Set of limit points of topologist's sine curve $S$ Let $S=\{(x,\sin(1/x)):x \in (0,1]\}$ be the topologist's sine curve. Find the limit points $\lim S$ of $S$.

I claim $\lim S = S \cup \{(0,y):y \in [-1,1]\}$. But, how do you show that any of these points is such? Certainly there is some nice lemma for this? All I can think to do is revert to analysis by showing that, for $s=(x,\sin(1/x)) \in S$, and any open ball $B=B(z,\epsilon)$ which contains $s$, there is a point $s'=(x',\sin(1/x')) \in S$ with $s' \in B$, but I'm having trouble finding an $x'$...
Because $\sin(1/x)$ is continuous, any point in $S$ will be a limit point of $S$.

For a point $z=(0,y)$ where $y \in [-1,1]$, you need to show that every deleted neighborhood $B(z,r) \setminus \{z\}$ contains a point of $S$. This is true due to the rapid oscillation of $\sin(1/x)$ near $x=0$. To explicitly find an $x$ that fits in this ball, take some $n$ so that $\frac{1}{2\pi n} &lt; r$. Then $\sin(1/x)$ takes on all values in $[-1,1]$ for $x \in\left(\frac{1}{2\pi(n+1)}, \frac{1}{2\pi n}\right)$, so it must cross inside the deleted neighborhood.
A picture may be better than many words. So $[-1,1]$ is also the limit points of  topologist's sine curve.