# Find the supremum and infimum of the following sets

Gold Member
Find the supremum and infimum of the following sets:

$$\begin{array}{l} A = Q \cap (\sqrt 2 ,\infty ) \\ B = \{ n + \sin n|n \in Z^ + \} \\ C = \{ 0.1,0.01,0.001...\} \\ \end{array}$$

From the definition of supremum, it is obvious that sup A does not exist, because for any rational number x/y in set A, (x+1)/y is also in A and is greater than x/y, hence there are no upper bounds.

Im a bit stuck with the infimum of A. I know that it exists because 0 is a lower bound of A, therefore an infumum must exist. I would be inclined to say that it is just sqrt(2), but not really sure how to justify it. Is it true that in every interval, there exists a rational number? If this is the case (which i believe it is), then for any rational number m, there will be some other rational number within the interval (sqrt(2), m), and hence m cannot be the lowest upper bound. Is this reasoning valid?

For set B, sup B will not exist because for any element x + sin x, (x+5) + sin (x+5) is greater and also within the set, hence no upper bound exists (since the sine function has a range small finite range), and inf B will just be 1+sin 1?

I had no troubles with C. I just said that sup C=0.1 and inf C=0.

Does that look right? Its mainly the first set i wasnt sure about, mainly the justification.

## Answers and Replies

D H
Staff Emeritus
Science Advisor
Is it true that in every interval, there exists a rational number? If this is the case (which i believe it is), then for any rational number m, there will be some other rational number within the interval (sqrt(2), m), and hence m cannot be the lowest upper bound. Is this reasoning valid?
Yes, it's valid reasoning, and no, it's not a valid proof -- until you can show that any non-trivial interval in the reals contains a rational. (OTOH, you might have been given this as a theorem in your text, in which case all you have to do is cite the relevant theorem). If you need to prove it, here is a starter hint: Show that for any real number r, the interval [r,r+1) contains an integer.

Gold Member
Yes, it's valid reasoning, and no, it's not a valid proof -- until you can show that any non-trivial interval in the reals contains a rational. (OTOH, you might have been given this as a theorem in your text, in which case all you have to do is cite the relevant theorem). If you need to prove it, here is a starter hint: Show that for any real number r, the interval [r,r+1) contains an integer.

Alright thanks.

Clearly, the width of the interval [r, r+1) is 1. Now ive never had a detailed look at the construction of the natural number and the integers, but i believe the correct terminology is that they are inductive sets by definition, which implies that for every k in the set of integers, k+1 is also in the set i.e. integers have a "distance" of 1 between them, so any continuous interval of width 1 will contain an integer? Like i mentioned, i haven't taken any classes on number theory (being an engineering student) so anything i know about the construction of real numbers is what i have briefly read through while searching for something else.