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Consider $A \subseteq \mathbb{R}^{n}$ closed set. I must show that if $A$ is not convex then there exist $x,y \in A$ such that \begin{align*} A \cap (x,y) = \emptyset, \text{ where } (x,y):= \{ (1-t)x + ty | t \in (0,1)\}. \end{align*}

Is it that trivial as I think? A set being not convex means that it does not contain the lines that make up a segment, so naturally its intersection with the convex set would be nothing.

Am I missing something?

algevristis
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  • If $A$ is not convex then there exist $x, y \in A$ such that the segment $(x, y)$ is not a subset of $A$. You still have to show that there exist $x', y' \in A$ such that the segment $(x', y')$ is completely in the complement of $A$. – I am not sure if that is what you have in mind with “as trivial as one thinks.” – Martin R May 16 '22 at 13:35
  • @MartinR are you willing to write a complete answer for me to gloss over? – algevristis May 16 '22 at 13:37

1 Answers1

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If $A$ is not convex then there exist $x, y \in A$ with $x \ne y$ such that the segment $(x, y)$ is not a subset of $A$. One still has to show that there exist $x', y' \in A$ with $x' \ne y'$ such that the segment $(x', y')$ is completely in the complement of $A$.

Here is a possible strategy: The set $$ T = \{ t \in (0, 1) \mid (1-t)x + t y \notin A \} $$ is not empty, and it is open because the complement of $A$ is open. $T$ is therefore a finite or countable union of disjoint open intervals (see for example here). Let $I = (t_1, t_2)$ be one of these open intervals.

Then $x' = (1-t_1)x + t_1 y \in A$, $y' = (1-t_2)x + t_2 y \in A$, and $(x', y') \cap A = \emptyset$.

Martin R
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