Exercise statement
Let 
be a non-empty subset of 
, let 
be an integer, and let 
be integers with the properties that 
and 
are upper bounds for , but 
and 
are not upper bounds for . Show that 
. This shows that the integer 
constructed in Exercise 5.5.2 is unique.
Hints
- Drawing a picture will be helpful.
How to think about the exercise
This is a simple exercise, so I don’t have much to say.
I want to comment on one part of the proof below, which is where I go from 
to 
. This is valid when are both integers—you can think of it as a generalization of Proposition 2.2.12(e). How would we prove this? If 
are integers, we can add 
to both sides to get 
. Now both sides are natural numbers, so we can apply Proposition 2.2.12(e) to get 
, and finally we can add 
to both sides to get 
as required. The other direction is similar, but we don’t need it for this exercise so I will omit it.
Model solution
Since is not an upper bound for , there exists some 
such that 
. Similarly, since is not an upper bound for , there exists some 
such that 
. Thus we have 
and 
using the fact that and are upper bounds for . This implies and 
(by multiplying through by 
, which we can do since is positive). Since are integers, we have and 
, which means as required.
Tao Analysis Solutions 
You can also use Order of Trichotomy on Real Numbers, i.e., m < m’, then 0<m’-m. Therefore, we have a positive integer/natural number. Hence 1 <= m’-m or m <= m’-1, hence m/n <= (m’-1)/n a contradiction. similarly for m'<m. Just posting this as an idea.
LikeLike
You could also argue using either m < m’ that by transitivity this means x < m’, not accounting for a supremum being part of the set x is part of.
LikeLike
I do not understand, what do you mean ” x<m’ “, how may that help us?
LikeLike