How would I prove the following?
Show that the assertion of the Bolzano-Weierstrass Theorem is equivalent to the Completeness Axiom for real numbers.
I know that the Nested Set Theorem implies the Bolzano-Weierstrass Theorem, and that the Completeness Axiom implies the Nested Set Theorem, so I understand how Completeness Axiom $\Rightarrow$ Bolzano-Weierstrass. How do I show that the Bolzano-Weiertsrass Theorem implies the Completeness Axiom?
Best Answer
You can try to prove Nested Interval Principle using Bolzano Weierstrass Theorem. Just choose one point from each interval in the sequence of nested intervals such that these points are distinct and then the chosen infinite set of such points possesses an accumulation point $c$. This $c$ must lie in all the nested intervals (why?)
Next show that Nested Interval Principle implies Completeness Axiom. Let $A$ be a set bounded above by $b$ and let $a$ be some number less than some element of $A$. Consider the interval $[a, b] $. If the midpoint $c=(a+b) /2$ is an upper bound for $A$ then choose $a_{1}=a,b_{1}=c$ otherwise choose $a_{1}=c,b_{1}=b$. Using same procedure get $a_{2},b_{2}$ from $a_{1},b_{1}$ and repeat the process indefinitely to get a sequence of nested intervals. There is a unique $c$ which lies in all these intervals and one can easily show that $c=\sup A$ (how?)