Prove that for fixed , there is exactly one such that .

First, we prove a lemma.

Lemma: 1 is the smallest element of .

* Proof. * Consider the set . Then, is an inductive set since and for all , . Thus, (since, by definition, is the set of elements common to every inductive set). But, for any , if , then and hence . Thus, 1 is the least element of .

Now, for the theorem of the exercise.

* Proof. * First, we prove existence.

Let . We know is non-empty since by I.3.12, Exercise #2 we know there exist integers and such that . This must be in our set . We also know is bounded above since is an upper bound by our definition of . Therefore, by the least upper bound axiom (Axiom 10, p. 25), we know exists.

Then, by Theorem I.32, we know that for any positive real number , there is some such that

Letting , we have

Therefore, since the supremum of is an upper bound, we know ; therefore, by definition of . But we already know , and so by definition of we must have . Therefore, we have found an integer such that

Now, for uniqueness. Suppose there are integers with this property. Then we have

Then, adding the terms of these inequalities we have,

Without loss of generality, assume . Then this inequality implies, . But since 1 is the smallest element of (from the lemma above) and is closed under subtraction, we must have , i.e., . Thus, any such is unique

I can’t understand how this can be correct let us say x is 5 then there are multiple ‘n’ that is valid with the equation such as:

4.9<5<5.9

4.8<5<5.8…..and so on. how is it possible for only one such 'n' to exist?

Read the proposition again. ‘n’ must be an integer, and in your examples 4.9, 4.8,… aren’t integers.

But S is a set of integers, not reals. How does the Theorem I.32 apply?

I guess that if the theorem applies to all sets of reals, then it has to apply to all sets of integers because integers are reals.

You missed a prime off one of your n primes. Maybe using m instead of n prime would make this proof more readable.

The statement `Then, by Theorem I.32, we know that for any positive real number h, there is some n in S such that n > sup S – h` is incorrect since S only contains integers and is not dense. consider the following example. take x = 4.7 and h = 0.1. sup S – h = 4.6. No n bigger than 4.6 exists

I think it’s also unnecessary, since if is any other integer less than , then , so .

It’s also possible to eliminate the appeal to the lemma: suppose is another integer such that , then , so , contradicting the hypothesis. Therefore, and , as required.

Hmm, the text of my comment got a little truncated. Here’s what I wanted to say:

Suppose is another integer such that . Then , so (since is the supremum). Hence, ; set (such a exists by Theorem I.2). Then . If , then , so , contradicting the hypothesis. So and .

Gott itt…😍

But does have to do with proving the uniqueness of n. In the first part of the proof, to prove that at least one n exists such that n \leq x < n+1, theorem I.32 is used, which is not valid since it is applied no S, which only has integers.

The user that posted the comment showed a counterexample.

Actually, it does work because in the S taken sup S is 4 and not x = 4.7.

This is because by definition of sup, if you take sup S to be 4.7, there is a number 4 which is less than 4.7 and greater/equal to all other elements of S.

You’re assuming that sup S = x, which is wrong.