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.

  • 2
    $\begingroup$ I agree with you, it seems like the difficulty of the problem has been swept under the carpet. $\endgroup$ – 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 $0<a<x$ 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 $0<a<x$.

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 $D\cap\overline{G}$.


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