Proof of the Theorem: Omega and D Relationship

[Terrina]

I need to prove the Theorem: If omega is equal to infinity then D =[0,infinity) and if omega<infinity, then D =[o,infinity].
Where omega=the least upper bound of D and D=the set of all distances that occur between points of the plane.

I'm really just not sure where to start, but I'm pretty sure it uses the following axioms:
*For any point (A) on a line (m), there exists a point (B) on (m) with 0<AB<omega.

and

*For any ray AB and any real number s with 0< s< omega, there is a point X in the ray AB with AX=s.


Any help would be great. Thanks!

 

[member 553083]

We can prove this theorem using the two axioms stated above. Let D = the set of all distances that occur between points of the plane. We want to show that if omega is equal to infinity, then D = [0,infinity), and if omega < infinity, then D = [0,omega]. First, we consider the case where omega is equal to infinity. By the first axiom, given any point A on a line m, there exists a point B on m such that 0 < AB < omega. Since omega = infinity, this means that for any point A on a line m, there exists a point B on m such that 0 < AB < infinity. This implies that every distance AB between points of the plane is greater than or equal to 0 and less than infinity. Thus, D = [0,infinity). Now, consider the case where omega < infinity. By the second axiom, given any ray AB and any real number s with 0 < s < omega, there is a point X in the ray AB with AX = s. Since omega is less than infinity, this means that for any ray AB and any real number s with 0 < s < omega, there is a point X in the ray AB with AX = s. This implies that every distance AB between points of the plane is greater than or equal to 0 and less than or equal to omega. Thus, D = [0,omega]. Therefore, if omega is equal to infinity then D =[0,infinity) and if omega<infinity, then D =[o,infinity], as desired.

 

[thepatientmental]

To prove this theorem, we will use a proof by contradiction. We will assume that the statement is false and show that it leads to a contradiction.

Assuming that omega is equal to infinity, we will show that D must be equal to [0,infinity).

By the first axiom, we know that for any point A on a line m, there exists a point B on m with 0<AB<omega. Since omega is equal to infinity, this means that for any point A on m, there exists a point B on m such that AB is infinite.

Now, let's consider the set D. Since D is the set of all distances that occur between points of the plane, this means that for any two points A and B on the plane, the distance AB must be in D.

But we just showed that for any point A on m, there exists a point B on m such that AB is infinite. This means that AB is also in the set [0,infinity), since it includes all values from 0 to infinity.

Therefore, we have shown that for any two points A and B on the plane, the distance AB is in [0,infinity). This means that D=[0,infinity), which is what we wanted to prove.

Now, let's assume that omega is less than infinity. We will show that D must be equal to [0,infinity].

By the second axiom, we know that for any ray AB and any real number s with 0<s<omega, there is a point X in the ray AB with AX=s.

Since omega is less than infinity, this means that there exists a point X on the ray AB such that AX is less than infinity. In other words, AX is a finite value.

Now, let's consider the set D. As before, for any two points A and B on the plane, the distance AB must be in D.

But we just showed that for any ray AB, there exists a point X on AB such that AX is finite. This means that for any two points A and B on the plane, the distance AB is also finite.

Therefore, we have shown that for any two points A and B on the plane, the distance AB is in the set [0,infinity]. This means that D=[0,infinity], which is what we wanted to prove.

Thus, we have proven the theorem: If omega