I can prove that nested interval property implies Cauchy convergence just by building intervals that become smaller and smaller and the diameter dwindles down to zero, intersecting in only one point.
But, how do I prove that the limit of the Cauchy sequence is the intersection of all these intervals?
Also, how do I go about Cauchy convergence implies Nested interval property?
I can build intervals using the definition of Cauchy but how do I prove that the intervals dwindle down to the limit of the cauchy sequence?
Best Answer
Assume that ${\mathbb R}$ possesses the NIP, and let $(x_n)_{n\geq1}$ be a Cauchy sequence in ${\mathbb R}$. We have to prove that this sequence converges to a point $\xi\in{\mathbb R}$. For the proof we have to construct a feasible sequence of nested intervals.
There is an increasing sequence $k\mapsto N_k$ $(k\geq1)$ such that $$|x_m-x_n|\leq {1\over k}\qquad\forall\>m, \>n\geq N_k\ .$$ Put $x_{n_k}=:y_k$, and define $$I_0:={\mathbb R},\qquad I_k:=\left[y_k-{1\over k}, \> y_k+{1\over k}\right]\ \cap\>I_{k-1}\quad(k\geq1)\ .$$ Since $n_k\geq n_{k-1}$ we are sure that $y_k=x_{n_k}\in I_{k-1}$, hence $I_k\ne\emptyset$. It follows that the $I_k$ form an admissible sequence of nested intervals; hence there is a $\xi\in \bigcap_k I_k$. Now let an $\epsilon>0$ be given. There is a $k$ with ${1\over k}<{\epsilon\over2}$. When $n\geq n_k$ then $x_n\in I_k$, and therefore $$|x_n-\xi|\leq |x_n-y_k|+|\xi-y_k|\leq{2\over k}<\epsilon\ .\qquad\square$$ Conversely: Assume that ${\mathbb R}$ is complete as a metric space, and consider a sequence of nested intervals $I_n=[a_n,b_n]$ with $\lim_{n\to\infty} (b_n-a_n)=0$. We have to prove that there is a $\xi\in{\mathbb R}$ with $\xi\in\bigcap_n I_n$.
Note that the $a_n$ form an increasing sequence, the $b_n$ a decreasing sequence. It follows that $a_m\leq b_n$ for all $m$ and $n$. Let an $\epsilon>0$ be given. There is an $n_0$ with $b_n-a_n<\epsilon$ for all $n\geq n_0$. It follows that $$|a_{n+p}-a_n|=a_{n+p}-a_n\leq b_n-a_n<\epsilon\qquad\forall n\geq n_0,\quad\forall p\geq0\ .$$ This shows that the $a_n$ form a Cauchy sequence, having a limit $\xi\in{\mathbb R}$. It is then easy to see that the $b_n$ converge to $\xi$ as well, and as $a_n\leq\xi\leq b_n$ for all $n$ we see that $\xi\in\bigcap_n I_n$ holds.$\qquad\square$