An open covering of the Topologist's sine curve.

Question

How does an open covering of of the topologist's sine curve look like?

I am asking since I want to show that it has a topological dimension 1.

Answer 1

Some hints.An open subset $T$ of $S=\{(0,0)\}\cup \{(x,\sin 1/x) :0<x<1\}$ is $S\cap T^*$ where $T^*$ is an open subset of the real plane $R^2$. If $(0,0)\in T$ then $T^*$ covers an open square $$V_r=\{(x,y)\in R^2 : |x|<r \wedge |y|<r\}$$ for some $r>0$. In particular when $T^*=V_r$ we have $$T=S\cap V_r= \{(0,0)\}\cup \{(x,\sin 1/x) : 0<x<r \wedge |\sin 1/y|<r \}.$$Also, when $0<a<b<1$ and $$S[a,b]= \{(x,\sin 1/x):x\in [a,b]\}$$ and $t>0$, we can cover $S[a,b]$ with a $finite$ family $F$ of open balls of $R^2$,each of radius $t$, each centered at a point of $S[a,b]$. If $t$ is small enough ( which depends on $a$ and $b$) then $$S\cap (\cup F)=\{(x,\sin 1/x):a-e_1<x<\min (1,1+e_2)\}$$ where $e_1>0<e_2$ and $\max (e_1,e_2)$ can be as small as we like.