Consider the set
$$
S = \left\{\left(x,\sin\frac 1x\right) : x\in(0,1]\right\}.
$$
Then $S=f((0,1])$ where $f(x) = \left(x,\sin\frac1x\right)$. Since $f$ is continuous and $(0,1]$ is path connected, it should follow that $S$ is path connected. But this runs counter to the well-known fact that the topologist's sine curve is not path connected. Is it the case that this set $S$ is path connected, it's just that the "topologist's sine curve" is actually $S\cup\{(0,0)\}$ (which most definitely is not path connected)?
Is the topologist’s sine curve (without boundary) path connected
general-topologypath-connected
Best Answer
Yes, $S$ is path connected. In my experience, the "topologist's sine curve" usually refers to $S\cup\{0\}\times[-1,1]$ (i.e., the closure of $S$ in $\mathbb{R}^2$), though I wouldn't be surprised if you sometimes see it used to refer to $S\cup\{(0,0)\}$ or (more rarely, in contexts where it is convenient to do so) to $S$.