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Coninuity Axiom

    This topic defines a Dedekind cut and proves the Dedekind Axiom implies the Archimedes Axiom. After introducing the measure of a segment and an angle, the triangular inequality and the Saccheri-Legendre Theorem are proven.

Axiom (Archimedes Axiom) If is any segment, any point, and any ray with vertex then for every point on there is a number such that when is laid off times on starting at , a point is reached such that and either or is between and

Axiom (Dedekind Axiom) Suppose there exists two subsets and of with the following properties:

    (i)
    
    (ii)
    
    (iii)
    
    (iv)
    
    (v) , and
        
    (vi)   .

Then there exists a unique point on such that one of the subsets is equal to a ray of with vertex and the other subset is equal to the complement.

Definition (Dedekind Cut Of A Line) The subsets and on a line with the properties in the Dedekind Axiom are called a Dedekind cut of the line

    In a similar manner a Dedekind cut for a ray can be defined and a proposition corresponding to the Dedekind Axiom (but for rays) can be proven from the Dedekind Axiom.

Proposition (Dedekind Axiom Implies Archimede Axiom) The Archimede Axiom is a consequence of the Dedekind Axiom and the Hilbert Axioms of Incidence, Betweenness, and Congruence.

    Proof. Given a segment and a point on line with ray of emanating from Let consist of and all points on reached by laying off copies of segment on starting from Let be the complement of in If is empty, then the Archimede axiom holds, so assume that is nonempty.
    We will show that and make a Dedekind cut of We are assuming is nonempty and clearly is non empty because we lay off at least one copy of Since is the complement of in we note that Obviously, since is the complmenet of
    Next, suppose two points are in with say then holds; to see this, let be between and Suppose could be reached by laying off some numbers of copies of starting at , then by the Betweenness Property, is also reached by contradicting Thus, Similarly,  when and are two points in it follows So we have a Dedekind cut of
    Let be the point of furnished by the Dedekind Axiom.
    Case 1. Suppose Then for some number can be reached by laying off copies of segment on starting from By laying off one more copy of we can reach a point in but by definition of that is not possible.
    Case 2. Suppose Lay off a copy of on the ray opposite to starting at obtaining a point lies on so Then for some number can be reached by laying off copies of Whence, can be reached by laying off one more copy of starting from contradicting Therefore, is empty which means that every point on can be reached by laying off some number of copies of so the Archimede axiom holds.

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