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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 $x'$...

JustAskin
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3 Answers3

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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.

angryavian
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  • OK, but this doesn't show that these are the only limit points of S. And I guess this also assumes we know $\sin(1/x)$ is continuous which may be well-known but not in the context of this problem (this question is from Chapter 2 of Adams-Franzosa, where it comes 2 chapters before continuity). – JustAskin Sep 29 '14 at 02:33
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    If you are looking at the standard topology (as Adams-Franzosa is), it is sufficient to use the $\epsilon-\delta$ definition of continuity from calculus. – Becca Winarski Feb 09 '15 at 13:34
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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.

Dom
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A picture may be better than many words. So $[-1,1]$ is also the limit points of topologist's sine curve.

enter image description here

Paul
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    While a picture can be useful, a picture is not a proof; thus maybe this would fit better as a comment rather than an answer? – JustAskin Sep 29 '14 at 03:18