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Can $\mathbb R^m$ be embedded in $\mathbb R^n$ with dense image？Here, $m$ is smaller than $n$. I think it is impossible, but I have difficulty proving it. Embedding is only homeomorphic with a subset, not diffeomorphic. So, Sard's Theorem is useless.

Some examples for low dimension densely immersed in high dimensional space like the dense one dimensional subgroup on the torus, but we all know it’s not an embed. And I do think there is a topological reason for this kind of thing can not happen, which don’t need smooth structure or group structure on it. So maybe some algebraic topology tools is helpful, but I have no idea about it.

Moishe Kohan
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wichard
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  • Sard would not help you as you only require the image to be dense (which very well can be a Lebesgue null set). – Severin Schraven May 22 '25 at 06:05
  • So, to be clear, you are asking whether there exists $m<n$ and a homeomorphism of $\mathbb R^m$ onto a subset of $\mathbb R^n$ which is dense in $\mathbb R^n$? – Joe May 22 '25 at 06:05
  • @Joe Yes, you’re right. – wichard May 22 '25 at 06:14
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    @Academic The injection is from $\mathbb{R}^m$ to $\mathbb{R}^n$, $ m<n$, not the converse direction – wichard May 22 '25 at 06:16
  • @wichard You’re right, apologies. – Academic May 22 '25 at 06:18
  • @SeverinSchraven Yes, you’re right. For diffeo Sard is also useless. But at that time it is an embedded submanifold, so use slice chart I can easily deduce a contradiction. – wichard May 22 '25 at 06:22

2 Answers2

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That's impossible. A dense and locally compact subset of Hausdorff space is open: Dense and locally compact subset of a Hausdorff space is open and then invariance of domain applies.

Another interesting question is whether there is a continuous injection $\mathbb{R}^m\to\mathbb{R}^n$ with dense image? Which is not necessarily an embedding (and therefore the image does not have to be locally compact). Similar to your torus example. I don't know the answer to this question.

freakish
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    A continuous injection with dense image should exist by just picking a countable dense subset and basically drawing a curve through them, I think. At any finite stage you have an embedding of a compact interval into $\mathbb{R}^n$, so the remainder should still be path-connected and thus you can extend to a larger interval while covering more of the remaining points in the countable dense set. – David Gao May 22 '25 at 07:12
  • @DavidGao so you are talking about space filling curves in an unbounded sense. Which afaik are not injective. Will think about it later. – freakish May 22 '25 at 07:26
  • And in fact this: https://math.stackexchange.com/questions/43096/is-it-true-that-a-space-filling-curve-cannot-be-injective-everywhere gives a negative answer (the title is misleading, look at the third answer) for $\mathbb{R}^1$ case. Based on Baire category. – freakish May 22 '25 at 07:31
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    Note that that answer is about mapping $\mathbb{R}$ injectively to a subset of $\mathbb{R}^n$ with nonempty interior, which, of course, is not possible. But a dense subset of $\mathbb{R}^n$ certainly needs not to have nonempty interior. The curve I’m suggesting only passes through a countable dense subset. There’s no reason it has to be space-filling (i.e., surjective). – David Gao May 22 '25 at 07:43
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    The construction is just by induction - write a countable dense subset of $\mathbb{R}^n$ by ${\cdots,x_{-1},x_0,x_1,\cdots}$. At stage $k$ we have an embedding $[-k,k]\to\mathbb{R}^n$ whose range covers ${x_{-k},\cdots,x_k}$. Then since the remainder should still be path-connected, we can extend to an embedding $[-k-1,k+1]\to\mathbb{R}^n$ whose range also covers $x_{-k-1}$ and $x_{k+1}$. Now take the union. – David Gao May 22 '25 at 07:49
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    And the construction of David Gao of a map $\mathbb R\to\mathbb R^{n-m+1}$ extends easily to continuous injection of $\mathbb R^m$ into $\mathbb R^n$ by preserving remaining coordinates. – Christophe Boilley May 22 '25 at 12:50
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A simple way of constructing a dense embedding of $\mathbb R$ into $\mathbb R^n (n>1)$ is to start with any compact embedded path $\gamma_0$ and extend it inductively on $k$ by choosing a nearest (to $\gamma_0$) point $p_k\in\mathbb R^n \setminus \gamma$ at distance $\geq \frac1k$ from the already existing path $\gamma_k$, and connecting the endpoint of $\gamma_k$ to $p_k$ by an embedded path in $\mathbb R^n \setminus \gamma$. This can even be done with smooth paths.

To get a continuous map $\mathbb R^m\to\mathbb R^n$, start with a map $\mathbb R\to \mathbb R^{n-m+1}$ as above, and take Cartesian product with $\mathbb R^{m-1}$.

Mikhail Katz
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