Let $\{U_i\}_{i=1}^{I}$ be a non-empty and finite collection of non-empty, disjoint, open, (and obviously bounded) subsets of $[0,1]^n$. Suppose also that $[0,1]^n=\cup_{i =1 }^{ I} \overline{U_i}$. Under what condition does there exist a continuous function $f:[0,1]^n\rightarrow [0,1]^I$ such that $$ x\in U_i \Leftrightarrow \|f(x)-e_i\|< \min_{\tilde{i}\neq i}\|f(x)-e_{\tilde{i}}\| ; $$ where $\{e_i\}_{i=1}^I$ is the standard basis of $\mathbb{R}^I$.

What I have tried:

  • In the case where each $U_i$ is convex: If $U_i$ is convex then there it is star-shaped so there is an $x_i\in U_i$ for which every $x \in U_i$ can be reached by a line segment emanating from $x_i$ and contained within $U_i$. Note, since $U_i$ is convex then such a line segment must stay within $U_i$. Thus, for any $x'\in \cup_{i\in I} (0,1)^N$, $ \|x-x_i\|<\min_{\tilde{i}\neq i} \|x-x_{\tilde{i}}\|. $ So we set: $$ f(x)= \left(\|x-x_i\| e_i \right)_{i=1}^I. $$ Is this reasoning correct?

  • For the general case, I was trying to use Urysohn's Lemma but alas, to no avail..


For $x \in U_i$ let $f(x) = r e_i$ where $r = \min_{y \notin U_i} |x - y|$, i.e. the radius of the largest open ball centered on $x$ and contained in $U_i$. If $x$ is not in any $U_i$ let $f(x) = 0$. Note that $|f(x) - f(y)| \leq |x - y|$

  • 2
    $\begingroup$ What's the $x$ on the right hand side? $\endgroup$ – Phillip Harris Jan 31 at 9:54

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