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

1. ## 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?

2. 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?

3. 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

4. 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