# 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 $l\in \overline{\mathbb{R}}$ , where $l=\underset{n\to \mathrm{\infty}}{lim}{x}_{n}$.

Now , assume that $$ . For every $\lambda \in (m,M)$ , we will construct
inductively two subsequences^{} ${x}_{{k}_{n}}$ and ${x}_{{l}_{n}}$ such that $\underset{n\to \mathrm{\infty}}{lim}{x}_{{k}_{n}}=\underset{n\to \mathrm{\infty}}{lim}{x}_{{l}_{n}}=\lambda $
and $$

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

$$ |

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 $$ . So , choose ${k}_{1}={n}_{1}-1$ , ${l}_{1}={n}_{1}$ .
Now, there is an ${N}_{2}>{k}_{1}$ such that :

$$ |

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 :

$$\underset{n\to \mathrm{\infty}}{lim}{x}_{{k}_{n}}=\underset{n\to \mathrm{\infty}}{lim}{x}_{{l}_{n}}$$ |

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 |