The irrational winding of the torus given by the map $f\colon\mathbb{R}\to T^2$ where $f(t)=(e^{it},e^{i\alpha t})$ for some irrational $\alpha$. Wikipedia mentions this is not a regular submanifold, and the answer here says that it is clear that the image $f(\mathbb{R})$ is not locally path connected.
I'm having trouble seeing this. Suppose $(e^{ip},e^{i\alpha p})$ is a point in the image, and $U\times V$ is an open neighborhood around this point. Can't we visualize $U$ and $V$ as containing arcs on $S^1$ around $e^{ip}$ and $e^{i\alpha p}$, respectively. If we vary $t$ a little around $p$, won't the coordinates in each circle just move a little around $e^{ip}$ and $e^{i\alpha p}$. Then we could restrict to small arcs around each coordinate and get a locally path connected nbhd in $U\times V$? What am I doing wrong?
Best Answer
The problem is that $U \times V$ won't just contain the arc that $(e^{ip}, e^{i \alpha p})$ lives on. Because $f(\Bbb{R})$ is dense in the torus, $U \times V$ contains lots of other arcs too, each of which forms a (local) path-component.