# Help understanding a proof

1. Dec 31, 2015

### A.MHF

1. The problem statement, all variables and given/known data

I'm reading Goldrei's Classic Set Theory, and I'm kind of stuck in the completeness property proof, here is the page from googlebooks:

https://books.google.com/books?id=1dLn0knvZSsC&pg=PA14&lpg=PA14&dq="first+of+all,+as+a+is+non+empty"&source=bl&ots=6fwW7jd8i4&sig=txERO1ZYpOKkJ_SVdY82LNMs3io&hl=en&sa=X&ved=0ahUKEwjmxLn_wIbKAhVX-mMKHdPaCSkQ6AEIHzAA#v=onepage&q="first of all, as a is non empty"&f=false

2. Relevant equations
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3. The attempt at a solution
Ok, I guess I understand the first part regarding α=UA.

What I don't get is why are we trying to prove that UA≠ℚ? Also, what does he mean when he says: "As s∈ℝ there is some rational q such that q∉s", how is that possible? if s is a real number, it should include all rational numbers, right? Unless what's meant by that is that there is a rational q that's greater than s, and thus doesn't belong to s?

2. Dec 31, 2015

### Samy_A

A real number is defined as a Dedekind left set.
See the definition of a Dedekind left set on page 9 of the book: it has to be a proper subset of ℚ.

It follows that a real number does not include all rationals.

3. Dec 31, 2015

### A.MHF

Yeah that's what I meant when I said unless q is greater than the real number s,since it's a Dedekind left set, it won't include it. But still, why are we trying to prove that UA≠ℚ?

4. Dec 31, 2015

### Samy_A

Because that's by definition a requirement for a real number (as defined in the book). It may look self-evident that UA≠ℚ (as A is bounded above), but at this stage in the book all you know about real numbers are the definition and some very basic properties. The author uses these to show that UA≠ℚ: a requirement for UA to be a real number.

5. Dec 31, 2015

### A.MHF

Thanks, that was helpful. Regarding the exercise (in the same page), I constructed this proof, what are your thoughts?
Assuming there was such a number s such that rs, for all r∈ℝ then since s∈ℝ, there is q such that q∉s.
Now suppose that p=q+1.
let q be the real number corresponding to q and p the real number corresponding to p.
q={x∈Q:x<q}
p={q∈Q:q<p}
now, q∉sqss⊆q (according to one of Dedekind's left set linear property)
Also qp (from their definitions)
s⊂q doesn't imply that s⊂p, therefore p≤s isn't satisfied and thus r≤s isn't satisfied as well.

6. Dec 31, 2015

### Samy_A

Yes, this is correct. You prove that p is a real number that doesn't satisfy ps, proving that ℝ has no upper bound.
A few minor remarks:
You should emphasize that ps.
There are some typo's in your post:
p={q∈Q:q<p} is confusing, as q has already been used. Better write p={x∈ℚ:x<p}.
And s⊂q does imply that s⊂p, not doesn't as you wrote.

7. Dec 31, 2015

### A.MHF

Oh I see.
So I should define p=q+1 in which the corresponding real number is p={x<ℚ:x<p} such that p≠s.
And
s⊂q → s⊂p, thus we proved that there is a a real number p such that it's greater than q yet doesn't satisfy p≤s, therefore ℝ has no upper bound.