# Bolzano’s theorem and the Archimedian property of R.

• Oct 1st 2010, 06:54 AM
calculuskid1
Bolzano’s theorem and the Archimedian property of R.
Use Bolzano’s theorem and the supremum to prove the Archimedian property of R.
I know bolzano's theorem shows that supA exists and the Arc.property of R shows that n>x.. but how do I use those two methods to prove one another?
• Oct 1st 2010, 03:41 PM
emakarov
Could you describe the Bolzano’s theorem and the Archimedian property more precisely? This is not Bolzano–Weierstrass theorem, is it? Or is it the extreme value theorem?
• Oct 1st 2010, 04:50 PM
calculuskid1
A⊂R is bounded, A≠0, and A⊂R is bounded above, SupA exists.
This is proven by bisection.
Thats all I have on his theorem
• Oct 1st 2010, 05:13 PM
Iondor
The archimedian property is: For any x, there exists a natural number n with n>x.

You can prove that R has the archimedian property this way:
Let x>0 and
A={n : n is a natural number and n<=x}
Then A is bounded above by x and 0\in A,
so sup A=s exists.
Since s is the least upper bound of A, s-1 is not an upper bound of A anymore.
So there exists n \in A with s-1<n
=>
s<n+1
Since s is an upper bound of A
=> n+1 \notin A
=>
x<n+1

i.e. m:=n+1 is a natural number with m>x.
qed