# Is the dimension given by Klee trick ever sharp?

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.

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 $$g$$.
• 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 $\mathbb{R}^{2n}\ne \mathbb{R}^n \times \mathbb{R}$. – Neil Hoffman Jan 17 '19 at 15:50
• 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. – IJL Jan 18 '19 at 16:04