# Upper Bound Confusion

1. Sep 12, 2009

### jgens

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.

2. Sep 12, 2009

### pep_i

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

3. Sep 12, 2009

### Tobias Funke

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?

4. Sep 12, 2009

### tiny-tim

Last edited by a moderator: Apr 24, 2017
5. Sep 12, 2009

### jgens

Tiny-tim: This is from Spivak's Calculus

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.