We begin this article with something more general. Let $P$ be a poset. A subset $A\subseteq P$ is said to be convex if for any $a,b\in A$ with $a\le b$ , the poset interval $[a,b]\subseteq A$ also. In other words, $c\in A$ for any $c\in P$ such that $a\le c$ and $c\le b$ . Examples of convex subsets are intervals themselves, antichains, whose intervals are singletons, and the empty set.

One encounters convex sets most often in the study of partially ordered groups. A convex subgroup $H$ of a po-group $G$ is a subgroup of $G$ that is a convex subset of the poset $G$ at the same time. Since $e\in H$ , we have that $[e,a]\subseteq H$ for any $e\le a\in H$ . Conversely, if a subgroup $H$ satisfies the property that $[e,a]\subseteq H$ whenever $a\in H$ , then $H$ is a convex subgroup: if $a,b\in H$ , then $a^{-1}b\in H$ , so that $[e,a^{-1}b]\subseteq H$ , which implies that $[a,b]=a[e,a^{-1}b]\subseteq H$ as well.

For example, let $G=\mathbb{R}^2$ be the po-group under the usual Cartesian ordering. $G$ and $0$ are both convex, but these are trivial examples. Let us see what other convex subgroups $H$ there are. Suppose $P=(a,b)\in H$ with $(a,b)\ne (0,0)=O$ . We divide this into several cases:

1. $ab>0$ . If $a>0$ , then $b>0$ ($P$ in the first quadrant), so that $O\le P$ , which means $[O,P]\subseteq H$ . If $a<0$ , then $b<0$ ($P$ in the third quandrant), so that $O\le -P$ . In either case, $H$ contains a rectangle ($[O,P]$ or $[O,-P]$ ) that generates $G$ , so $H=G$ .
2. One of $a$ or $b$ is $0$ . Suppose $a=0$ for now. Then either $0<b$ so that $[O,P]\subseteq H$ or $b<0$ so that $[O,-P]\subseteq H$ . In either case, $H$ contains a line segment on the $y$ -axis. But this line segment generates the $y$ -axis. So $y$ -axis $\subseteq H$ . If $H$ is a subgroup of the $y$ -axis, then $H$ =$y$ -axis.

   Otherwise, another point $Q=(c,d)\in H$ not on the $y$ -axis. We have the following subcases:

   1. If $cd>0$ , then $H=G$ as in the previous case.
   2. If $cd< 0$ , say $d<0$ (or $0<c$ ), then for some positive integer $n$ , $0<d+nb$ , so that $O\le Q+nP$ , and $H=G$ as well. On the other hand, if $c<0$ (or $0<d$ ), then $-Q$ returns us to the previous argument and $H=G$ again.
   3. If $d=0$ (so $c\ne 0$ ), then either $O\le P+Q$ (when $0<c$ ) or $O\le P-Q$ (when $c<0$ ), so that $H=G$ once more.

   A similar set of arguments shows that if $H$ contains a segment of the $x$ -axis, then either $H$ is the $x$ -axis or $H=G$ . In conclusion, in the case when $ab=0$ , $H$ is either one of the two axes, or the entire group.

3. $ab<0$ . It is enough to assume that $0<a$ and $b<0$ (that $P$ lies in the fourth quadrant), for if $P$ lies in the second quadrant, $-P$ lies in the fourth.

   Since $O,P\in H$ , $H$ could be a subgroup of the line group $L$ containing $O$ and $P$ . No two points on $L$ are comparable, for if $(r,s)<(t,u)$ on $L$ , then the slope of $L$ is positive $$0<\frac{u-s}{t-r},$$ a contradiction. So $L$ , and hence $H$ , is an antichaine. This means that $H$ is convex.

   Suppose now $H$ contains a point $Q=(c,d)$ not on $L$ . We again break this down into subcases:

   1. $Q$ is in the first or third quandrant. Then $H=G$ as in the very first case above.
   2. $Q$ is on either of the axes. Then $H=G$ also, as in case 2(b) above.
   3. $Q$ is in the second or fourth quadrant. It is enough to assume that $Q$ is in the same quadrant as $P$ (fourth). So we have $0<c$ and $d<0$ . Since $L$ passes through $P$ and not $Q$ , we have that $$\frac{a}{c}\ne \frac{b}{d}.$$ Let $0<r=a/c$ and $0<s=b/d$ and assume $r<s$ . Then there is a rational number $m/n$ (with $0<m,n$ ) such that $$r<\frac{m}{n}<s.$$ This means that $na< mc$ and $nb<md$ , or $nP<mQ$ . But $nP,mQ\in H$ , so is $R=mQ-nP\in H$ , which is in the first quadrant. This implies that $H=G$ too.
   In summary, if $H$ contains a point in the second or fourth quadrant, then either $H$ is a subgroup of a line with slope $<0$ , or $H=G$ .

If the Euclidean plane is equipped with the lexicographic ordering, then the story is quite different, but simpler. If $H$ is non-trivial, say $P=(a,b)\in H$ , $P\ne O$ . If $0<a$ , then $(c,d)\le (a,b)$ for any $c< a$ regardless of $d$ . Choose $Q=(c,d)$ to be in the first quadrant. Then $[O,Q]\subseteq H$ , so that $H=G$ . If $a<0$ , then $-P$ takes us back to the previous argument. If $a=0$ , then either $[O,P]$ (when $0<b$ ), or $[O,-P]$ (when $b<0$ ) is a positive interval on the $y$ -axis. This implies that $H$ is at least the $y$ -axis. If $H$ contains no other points, then $H=y$ -axis. In summary, the po-group $\mathbb{R}^2$ with lexicographic order has the $y$ -axis as the only non-trivial proper convex subgroup.

Bibliography

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G. Birkhoff Lattice Theory, 3rd Edition, AMS Volume XXV, (1967).