Finding the closure of set $S=\left \{(x,sin(\frac{1}{x} )) ;0<x\leq 1\right \}$
I know that $\bar{S}=(\left \{  0\right \}\times[-1 ,1])\cup     S$
How could I prove it ?
 A: Suppose $(x_n, \sin(\frac{1}{x_n}))$ is a sequence in $S$ that converges (in the plane) to $(p,q)$. This means that $x_n \to p$ and $\sin(\frac{1}{x_n}) \to q$.
But $x_n \in (0,1]$ so $p \in \overline{(0,1]}=[0,1]$. If $p >0$ we know by continuity of the sine and the $\frac{1}{x}$ function that $q_n\to\sin( \frac1{p})=q$ (as limits are unique) so that $(p,q) \in S$. Otherwise $p=0$ and we can only say that $q \in [-1,1]$ as all values $\sin(\frac{1}{x_n}) \in [-1,1]$ and $[-1,1]$ is closed. 
This shows that $S \subseteq \overline{S} \subseteq S\cup (\{0\}\times [-1,1])$.
On the other hand, if $(0,q)$ is in $\{0\}\times [-1,1]$, then pick $p>0$ such that $\sin(p) = q$. Then $p_n=\frac{1}{p+2n\pi}$, for $n \in \mathbb{N}$ is in $(0,1]$ for $n\ge 1$ and $p_n \to 0$ and $\sin(\frac{1}{p_n}) = \sin(p+2n\pi) = \sin(p)=q$ for all those $n$, and $(p_n,q)=(p_n, \sin(\frac{1}{p_n})) \in S$ and this sequence converges to $(0,q)$ and so $(0,q) \in \overline{S}$. So we have now shown: 
$$S\cup (\{0\}\times[-1,1]) \subseteq \overline{S} \subseteq S \cup (\{0\}\times[-1,1])$$
and we have equality.
A: Let $x$ be any point in $L = \{0\}×[-1,1]$.
Show for all open $U$ nhood $x$, $U \cap S$ is not empty.
Thus $L$ subset $\overline S$ and $L \cup S$ subset $\overline S$.  
To show there are no other points $p$ in $\overline S$ find an open ball about $p$ that misses $S$.
