# When is an interval (order theory) = line segment (in $\mathbb{R}^n$)?

If $(X,\leq)$ is any poset and $a,b\!\in\!X$, then we define the interval as $$[a,b]_\leq:=\{x\!\in\!X;\;a\!\leq\!x\!\leq\!b\}.$$

If $a,b\!\in\!\mathbb{R}^n$, then we define the line segment as $$[a,b]_{\mathbb{R}^n}:=\{a\!+\!t(b\!-\!a);\;t\!\in\![0,1]\}.$$

A subset $A\!\subseteq\!X$ is called convex when $\forall a,a'\!\in\!A\!:[a,a']_\leq\!\subseteq A$.

A subset $A\!\subseteq\!\mathbb{R}^n$ is called convex when $\forall a,a'\!\in\!A\!:[a,a']_{\mathbb{R}^n}\!\subseteq A$.

Question: how can I define a partial ordering $\leq$ on $\mathbb{R}^n$, such that $\forall a,b\!\in\!\mathbb{R}^n\!:[a,b]_\leq=[a,b]_{\mathbb{R}^n}$?

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You can't, for $n\geq 2$. The endwise joining of intervals $[a,b]_{\leq}$ and $[b,c]_{\leq}$ is again an interval $[a,c]_{\leq}$, but even if $[a,b]_\leq=[a,b]_{\mathbb{R}^n}$ and $[b,c]_\leq=[b,c]_{\mathbb{R}^n}$, their union need not be a line segment if $b$ doesn't lie on a line through $a,c$.

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Of course, you're right. Too bad. thanks –  Leon May 13 '12 at 22:06
Glad I could help. –  Cameron Buie May 13 '12 at 22:18