The Klee Trick allows one to find an $\mathbb{R}^m$ where two embeddings of same compact metric space have homeomorphic complements. More precisely, given two embeddings of a compact metric space $K$ into $\mathbb{R}^n$, $f_{n,1}$, $f_{n,2}$, we can construct two embeddings of $K$ into $\mathbb{R}^{2n}$ such that the images of $K$ are equivalent under a homeomorphism of $\mathbb{R}^{2n}$. (The trick itself produces an isotopy of the the two embeddings in $\mathbb{R}^{2n}$.)

However, I was wondering if there is an example of a compact metric space $K$ and a pair of embeddings $f_{n,1},f_{n,2}:K \rightarrow \mathbb{R}^n$ such that the embeddings are not equivalent under a homeomorphism of $\mathbb{R}^{n+m}=\mathbb{R}^n \times\mathbb{R}^m$ for all $m < n$.

To clarify with a non-example, any two (tamely) embedded knots in $\mathbb{R}^3$ will be isotopic in $\mathbb{R}^4=\mathbb{R}^3 \times\mathbb{R}$, and so for embeddings of $S^1$ into $\mathbb{R}^3$, the isotopy obtained from the Klee trick is not optimal.


1 Answer 1


Are the embeddings required to be isometric embeddings? If not, then what about including three points into $\mathbb{R}$ in two ways, so that the middle point of the three changes? More explicitly, let $K=\{a,b,c\}$ and define $f(a)=0$, $f(b)=1$, $f(c)=2$, whereas $g(a)=1$, $g(b)=0$, $g(c)=2$. There isn't any self-homeomorphism of $\mathbb{R}$ whose composition with $f$ is equal to $g$.

  • $\begingroup$ Thanks! This is probably the simplest example of what I was looking for. Its also fairly satisfying to see the isotopy in $\mathbb{R}^2$ of the two embeddings $(f(x),0)$ and $(g(x),0)$. However, it would be nice to have an example of where $n\ne 1$ so $\mathbb{R}^{2n}\ne \mathbb{R}^n \times \mathbb{R}$. $\endgroup$ Jan 17, 2019 at 15:50
  • $\begingroup$ Mea Culpa: I misunderstood your answer and I voted it down. Now, the only way to remove my vote down and vote it up was to edit your answer so I had to do it. I just replaced whereas with where and then where back with whereas. Sorry for that. I hope you appreciate my effort and honesty :) $\endgroup$ Jan 17, 2019 at 23:45
  • $\begingroup$ I couldn't think of a way to make any higher-dimensional examples, but my guess is that the place to look is when $K$ is either $n$ or $n-1$-dimensional. $\endgroup$
    – IJL
    Jan 18, 2019 at 16:04
  • $\begingroup$ Piotr: thanks for your message and correcting yourself with an edit. $\endgroup$
    – IJL
    Jan 18, 2019 at 16:08

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