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Definition 1

A subset $A$ of $\Bbb{R}$ is convex iff for any $a,b\in A$ and $x\in\Bbb{R}$ such that $a\le x\le b$ it results that $x\in\ A$.

Definition 2

A real interval is a set of real numbers lying between two numbers.

Clearly the intervals are convex sets. But how can I conclude that convex sets with more than one point are intervals? It seems to me that it is evident: however I can't explain to me this formally. Could someone help me, please?

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2 Answers 2

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If $A$ is order convex and non-empty in $\Bbb R$, let $m := \inf A \le M:=\sup A$, where the inf and sup are taken in $\overline{\Bbb R}=[-\infty, +\infty]$.

Then if $m=M$ we must have that $A=\{p\}=[p,p]$ a degenerate interval. So we can assume $m < M$. If $m < x < M$ we have some $a_0 \in A$ such that $a_0 < x$ (or $x$ would be a larger lowerbound for $A$ than $\inf A$, which cannot be) and we have $a_1 \in A$ with $x < a_1 < M$ (or $x$ would be a smaller upperbound for $A$ than $\sup A$, which cannot be) and then by order convexity of $A$, we have $x \in A$. So $(m,M) \subseteq A$. Clearly, as $m,M$ is a lower resp. upperbound we have that $A \subseteq [m,M]$ so $A=[m,M], [m,M), (m,M), (m,M]$ depending on whether $m,M \in A$ (4 options).

So order convex subsets of $\Bbb R$ are empty, $\Bbb R$, singletons, or real intervals of 4 types.

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Hints: Let $m=\inf A$ and $M=\sup A$. (It is possible that $m=-\infty$ and/or $M=\infty$). If $m=M$ then $A$ is a singleton. Now assume that $m<M$. If $m<x<M$ then there exist $y,z \in A$ such that $m<y<x<z<M$. [This follows from definitions of infimum and supremum]. Hence $x \in A$. It follows that $A$ is an interval with end points $m$ and $M$. The end points may or may not belong to $A$.

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  • $\begingroup$ Sorry, but I don't understand completely your proof. So if $A$ is convex and $y,z\in A$ and $y<x<z$ then $x\in A$. So if $x=ty+(1-t)z$ for some $t\in (0,1)$, then how can I conclude that $A$ is an interval? $\endgroup$
    – Antonio Maria Di Mauro
    Commented Apr 9, 2020 at 10:08
  • $\begingroup$ @AntonioMariaDiMauro I have proved that all points strictly between $m$ and $M$ belong to $A$. Obviously any number less than $m$ or greater than $M$ cannot be in $ A$. What does that tell you about $A$? $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 9, 2020 at 10:14
  • $\begingroup$ Okay. However it seems to me that to prove this it is not necessary to write $x$ as convex combination of $y$ and $z$: indeed for any $y,z\in A$ such that $m<y<z$ and for any $x\in\Bbb{R}$ such that $y<x<z$ we can conclude that $x\in A$, since $A$ is convex; right? $\endgroup$
    – Antonio Maria Di Mauro
    Commented Apr 9, 2020 at 10:18
  • $\begingroup$ Well, my definition of a convex set is a set $A$ such that $tx+(1-t)y \in A$ whenever $x,y \in A$ and $0 <t<1$. This is the standard definition. What is your definition ? $\endgroup$
    – Kavi Rama Murthy
    Commented Apr 9, 2020 at 10:20
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    $\begingroup$ @AntonioMariaDiMauro your notion is called "order convex", convex alone is more often the linear space/geometric notion. There is in fact a general theory of convexity structures (very much like a topology, but with "convex" instead of "open" as a basic notion), and convexity preserving maps (analogues of continuity), even "separation axioms" for them. There is also "metrically convex", which is studied a lot, also an instance of a convex structure. For subsets of $\Bbb R$ all 3 notions coincide. $\endgroup$
    – Henno Brandsma
    Commented Apr 9, 2020 at 11:53

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