The [Klee Trick](http://books.google.com/books?id=qYqd12vt2x8C&pg=PA66&lpg=PA66&dq=The%20Klee%20Trick&source=bl&ots=dh9LHDkfNr&sig=ZMinrf9nTN8C4FtpoKxeK6qC-Bw&hl=en&ei=dbv5TNqWPIGKlwfT3s3iBw&sa=X&oi=book_result&ct=result&resnum=5&ved=0CDMQ6AEwBA#v=onepage&q=The%20Klee%20Trick&f=false) 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. 



Is the dimension given by Klee trick ever sharp?
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 $g$. 


