Can This Proof Verify the Existence of Integers m and n for Any Real Number x?

[Physics2341313]

1. Question I 3.12 from Apostle's calculus volume 1
If x is an arbitrary real number, prove that there are integers m and n such that m<x<n



2. Theorem I.27:Every nonempty set S that is bounded below has a greatest lower bound; that is, there is a real number such that L= infS



3. Suppose x$\in$R and belongs to a nonempty set S with no maximum element. It follows that ∃B$\in$Z$\stackrel{+}{}$ such that B is an upper bound for S. Let n=B, then we have n>x. Similarly, let -S denote the set of negatives of numbers in S. Suppose that -B$\in-Z$$\stackrel{+}{}$ of -S, then by theorem I.27 -B is a lower bound of -S. Let m=-B. , such that m<x<n.

 

[Fredrik]

Suppose x$\in$R and belongs to a nonempty set S with no maximum element.

You have stated this as if you're making two assumptions about x. You should only make one: $x\in\mathbb R$. If you need to use some other statement about x, you need to be able to prove that it follows from your one and only assumption about x.

It follows that ∃B$\in$Z$\stackrel{+}{}$ such that B is an upper bound for S.

It doesn't. Consider e.g. x=1 and S the set of integers (which doesn't have a maximum element).

 

[Physics2341313]

I thought an inductive set could have an upper bound but not a greatest upper bound / supremum because for every x in the set x+1 is also in the set?

and to fix my assumption issue could i do something along the lines of- Suppose S is a nonempty set of real numbers and suppose there is a number B such that x$\leq$B so B could be an upper bound and then go an to show the lower bound as well equating m and n, the integers, as a subset of the real numbers?

 

[Fredrik]

I thought an inductive set could have an upper bound but not a greatest upper bound / supremum because for every x in the set x+1 is also in the set?

The set of natural numbers is an inductive set, but it doesn't have an upper bound in $\mathbb R$. This would contradict the result that you're trying to prove.

and to fix my assumption issue could i do something along the lines of- Suppose S is a nonempty set of real numbers and suppose there is a number B such that x$\leq$B so B could be an upper bound and then go an to show the lower bound as well equating m and n, the integers, as a subset of the real numbers?

If you let S be an arbitrary non-empty subset of $\mathbb R$ and let B be an arbitrary real number such that $x\leq B$, then what set would B be an upper bound of? Certainly not S. S is arbitrary, so we could have $S=\{B+1\}$. B is an upper bound of {x}, but that's not very useful.

Edit: I'll give you a hint about one way to solve the problem. The set $S=\{n\in\mathbb Z|n<x\}$ is useful.