Exercise 16.7

Proof. As a counterexample consider $X=\mathbb{Q}$ and the proper subset $Y=\left\{x\in \mathbb{Q}\mid x^2<2\right\}$. We claim that $Y$ is convex but not an interval or a ray.

First, consider $a,b\in Y$ where $a<b$, thus $a^2,b^2<2$. Also consider $x\in (a,b)$ so that $a<x<b$. If $x\ge 0$ then $0\le x<b$ so that $x^2<b^2<2$. If $x<0$ then $a<x<0$ so that $2>a^2>x^2$. Thus in either case $x^2<2$ so that $x\in Y$. Since $x$ was arbitrary, this shows that $(a,b)\subset Y$ so that $Y$ is convex since $a$ and $b$ were arbitrary.

Now, clearly, $Y$ cannot be a ray with no lower bound since then there would be an $x$ in the ray where $x<-2$ so that $x^2>4>2$ and hence $x\notin Y$. Similarly, $Y$ cannot be a ray with no upper bound since then the ray would contain an $x>2$ so that $x^2>4>2$ and thus $x\notin Y$. So suppose that $Y=[a,b]$ for some $a,b\in X=\mathbb{Q}$ where $a\le b$. Now, it cannot be that $b^2=2$ since then $b=\sqrt{2}$, which is not rational. Similarly, it cannot be that $a^2=2$ for the same reason.

Case: $b^2<2$. Then there is a rational $p$ where $b<p<\sqrt{2}$ since the rationals are order-dense in the reals. Let $x=\max <!--\; FUNCTION\; APPLICATION\; -->(0,p)$ so that $b<p\le x$ and hence $x\notin [a,b]$. However, if $0<p$ then $x=p$ so that $x^2=p^2<2$, and if $0\ge p$ then $x=0$ so that $x^2=0<2$. Thus either way $x\in Y$ and $x\notin [a,b]$, which shows that $Y$ cannot be $[a,b]$.

Case: $b^2>2$. Then $\sqrt{2}<b$ since $0<2<b^2$. If $\sqrt{2}<a$ then clearly for any $x\in [a,b]$ we have that $0<\sqrt{2}<a\le x$ so that $2<x^2$ and hence $x\notin Y$. If $a<\sqrt{2}$ then there is a rational $p$ such that $a<\sqrt{2}<p<b$ since the rationals are order-dense in the reals. Hence $2<p^2$ so that $p\notin Y$. Either way, there is an $x\in [a,b]$ where $x\notin Y$ so that $Y$ cannot be $[a,b]$.

Similar arguments show that neither $Y=(a,b)$, $Y=[a,b)$, nor $Y=(a,b]$ for $a,b\in X=\mathbb{Q}$ and $a<b$. Hence $Y$ cannot be an interval. Thus $Y$ is convex but neither an interval nor a ray in $X$. This shows the desired result. □