The author of my calculus book defines an "almost upper bound" as follows: A number [itex]x[/itex] is an almost upper bound for the set [itex]A[/itex] if there are only finitely many number [itex]y \in A[/itex] with [itex]y \geq x[/itex]. He then asks the reader to prove that if [itex]A[/itex] is a bounded infinite set, then the set [itex]B[/itex] of all almost upper bounds for [itex]A[/itex] is non-empty and bounded below. This seemed simple enough but I got confused when I thought about it in terms of a concrete example. Here are my thoughts . . . Let [itex]A = (0,1)[/itex] and [itex]x[/itex] be an almost upper bound for [itex]A[/itex]. Clearly [itex]x[/itex] can be written in the form [itex]x = 1 - \varepsilon[/itex] where [itex]0 < \varepsilon < 1[/itex]. Since [itex]x[/itex] is an almost upper bound, there should only be finitely many numbers [itex]y \in A[/itex] with [itex]y \geq x[/itex]. However, since the infinite sequence of numbers [itex]1-\varepsilon/2, 1-\varepsilon/3, 1-\varepsilon/4, . . . [/itex] are all in [itex]A[/itex] and greater than [itex]x[/itex] this is a contradiction. Could someone help me figure out where I went wrong in my thinking? Thanks.