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Elementary Continuity Principle

We will now take up the Axioms of Continuity. First, we will prove the two Continuity Principles and then discuss some of the different uses of the Continuity Axioms in our work.

First, we shall need the famous Triangle Inequality. It is usually proved after we have given a measure to line segments, but that is not necessary.

Proposition 10.1:[Triangle Inequality]   If A, B, and C are three noncollinear points, then AC<AB+BC, where the sum is segment addition.

 figure2342
Figure 11.1: Triangle inequality 

Proof: There is a unique point D such that tex2html_wrap_inline12520 and tex2html_wrap_inline14756 by the first congruence axiom. Then tex2html_wrap_inline14758 by Proposition 8.1. Now, tex2html_wrap_inline14760 and since tex2html_wrap_inline14756 we have that tex2html_wrap_inline14764. By Proposition 7.7 tex2html_wrap_inline14766 is between the rays tex2html_wrap_inline14768 and tex2html_wrap_inline14770. Then, by definition, tex2html_wrap_inline14772. By Proposition 8.12 and what we have just shown, tex2html_wrap_inline14774. Thus, by Proposition 10.5 AD>AC, and we are done.

Theorem 10.1:[Elementary Continuity Principle]  If one endpoint of a segment is inside a circle and the other outside, then the segment intersects the circle.

Proof: Let tex2html_wrap_inline14778 be a circle centered at O with radius OR, so that tex2html_wrap_inline14784. Let AB be the above segment. Let tex2html_wrap_inline14788 and tex2html_wrap_inline14790. Let
eqnarray2369
Note that neither of tex2html_wrap_inline12652 or tex2html_wrap_inline12654 is empty. We wish to show that these two sets form a Dedekind cut of the segment AB. There are then several things to show.

  1. If tex2html_wrap_inline14798 then by Proposition 8.4 we must have that tex2html_wrap_inline14800. This puts X in one of the two sets. Therefore, tex2html_wrap_inline14804.
  2. Clearly, tex2html_wrap_inline14806.
  3. We have that tex2html_wrap_inline14808 and tex2html_wrap_inline14810.
  4. Let tex2html_wrap_inline14812 and tex2html_wrap_inline14814. We need to show then that tex2html_wrap_inline14816. If tex2html_wrap_inline14818 then tex2html_wrap_inline14820. tex2html_wrap_inline14822 is exterior to tex2html_wrap_inline14824 so tex2html_wrap_inline14826. Thus, by Proposition 10.5 OC<OX. Since OX<OR by Proposition 8.4 OC<OR and tex2html_wrap_inline14816.

    Now, assume that tex2html_wrap_inline14836. Then, for the first case, assume that OA<OX. Hence, tex2html_wrap_inline14840. Again, tex2html_wrap_inline14842 which implies OC<OX<OR. Thus, tex2html_wrap_inline14816. If OA>OX reverse the labels in the previous argument and we again find that tex2html_wrap_inline14816.

Thus, tex2html_wrap_inline12652 and tex2html_wrap_inline12654 form a Dedekind cut of AB. Let M be the cut point.

CLAIM: tex2html_wrap_inline14860.

Assume OM<OR. This puts tex2html_wrap_inline14864 and so tex2html_wrap_inline14866. Now, there is a segment which is less than the difference between OM and OR. Let M' be a point outside tex2html_wrap_inline14778 and on AB so that MM' is congruent to that segment. Thus, we have that tex2html_wrap_inline14880. By the Triangle Inequality we have that
eqnarray2380
But, tex2html_wrap_inline14880 so that tex2html_wrap_inline14884: a contradiction.

Assuming that OM>OR we use the same technique to arrive at a contradiction. Thus, tex2html_wrap_inline14860 and tex2html_wrap_inline14890.

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Next: Measure of Angles and Up: Theorems of Continuity Previous: Theorems of Continuity

david.royster@uky.edu