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Given an ordered set $X$, say a subset $Y$ of $X$ is convex in $X$ if for each pair of points $a<b$ of $Y$, the entire interval $(a,b)$ of points of $X$ lies in $Y$. An interval is of the form $(a,b),(a,b],[a,b),[a,b]$, and a ray is of the form $(a,+\infty),(-\infty,a),[a,+\infty),(-\infty,a]$.

Let $X$ be an ordered set. If $Y$ is a proper subset of $X$ that is convex in $X$, does it follow that $Y$ is an interval or a ray in $X$?

Well, suppose $Y$ has a smallest element $a$ and a largest element $b$. Then $[a,b]\in Y$, and so $Y=[a,b]$ and $Y$ is an interval.

What if $Y$ has no smallest element? So for each element $a\in Y$, there exists an element $b\in Y$ such that $b<a$. What then?

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Paul S.
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1 Answers1

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HINT: Let $X$ be the set of non-zero real numbers with the usual order, and let $Y$ be the set of positive real numbers.

Brian M. Scott
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  • I realize that (0,infinity) is not a ray as Munkres defines it since 0 is not an element of X, however, why does he define it that way? It seems like we should be free to use elements outside the set to help us simply describe a set we want to talk about. – H_1317 Mar 02 '18 at 04:45
  • @H_1317 Same reason end points of open intervals in X are elements of X, I think. Or do you ask about open intervals too? Btw, what then of closed rays, closed intervals or half-open intervals? – BCLC Sep 11 '18 at 07:54
  • @H_1317: (I apparently missed your question back in 2018; an answer may still be useful to some reader.) Let $$X_1=\big({0}\times Y\big)\cup\big(\Bbb R\times{0}\big),,$$ and define a linear order $\preceq$ on $X_1$ by letting $\langle x,0\rangle\preceq\langle y,0\rangle$ iff $x\le y$, $\langle 0,x\rangle\preceq\langle 0,y\rangle$ iff $x\le y$, and $\langle 0,x\rangle\preceq\langle y,0\rangle$ for all $x,y\in\Bbb R$. ... – Brian M. Scott Oct 20 '21 at 21:35
  • ... Let $Y_1=\Bbb R\times{0}$, and let $h:X_1\to X$ be defined as follows: $h(\langle 0,x\rangle)=-\frac1x$, and $h(\langle x,0\rangle=e^x$. Then $h$ is an order-isomorphism between $\langle X_1,\preceq\rangle$ and $\langle X,\le\rangle$ that sends $Y_1$ to $Y$, so from an order-theoretic point of view the structure $\langle X_1,Y_1,\preceq\rangle$ is indistinguishable from the structure $\langle X,Y,\le\rangle$. In particular, $Y_1$ is a convex subset of $X_1$. And when I present the example this way, there is no handy point that you could use to make $Y_1$ a ray. The point $0$ works ... – Brian M. Scott Oct 20 '21 at 21:47
  • ... in the original presentation only because I embedded the order in $\Bbb R$ in a very nice way. When I embed it in $\Bbb R^2$ as I did above, there is no such point, yet it is for all practical purposes the same linear order with the same convex subset. In short, when answering the question of whether $Y$ is a ray in $X$, you are working strictly in $X$ and have no access to anything outside of $X$. – Brian M. Scott Oct 20 '21 at 21:50
  • @BrianM.Scott What about $X = \mathbb{Q}$ with usual order inherited from $\mathbb{R}$, and $Y = (0, \pi) \cap \mathbb{Q}$? – Kalas678 Oct 22 '21 at 18:44
  • @Sushant: Very good example: in that case $Y$ is convex but is neither an interval nor a ray in $X$. – Brian M. Scott Oct 22 '21 at 22:42