus G=f^{-1}(Y\setminus F)$ is then an open subset of $X$, so $G=X\setminus(X\setminus G)$ is a closed subset of $X$.

*Using limit points : Let $(x_n)_{n\in\Bbb N}\in G^{\Bbb N}$ a convergent sequence with limit point $x$. You want to prove that $x$ is in $G$. By the definition of $G$, you know that $f(x_n)$ is in $F$ for all $n\in\Bbb N$. By continuity, you (should) know that $\lim f(x_n)=f(\lim x_n)=f(x)$. Since $F$ is closed, it implies that $f(x)$ is in $F$. Again, by the definition of $G$, it means that $x$ is in $G$, so $G$ is closed.
