I know that this question has been asked before, but it is still not clear to me why the topologist sine curve is not locally connected.

The topologist's sine curve: Let $S$ denote the following subset of the plane. $$S = \{ (x, \sin(1/x)) \mid 0 < x \le 1\}.$$ The set $\bar{S}$ is called the topologist's sine curve, which equals the union of $S$ and the vertical interval $0 \times [-1,1]$.

My attempt was:

Fix a point $p = (0, t)$ where $t > 0$, and consider a neighborhood $N = B(p,\delta)\cap\bar S$, where $B(p,\delta)$ denotes a ball centered at $p$ with radius $\delta$. Then, I want to show that every open subset of $N$ is disconnected, so there can't be connected neighborhood of $p$ contained in $N$.

But, I am having a trouble with establishing that every open subset of $N$ is disconnected.

How can I prove the claim?


1 Answer 1


Let's stick with a particular point on the interval $0 \times [-1, 1]$, say $p = (0, 0)$. Notice that there is no need to make your neighborhoods balls centered at $p$; we could consider open squares instead (they also form a basis for the topology of the plane). Let $U_\epsilon := (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$ be some open square centered at $p$ (where $\epsilon > 0$). Then $U_\epsilon \cap \overline{S}$ consists of $0 \times (-\epsilon, \epsilon)$ and the graph of the function $\sin(1/x)$ restricted to the domain $D_\epsilon:=\{x \in (0, \epsilon) : |\sin(1/x)| < \epsilon\}$. You should be picturing a bunch of very short curve segments which are almost vertical. We can choose $\epsilon$ small enough that $D_\epsilon$ does not contain any $x$ such that $\sin(1/x) = 1$.

Now let $V$ be some nonempty open subset of $U_\epsilon$ containing $p$. It contains $U_{\epsilon'}$ for some smaller $\epsilon' > 0$. Then there exists some $x_0 \in (0, \epsilon')$ such that $\sin(1/x_0) = 1$ and $(x_0, \infty) \cap D_{\epsilon'} \neq \emptyset$ (this should be easy to see; there is a sequence of such $x_0$ which converges to $0$). It follows that $D_{\epsilon'} = (D_{\epsilon'} \cap (0, x_0)) \cup (D_{\epsilon'} \cap (x_0, \infty))$, i.e. it is disconnected.

I claim that we can use this information to prove that $V \cap \overline{S}$ is disconnected. The idea is to look at the intersections of this set with $(-\infty, x_0) \times \mathbb{R}$ and with $(x_0, \infty) \times \mathbb{R}$. Note that neither of these intersections is empty since, in either case, we can take an appropriate value of $x \in D_{\epsilon'}$ and note that $(x, \sin(1/x)) \in V$. Secondly, these open sets do indeed cover $V \cap \overline{S}$ since $(x_0, 1)$ is the only point in $\overline{S}$ whose $x$-coordinate is $x_0$, and $V$ contains no point whose $y$-coordinate is $1$. So we conclude that $V \cap \overline{S}$ is disconnected.

  • $\begingroup$ I do not understand why you want to show that "this open set is actually disconnected". To show that the curve is not locally connected, I need to show that every neighborhood of $p$ containned in $U_\epsilon$ is disconnected, which I do not think follows from showing that $U_\epsilon $ is disconnected $\endgroup$
    – user74261
    Jun 6, 2015 at 22:28
  • $\begingroup$ Oops! Of course. Luckily, my argument is not too far from showing this. I'll edit it. $\endgroup$
    – Alex G.
    Jun 6, 2015 at 22:33
  • $\begingroup$ Could you also tell me why showing $D$ is disconnected implies $U_\epsilon$ is disconnected? I am having a trouble seeing that $\endgroup$
    – user74261
    Jun 6, 2015 at 22:37
  • $\begingroup$ I hope my edit answers your question now. $\endgroup$
    – Alex G.
    Jun 6, 2015 at 22:48
  • $\begingroup$ The set $V$ is some non-empty open subset containing $p$, I suppose? $\endgroup$
    – user74261
    Jun 7, 2015 at 0:06

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