Continuity 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
![continuity axiom _gr_17.gif]](pages/continuity-axiom/Images/continuity-axiom_gr_17.gif)
Axiom (Dedekind Axiom) Suppose there exists two subsets
and
of
with the following properties:
(i)
![continuity axiom _gr_22.gif]](pages/continuity-axiom/Images/continuity-axiom_gr_22.gif) (ii)
![continuity axiom _gr_23.gif]](pages/continuity-axiom/Images/continuity-axiom_gr_23.gif) (iii)
![continuity axiom _gr_24.gif]](pages/continuity-axiom/Images/continuity-axiom_gr_24.gif) (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.
Cite this as: Continuity Axiom Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/continuity-axiom.html
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