Prove that $\overline{X}=X\cup Y$ where $X=\{(x,\sin(1/x));\;x>0\}$ and $Y=\{(0,y);\;|y|\leq1\}$ Let $X=\left\{\left(x,\sin\frac{1}{x}\right)\in\mathbb{R}^2;\;x>0\right\}$ and $Y=\left\{\left(0,y\right)\in\mathbb{R}^2;\;-1\leq y\leq1\right\}$. I would like to prove that $\overline{X}=X\cup Y$. I've proved that $X\cup Y\subset\overline{X}$ but I don't know how obtain the converse inclusion. Could someone help me?
Thanks.
 A: It may be helpful to note that $X$ is closed relative to $U:=(0,\infty)\times\Bbb R$ since it is the graph of $\sin(1/x):(0,\infty)\to\Bbb R$, which is continuous, and $\Bbb R$ is a Hausdorff space. This tells you that $X=\text{cl}_U(X)=\text{cl}(X)\cap U$. So each limit point of $X$ outside of $X$ itself must be in $(\infty,0]\times\Bbb R$. Since $\Bbb R^2-([0,\infty)\times[-1,1])$ is open, a limit point of $X$ can only be in $([0,\infty)\times[-1,1])-U,$ and this is precisely $Y$.
A: There are several characterizations of the closure of a set. In metric spaces, I find useful:

Given a set $X$, $x\in \overline{X}$ if and only if, there is a sequence $\left(x_n\right)_{n\in\Bbb N}$ such that $x_n\to x$.

So let $(x,y) \in \overline X$ (the $X$ in your question), and let $\left(\left(x_n,\sin\frac1{x_n}\right)\right)$ a sequence in $X$ which converges to $(x,y)$. Then
$$x_n\to x,\quad\text{and}\quad \sin\frac1{x_n}\to y.$$
There are two cases: $x=0$ or $x\gt 0$.
If $x\gt 0$, you know that the map $t\mapsto \sin(1/t)$ is continuous on $(0,\infty)$ and then
$$(x,y)=\lim_{n\to\infty} \left(x_n,\sin\frac1{x_n}\right)=\left(x,\sin\frac1x\right)\in X.$$
If $x=0$, since
$$-1\leq \sin\frac1{x_n}\leq 1\qquad \forall n$$
it must be that
$$-1\leq y\leq 1,$$
so
$$(x,y)\in \{0\}\times [-1,1].$$
This proves
$$\overline{X}\subseteq X\cup \{0\}\times [-1,1].$$
