Let $U$ and $V$ be open subsets of $R^m$ and $R^n$ respectively, where $m<n$. Let $f:U\rightarrow V$ be a smooth map. It is possible that $f(U)$ is dense in $V$?
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2$\begingroup$ Yes. Consider the map $f:\mathbb{R}\rightarrow \mathbb{T}:=\mathbb{R}^2/\mathbb{Z}^2$ given by $f(t)=(t,\alpha t)$, with $\alpha $ irrationnal. It is well known that its image is dense. Now identify the square $V=(0,1)\times (0,1)$ in $\mathbb{R}^2$ with its image in $\mathbb{R}^2/\mathbb{Z}^2$, and put $U:=f^{-1}(V)$. Then $f(U)$ is dense in $V$. $\endgroup$– abxNov 7, 2016 at 7:00
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$\begingroup$ Another example, $f:\mathbb{R}→]−1,1[^3$ such that $x\mapsto {x^2\over x^2+1}(\sin x,\sin \sqrt{2}x,\sin \sqrt{3}x)$ $\endgroup$– Pietro MajerNov 7, 2016 at 19:10
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Yes, at least as stated.
Let $U = \cup_{k \in \mathbb{Z}} (2k, 2k+1)$, an open subset of $\mathbb{R}$. And let $V = (0,1)^2$. Choose a bijection $\phi: \mathbb{Z} \to \mathbb{Q} \cap (0,1)$. Define
$$ f(x) = (\phi(k), x - 2k), \qquad x \in (2k, 2k+1).$$
(That $U$ is disconnected is inessential; I wrote it this way for simplicity of the formula.)
answered Nov 7, 2016 at 7:01
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1$\begingroup$ Nice... Then why not taking $f$ constant on any component of $U$, with $f_k(U_k)=y_k$ and $\{y_k\}_k$ dense in $V$ $\endgroup$ Nov 7, 2016 at 8:23
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$\begingroup$ @PietroMajer: haha, that's even simpler. $\endgroup$ Nov 7, 2016 at 12:34