Proof of Baroni’s theorem

Let $m=\inf A^{\prime}$ and $M=\sup A^{\prime}$ . If $m=M$ we are done since the sequence is convergent and $A^{\prime}$ is the degenerate interval composed of the point $l\in\mathbb{\overline{R}}$ , where $\displaystyle l=\lim_{n\rightarrow\infty}x_{n}$ .

Now , assume that $m<M$ . For every $\lambda\in(m,M)$ , we will construct inductively two subsequences $x_{k_{n}}$ and $x_{l_{n}}$ such that $\displaystyle\lim_{n\rightarrow\infty}x_{k_{n}}=\lim_{n\rightarrow\infty}x_{l_% {n}}=\lambda$ and $\displaystyle x_{k_{n}}<\lambda<x_{l_{n}}$

From the definition of $M$ there is an $N_{1}\in\mathbb{N}$ such that :

\lambda<x_{N_{1}}<M

Consider the set of all such values $N_{1}$ . It is bounded from below (because it consists only of natural numbers and has at least one element) and thus it has a smallest element . Let $n_{1}$ be the smallest such element and from its definition we have $x_{n_{1}-1}\leq\lambda<x_{n_{1}}$ . So , choose $k_{1}=n_{1}-1$ , $l_{1}=n_{1}$ . Now, there is an $N_{2}>k_{1}$ such that :

\lambda<x_{N_{2}}<M

Consider the set of all such values $N_{2}$ . It is bounded from below and it has a smallest element $n_{2}$ . Choose $k_{2}=n_{2}-1$ and $l_{2}=n_{2}$ . Now , proceed by induction to construct the sequences $k_{n}$ and $l_{n}$ in the same fashion . Since $l_{n}-k_{n}=1$ we have :

\lim_{n\rightarrow\infty}x_{k_{n}}=\lim_{n\rightarrow\infty}x_{l_{n}}

and thus they are both equal to $\lambda$ .

Title	Proof of Baroni’s theorem
Canonical name	ProofOfBaronisTheorem
Date of creation	2013-03-22 13:32:33
Last modified on	2013-03-22 13:32:33
Owner	mathwizard (128)
Last modified by	mathwizard (128)
Numerical id	5
Author	mathwizard (128)
Entry type	Proof
Classification	msc 40A05