> real numbers. In an open interval like (0, 1), there is no single real number which is less than one but is greater than all other real numbers less than one.
The rational numbers are like this too.
All the rational numbers between 0 and 1 also increase arbitrarily close to 1 with no largest element. You can prove this by simple contradiction: For every rational number x < 1, there is another rational number y = (x+1)/2 with x < y < 1. So the set of rational numbers less than 1 also has no largest element.
Actually real numbers can be defined (axiomatized) in terms of the least upper bound property -- every set of real numbers that's bounded above has a least upper bound. So you could actually have a set S of rational numbers that gets arbitrarily close to something like sqrt(2) from below. S then has no rational least upper bound -- for every rational number x greater than or equal to every element of S, there is another rational number y greater than or equal to every element of S but smaller than x.
Note that a set of real numbers need not contain its upper bound -- as you noted, a bounded open interval doesn't contain its upper bound.
The rational numbers are like this too.
All the rational numbers between 0 and 1 also increase arbitrarily close to 1 with no largest element. You can prove this by simple contradiction: For every rational number x < 1, there is another rational number y = (x+1)/2 with x < y < 1. So the set of rational numbers less than 1 also has no largest element.
Actually real numbers can be defined (axiomatized) in terms of the least upper bound property -- every set of real numbers that's bounded above has a least upper bound. So you could actually have a set S of rational numbers that gets arbitrarily close to something like sqrt(2) from below. S then has no rational least upper bound -- for every rational number x greater than or equal to every element of S, there is another rational number y greater than or equal to every element of S but smaller than x.
Note that a set of real numbers need not contain its upper bound -- as you noted, a bounded open interval doesn't contain its upper bound.