Ok, so I have 2 problems i've been working on and I want to see if I'm anywhere close to getting it right. Pretty new to the whole abstract math scene.
The problems are defined as: X,≤ is a totally ordered set which is not empty. a,b ∈ X and a < b, the open interval (a,b) is defined by
(a, b) = {x ∈ X : a < x < b}
1) Show that if c < a and b < d then (a,b)⊆(c,d)
2) Assume that {(an, bn)}n∈N is a growing set of open intervals i.e. an+1 ≤ and bn ≤ bn+1. Assume that there are elements c, d ∈ X such that c ≤ an og bn ≤ d for all n. Show that if (X, ≤) has the least upper bound property then S n∈N (an, bn) is an open interval. Show by example that this conclusion do not have to be right if (X, ≤) lacks the least upper bound property.
I'll try and keep this short. Now for 1) I started by showing that (a,b)R(c,d) is reflexiv, antisymmetric and transative. Reflexive is pretty straight forward. for antisymmetric I used
(a, b) = {x ∈ X : a < x < b}
(c, d) = {x ∈ X : c < x < d}
So for a value x (a,b)=(c,d) (first trace of doubt here)
Transitive
c < a and b < d
e < c and d < f
e < a and b < f
Is this enough? In my mind showing the relation is pretty much stating that they are both sets in X, and given that c < a and b < d then (c,d) then (a,b)⊆(c,d).
For 2) I have that for (X,≤) you have the set S = {b ≤ b2 ≤ ... bn+1} then d carries all the properties it requires to be upper bound of S and using that I can show that bn+1 carries the properties of the least upper bound.
for the last part of the question I assumed that if n goes towards infinity then there can be no upper bound our least upper bound properties.
Sorry for the length of this, I tried to keep it as short as possible by skipping a few steps. I just want to find out if I'm getting anywhere closer at all to cracking the code here.
If I understand the question I think you are right on both of these.
1) if x is in (a,b) then
x > a; a > c so by transitivity x > c.
x < b; b < d so by transitivity x < d.
So c < x < d. so x is in (c, d) for all points of x in (a, b). So (a, b) is a subset of (c, d).
2) Your notations and typing was a bit hard to read but I'm assuming the question is asking:
If $(a_n, b_n);$ n in N so that $a_0 > a_1 > ... $ and $b_0 < b_1 < ... $ which inductively by the above problem mean the are nested subsets (or "growing as you put it"). Assume there are c and d where c <= $a_n$ and d >= $b_n$ for all $a_n$ and $b_n$. Show that if X has the least upper bound property then the union of all ($a_n$, $b_n$) is an open interval (a, b) for some numbers a and b. Show this need not be true if X doesn't have the least upper bound property.
Is this right?
You are on the right track Since c <= $a_n$ for all $a_n$ {$a_n$} is bounded bellow and has an greatest lower bound. Call it a. Similarly since d >= $b_n$ for all $b_n$, {$b_n$} is bounded above and has a least upper bound. Call it b.
As all $a_n$ >= a an all $b_n$ <= b, ($a_n$, $b_n$)⊆(a,b) and the Union of all such ($a_n$, $b_n$) is a subset of (a, b).
If x is in (a, b), the a < x < b. As a is inf{$a_n$}, x is not a lower bound of {$a_n$} so there is an a <= $a_j$ < x. Likewise there exists a x < $b_k$ < b. Let n = max(j, k). So a <= $a_n$ <= $a_j$ < x < $b_n$ <= $b_j$ <= b. So x is in ($a_n, $b_n). So (a, b) is a subset of the union.
So (a, b) is the union.
Okay... so why wouldn't that be true if X doesn't have the least upper bound property? Well, there might not be any a and b that are extreme bounds. For example; If X is the rational numbers (which doesn't have the least upper bound property) and {$a_n$} are a decreasing sequence of rational number heading toward $\sqrt{2}$ and {$b_n$} are a sequence of rational numbers heading toward $\sqrt{3}$, then the union will be all the rational numbers between $\sqrt{2}$ and $\sqrt{3}$ but this is not an open interval for any (a, b) as there are no possible rational endpoints.