jgens
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The author of my calculus book defines an "almost upper bound" as follows: A number x is an almost upper bound for the set A if there are only finitely many number y \in A with y \geq x.
He then asks the reader to prove that if A is a bounded infinite set, then the set B of all almost upper bounds for A 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 A = (0,1) and x be an almost upper bound for A. Clearly x can be written in the form x = 1 - \varepsilon where 0 < \varepsilon < 1. Since x is an almost upper bound, there should only be finitely many numbers y \in A with y \geq x. However, since the infinite sequence of numbers 1-\varepsilon/2, 1-\varepsilon/3, 1-\varepsilon/4, . . . are all in A and greater than x this is a contradiction.
Could someone help me figure out where I went wrong in my thinking? Thanks.
He then asks the reader to prove that if A is a bounded infinite set, then the set B of all almost upper bounds for A 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 A = (0,1) and x be an almost upper bound for A. Clearly x can be written in the form x = 1 - \varepsilon where 0 < \varepsilon < 1. Since x is an almost upper bound, there should only be finitely many numbers y \in A with y \geq x. However, since the infinite sequence of numbers 1-\varepsilon/2, 1-\varepsilon/3, 1-\varepsilon/4, . . . are all in A and greater than x this is a contradiction.
Could someone help me figure out where I went wrong in my thinking? Thanks.