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 $x'$...
 A: 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 $\frac{1}{2\pi n} < r$. 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: Pick any point in $x\in\{0\}\times [-1,1]$. Then for any $\epsilon>0$ the graph of $\sin(1/y)$ oscillates infinitely many times for $y<\epsilon $, so for some $x'$, $\sin(1/x')=x$, and so $d((0,x),(x',x))<\epsilon $. Therefore, all of $x\in\{0\}\times [-1,1]$ are limit points.
That these are the only limit points is due to the fact that away from 0 (in say $[\epsilon,1]$) the curve is compact.
A: I know this answer is really late, but to answer your question of how to find an $x^\prime$. Let $U = B((0,y),\epsilon)$ where $y\in [-1,1], \epsilon >0$  be a neighborhood of $(0,y)$. Let $n\in N$ such that $n > \frac{1}{2\pi\epsilon}$. Then
$$x^\prime = \frac{1}{\arcsin(y)+2\pi n} < \epsilon$$
Notice that $\frac{1}{\sin(x^\prime)} = y$. Thus $\left(x^\prime,\frac{1}{\sin(x^\prime)}\right)=(x^\prime,y).$ We have $0\lt x^\prime\lt\epsilon$. So $(x^\prime,y)$ is going to be an element of the deleted neighborhood.
A: A picture may be better than many words. So $[-1,1]$ is also the limit points of  topologist's sine curve.

