# Path components of Topologist's Sine Curve

Let $$G:=\left\{(x,\sin\left(\frac{1}{x}\right):x>0\right\}$$, $$A:=\{0\}\times [-1,1]$$. We know that the closure of $$G$$ is $$\bar{G}=G\cup A$$. We want to show that $$\bar{G}$$ isn't path-connected.

Suppose it was path-connected, then there is a continuous function $$\gamma:[0,1]\to \bar{G}$$ such that $$\gamma(0)\in A$$ and $$\gamma(1)\in G$$.

The proof I've read now uses the following argument:

Let $$t_0 \in \gamma^{-1}(A)$$. Choose a small open disk $$D\subseteq \mathbb{R}^2$$ centred at $$\gamma(t_0)$$. Then $$D\cap \bar{G}$$ has infinitely many path-components, one of which is $$D\cap A$$.

It makes sense that $$D\cap \bar{G}$$ has infinitely many path components, but why can we be sure that one of those equals $$D\cap A$$? For me, this claim is as "obvious" as the fact that $$\bar{G}$$ is not path-connected.

• I agree with you, it seems like the difficulty of the problem has been swept under the carpet. – nicomezi Dec 16 '19 at 7:42

Specifically, let us suppose that $$D$$ has radius at most $$1$$. In particular, then, either $$D$$ contains no points with second coordinate $$1$$ or $$D$$ contains no points with second coordinate $$-1$$; let us assume we are in the first case (the second is similar). Observe now that if $$(x,y)\in D\cap G$$ then there exists $$a$$ such that $$0 and $$\sin(1/a)=1$$ (just choose $$a=1/b$$ where $$b>1/x$$ has the form $$2\pi k+\pi/2$$ for some integer $$k$$). Then $$(a,\sin(1/a))\not\in D$$ and thus $$D\cap\overline{G}$$ contains no point whose first coordinate is $$a$$. There can thus be no path from a point of $$D\cap A$$ to $$(x,y)$$ in $$D\cap\overline{G}$$, since the first coordinate of such a path would have to pass through $$a$$ since $$0.
Thus, there is no path in $$D\cap\overline{G}$$ from a point of $$D\cap A$$ to a point of $$D\cap G$$. On the other hand, $$D\cap A$$ is path-connected, since it just has the form $$\{0\}\times I$$ for some interval $$I$$. So, $$D\cap A$$ is a path component of $$D\cap\overline{G}$$.