Section C.1 Intervals

Every interval listed above has the stated property. Conversely, let \(I\subseteq\R\) have this property. If \(I=\emptyset\text{,}\) then \(I=(0,0)\text{;}\) if \(I=\{a\}\text{,}\) then \(I=[a,a]\text{.}\) Thus, we may assume that \(I\) is nondegenerate.

First suppose that \(I\) is bounded below but not above. Let \(a=\inf I\text{.}\) If \(z>a\text{,}\) then \(z\) is not a lower bound of \(I\text{,}\) so there is \(x\in I\) with \(x<z\text{.}\) Since \(I\) is not bounded above, there is also \(y\in I\) with \(z<y\text{.}\) Since \(x<z<y\text{,}\) the assumed property gives \(z\in I\text{.}\) Thus \((a,\infty)\subseteq I\text{.}\) On the other hand, no element of \(I\) can be less than \(a\text{.}\) Hence \(I=(a,\infty)\) or \(I=[a,\infty)\text{,}\) according as \(a\notin I\) or \(a\in I\text{.}\)

The case in which \(I\) is bounded above but not below is similar: if \(b=\sup I\text{,}\) then \(I=(-\infty,b)\) or \(I=(-\infty,b]\text{,}\) according as \(b\notin I\) or \(b\in I\text{.}\)

If \(I\) is neither bounded below nor bounded above, then for every \(z\in\R\) we can choose \(x,y\in I\) with \(x<z<y\text{.}\) The assumed property gives \(z\in I\text{,}\) so \(I=\R\text{.}\)

Finally, suppose that \(I\) is bounded both below and above. Let \(a=\inf I\) and \(b=\sup I\text{.}\) Since \(I\) is nondegenerate, we have \(a<b\text{.}\) If \(a<z<b\text{,}\) then \(z\) is neither a lower bound nor an upper bound of \(I\text{.}\) Thus, there exist \(x,y\in I\) with \(x<z<y\text{,}\) and hence \(z\in I\text{.}\) Therefore \((a,b)\subseteq I\text{.}\) Also, every element of \(I\) lies in \([a,b]\text{.}\) Consequently, \(I\) is one of \((a,b)\text{,}\) \([a,b]\text{,}\) \((a,b]\text{,}\) or \([a,b)\text{,}\) according to whether \(a\) and \(b\) belong to \(I\text{.}\)

Choose \(a,b\in I\) with \(a<b\text{.}\) By Proposition C.1.1, \((a,b)\subseteq I\subseteq\R\text{.}\) The map \(x\mapsto(x-c)/r\text{,}\) where \(c=(a+b)/2\) and \(r=(b-a)/2\text{,}\) is a bijection \((a,b)\to(-1,1)\text{.}\) Also, \(t\mapsto\tan(\pi t/2)\) is a bijection \((-1,1)\to\R\text{.}\) Hence \(|\R|=|(a,b)|\le|I|\le|\R|\text{,}\) so \(|I|=|\R|\text{.}\)

Let \(I,J\) be intervals. If \(I\cap J\) has fewer than two points, then it is an interval. Otherwise, let \(x,y\in I\cap J\) with \(x<y\text{.}\) Since \(I\) and \(J\) are intervals, \([x,y]\subseteq I\) and \([x,y]\subseteq J\text{.}\) Thus, \([x,y]\subseteq I\cap J\text{,}\) and Proposition C.1.1 shows that \(I\cap J\) is an interval.

Let \(I,J\) be overlapping intervals, and choose \(a\in I\cap J\text{.}\) Let \(x,y\in I\cup J\) with \(x<y\text{.}\) If \(x\) and \(y\) belong to the same one of the intervals, then \([x,y]\) is contained in that interval and hence in \(I\cup J\text{.}\) Otherwise, after relabeling, we may assume that \(x\in I\setminus J\) and \(y\in J\setminus I\text{.}\) If \(a<x\text{,}\) then \(x\in[a,y]\subseteq J\text{,}\) a contradiction. If \(y<a\text{,}\) then \(y\in[x,a]\subseteq I\text{,}\) also a contradiction. Therefore, \(x\le a\le y\text{.}\) Consequently, \([x,y]=[x,a]\cup[a,y]\subseteq I\cup J\text{.}\) By Proposition C.1.1, \(I\cup J\) is an interval.