The Klee Trick for subsets of $\mathbb{R}^3$

Update: The lead paragraph has been changed to reflect the solution to a related question.

I asked the question Is dimension given by the Klee trick ever sharp? and it has been answered in the affirmative using two embeddings of a $$0$$-dimensional metric spaces in $$\mathbb{R}^1$$.

In the interests of trying to better understand the Klee Trick, I thought I might ask a slightly more concrete question along those lines.

Given a metric space $$K$$ that embeds into $$\mathbb{R}^n$$, define a $$Klee_n(K)$$ as the minimum value for $$m$$ such that for any two embeddings $$f_1, f_2 : K \rightarrow \mathbb{R}^n$$ and any embeddings $$g_1,g_2: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ there exists with a homeomorphism $$h:\mathbb{R}^m \rightarrow \mathbb{R}^m$$ such that $$h(g_1(f_1(K)))=g_2(f_2(K)).$$

(The Klee trick says $$Klee_n(K) \leq 2n$$.)

For example, $$Klee_3(\mathbb{S}^1)=4$$, since if $$g_1, g_2$$ correspond to embeddings of distinct (hyperbolic) knots then we see $$Klee_3(\mathbb{S}^1)> 3$$, but any two embeddings of $$\mathbb{S}^1$$ into $$\mathbb{R}^4$$ can be unknotted via an isotopy.

The other question asked if there was a metric space $$K$$ such that $$Klee_n(K)=2n.$$ However, for this question, is there a $$K$$ such that $$Klee_3(K)=5$$?

• Just to be sure, is it true that $Klee_1([0,1]) =1$? – Mathieu Baillif Sep 19 '14 at 7:19
• @MathieuBaillif Yes. $Klee_1([0,1])=1$. – Neil Hoffman Sep 23 '14 at 1:24