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?

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- Oct 1st 2010, 07:54 AMcalculuskid1Bolzano’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, 04:41 PMemakarov
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, 05:50 PMcalculuskid1
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, 06:13 PMIondor
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