In reference to Showing a subset of the torus is dense, the responders helped show the poster that the image set $f(\mathbb{R})$ is dense in the torus. But, it's not immediately clear to me why the image set is not an embedded submanifold. If $f(\mathbb{R})$ is an embedded submanifold, we must have that it's
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a smooth manifold in the subspace topology and
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that the inclusion map from $f(\mathbb{R})$ to $T^2$ is a smooth embedding.
It's visually clear to me that under the subspace topology, $f(\mathbb{R})$ is not locally Euclidean (as it's not locally path connected). But, I can't seem to formalize this or any argument, using that $f(\mathbb{R})$ is dense in $T^2$, which says that $f(\mathbb{R})$ fails 1) or 2). Thanks for any help!
Best Answer
For convenience, let's call $f(\mathbb{R}) = A$. It's clear that $A$ is not locally path connected. As you point out, if $A$ were an embedded submanifold, it would be a smooth manifold in the subspace topology. If it were a smooth manifold in the subspace topology, every $a\in A$ would have a neighborhood homeomorphic to $\mathbb{R}$. In particular, $A$ would be locally path-connected. However, it's not, so it can't be a submanifold.
Alternatively, if $A$ were a submanifold, every $a\in A$ would have a neighborhood $U$ in $T^2$ with a coordinate chart taking $a$ to zero, $U$ to $\mathbb{R}^2$, and $U\cap A$ to the $x$-axis. This cannot happen, for every open ball $U_\epsilon$ of $a$ in $T^2$ (for $\epsilon$ sufficiently small) has that $U\cap A$ is disconnected (with countably many components).