# Proof that topologist sine curve is not locally connected

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?

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.
• 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 Jun 6, 2015 at 22:28
• Could you also tell me why showing $D$ is disconnected implies $U_\epsilon$ is disconnected? I am having a trouble seeing that Jun 6, 2015 at 22:37
• The set $V$ is some non-empty open subset containing $p$, I suppose? Jun 7, 2015 at 0:06