Re: Arkhimedes theorem and cardianlity

From: Shmuel (Seymour J.) Metz (spamtrap_at_library.lspace.org.invalid)
Date: 01/10/05 

Date: Mon, 10 Jan 2005 18:04:06 -0500

In <200501101208.j0AC86s31221@proapp.mathforum.org>, on 01/10/2005
   at 01:04 PM, not.adomain@foo.com (joccis) said:

>The corollary, a-e < x < a < y < a+e : a belongs to R and e>0 and x,y
>belong to Q, is a direct consequence of the Arkimede's theorem.

Are you trying to say that for any a \in R and any e>0 there exist x
and Y \in Q such that a-e < x < a < y < a+e? If so, that is indeed a
consequence of the Axiom of Archimedes. However, x and y are not
unique.

>Clearly we have a relation R->Q.

No, a relation R:QxQ, given as {(x,(a,b)), X \in R, a \in Q, b \in Q|
a<x<b}

>At first glance this looks very much lika a bijection,

No, it doesn't even look like a function, and it is not 1- in either
direction.

>could someone explain why the relation actually is not a bijection
>allowing us to list the real numbers.

See above; not only is it not a bijection, but for given rational a
and b the inverse image of {(a,b)} is uncountable. 

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Shmuel (Seymour J.) Metz, SysProg and JOAT  <http://patriot.net/~shmuel>
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