- Show that the topologist’s sine curve is connected but not path-connected.
Solution
1. Let\[G=\{(x,\sin(1/x)):0<x\le1\},\qquad T=G\cup(\{0\}\times[-1,1]).\]The set \(T\) is connected.
The interval \((0,1]\) is connected, so its continuous image \(G\) is connected. Its closure is exactly \(T\), and the closure of a connected subset is connected.
2. There is no path in \(T\) from the vertical segment \(\{0\}\times[-1,1]\) to \(G\).
Suppose \(\gamma(t)=(x(t),y(t))\) were such a path, with \(x(0)=0\) and \(x(1)>0\). Let \[t_0=\sup\{t\in[0,1]:x(t)=0\}.\] Then \(t_0<1\), \(x(t_0)=0\), and \(x(t)>0\) for \(t>t_0\). For any \(t_1>t_0\), continuity of \(x\) implies its image on \([t_0,t_1]\) contains every value between \(0\) and \(x(t_1)\). Hence for all sufficiently large \(n\) there are times tending to \(t_0\) at which \[x=\frac1{\pi/2+2\pi n}\quad\text{and}\quad x=\frac1{3\pi/2+2\pi n}.\] At these times \(y=1\) and \(y=-1\), respectively. The times must tend to \(t_0\): on every compact interval \([t_0+\delta,t_1]\), the positive continuous function \(x\) has a positive minimum. Thus \(y(t)\) cannot converge to the single value \(y(t_0)\), contradicting continuity of \(\gamma\).
3. Therefore the topologist’s sine curve is connected but not path-connected.
Use 1 and 2.