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A face of a closed convex set $X\subseteq\mathbb{R}^n$ is defined to be a set $F\subseteq X$ such that:

  1. $F$ is convex.
  2. Every line segment from $X$ whose interior meets $F$ is contained in $F$.

Is condition 1 needed? Is there an example where condition 2 is satisfied but condition 1 is not? Thank you.

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  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Aug 28, 2018 at 11:06

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Condition 1 is needed. Take a closed tetrahedron $X$, and let $F$ be the union of two faces of $X$. Then Condition 2 is satisfied, but Condition 1 is not.

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    $\begingroup$ Or even take $F$ to be the two endpoints of an interval in $\mathbb{R}$. $\endgroup$
    – Tobias Fritz
    Commented Aug 27, 2018 at 19:14
  • $\begingroup$ @TobiasFritz: Yes, $X=[a,b]$ and $F=\{a,b\}$ seems to be the simplest example satisfying Condition 2 but not Condition 1. $\endgroup$
    – GH from MO
    Commented Aug 27, 2018 at 19:16
  • $\begingroup$ Of course, you're right. When looking for an example, I was looking for a non-convex subset of a face, and it did not come to me to look for a union of faces. Thanks! $\endgroup$
    – Tom Werner
    Commented Aug 28, 2018 at 13:06
  • $\begingroup$ @TomWerner: OK. If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Aug 28, 2018 at 21:55

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