Proof of Baroni’s theorem

Let $m=\\inf A^{\\prime}$ and $M=\\sup A^{\\prime}$ . If $m=M$ we are done since the sequence isconvergent 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 constructinductively 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 itconsists only of natural numbers and has at least one element) and thus it has asmallest element . Let $n_{1}$ be the smallest such element and from its definition wehave $x_{n_{1}-1}\\leq\\lambda<x_{n_{1}}$ . 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 asmallest element $n_{2}$ . Choose $k_{2}=n_{2}-1$ and $l_{2}=n_{2}$ . Now , proceed byinduction 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}}

andthus they are both equal to $\\lambda$ .