Convexity of set $\{x\mid\operatorname{dist}(x,S) \leq \operatorname{dist}(x,T)\}$

Question

Is it possible to prove analytically the convexity of the set: $\{x\mid {\rm dist}(x,S) \leq {\rm dist}(x,T)\}$ where $S,T \subseteq \Bbb R^n$, and ${\rm dist}(x,S) = \inf\{\|x-z\|_2 \mid z \in S\}$?

Thanks in advance.

This is not true. For example take $S={ (x,0) : x \in \mathbb{R} }$ and $T={ (0,1), (0,-1) }$ subsets of $\mathbb{R}^2$. Then ${ x | d(x,S) \leq d(x,T) }$ is not connected, so it cannot be convex. Maybe this is true with some further hypothesis on $S,T$ (for example they are connected, or convex)? — Crostul, Nov 29 '14 at 18:19

score 2 · Answer 1 · edited Nov 29 '14 at 18:40

2

In general this set is not convex.

For example, in $\mathbb{R}$, $S =\{−1, 1\}, T = \{0\}$, we have $\{x\mid \operatorname{dist}(x,S)\le \operatorname{dist}(x,T)\}=\{x\in \mathbb{R}\mid x\le -\frac{1}{2} \text{ or } x\ge \frac{1}{2}\}$, which clearly is not convex.

edited Nov 29 '14 at 18:40

Michael Hardy

1

answered Nov 29 '14 at 18:19

John

13,204

Thanks for the reply. How you come up with -1/2 and 1/2? Could you please clarify:)? – Thoth Nov 29 '14 at 19:17
@Thoth $S$ is two points. You can draw $S$ and $T$ on a line. – John Nov 30 '14 at 03:26

Convexity of set $\{x\mid\operatorname{dist}(x,S) \leq \operatorname{dist}(x,T)\}$

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