Understanding Completeness Property Proof

[A.MHF]

Homework Statement

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

Homework Equations

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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?

 

[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.

 

[A.MHF]

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.

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?

 

[Samy_A]

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?

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.

 

[A.MHF]

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.

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 r≤s, 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∉s→q⊄s→s⊆q (according to one of Dedekind's left set linear property)
Also q≤p (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.

 

[Samy_A]

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 r≤s, 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∉s→q⊄s→s⊆q (according to one of Dedekind's left set linear property)
Also q≤p (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.

Yes, this is correct. You prove that p is a real number that doesn't satisfy p≤s, proving that ℝ has no upper bound.
A few minor remarks:
You should emphasize that p≠s.
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.

 

[A.MHF]

Yes, this is correct. You prove that p is a real number that doesn't satisfy p≤s, proving that ℝ has no upper bound.
A few minor remarks:
You should emphasize that p≠s.
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.

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.