What is the Difference Between Almost Upper Bounds and Upper Bounds in Calculus?

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.
 
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I was very confused.when i first came acrosse this.best tthing to do is try and remeber and "swat for yur exmas" it worked for me
 
Why must x be in A? Wouldn't 1 be an almost upper bound of that set, since there are 0 elements in A greater than or equal to 1?
 
Tiny-tim: This is from Spivak's Calculus

Thanks for the explanation guys! I guess I just wasn't thinking about this the right way.
 
For an interval like that, an "almost upper bound" is exactly the same as an upper bound.

If the set were, say (0, 1)\cup \{2\} any number larger than or equal to 2 would be an upper bound but any number larger than or equal to 1 would be an "almost upper bound". Both sets are non-empty and bounded below.

For a finite set finite set all numbers are "almost upper bounds"- which is why that theorem specifies that the set must be infinite.
 
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