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  • Show that if $(a_n)_{n\in \mathbb{N}},(b_n)_{n\in \mathbb{N}}$ are two sequences adjacentes then the parts $A=\{a_n,\ n\in\mathbb{N}\}$ and $B=\{b_n,\ n\in\mathbb{N}\}$ of $\mathbb{R}$ are adjcentes.

  • The converse is it true ? we can check these two sequences here : $$a_n=\dfrac{(-1)+(-1)^n}{2}\quad \text{ and }\quad b_n=\dfrac{(1)+(-1)^n}{2}$$

This question is related to that one Show that sets of real numbers $A, B$ are adjacent iff $\sup A = \inf B$

i can't find the right term used in english but here is :

The definition of adjecent sequences: $(a_n)$ and $(b_n)$ are adjacent if one of the two sequences is decreasing, one is increasing, and $\lim_{n\to\infty} a_n-b_n = 0$

for the converse i think it is false because $b_n-a_n$ doesn't tends to $0$ when $n \to \infty$

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  • $\begingroup$ What is the definition of adjecent sequences? In the link, I only see the definition of adjacent sets... $\endgroup$
    – 5xum
    Jan 7 '15 at 8:54
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    $\begingroup$ @5xum: $(a_n)$ and $(b_n$) are adjacent if one of the two sequences is decreasing, one is increasing, and $\lim_{n\to\infty} a_n-b_n = 0$. I can't find the term used in English, but this is a term used in French (suites adjacentes). $\endgroup$ Jan 7 '15 at 8:57
  • $\begingroup$ me tow i can't find the right term in english $\endgroup$
    – Educ
    Jan 7 '15 at 9:05
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Assume $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are adjacent. Without loss of generality, let's assume that $a_n$ is decreasing.

  1. Let's prove that for every $a\in A$, $b\in B$, we have $a\geq b$ by contradiction. If there exist such $a\in A, b\in B$ that $a< b$, then there exist such $n,m\in\mathbb N$ that $a=a_n, b=b_m$ and $a_n < b_m$. But, because $(a_n)$ is increasing and $(b_m)$ is increasing, taking $N=\max\{m,n\}$ will mean that for all $k>N$, we have $$a_k - b_k < a_N - b_N < 0$$ which also means that the limit of $a_k-b_k$ is not equal to zero, which is a contradiction.
  2. Let's now, for a given $\epsilon$, find such $a\in A$ and $b\in B$ that $b-a\leq \epsilon$. Because the limit of $a_n-b_n$ is equal to $0$, there exists such a $N$ that for all $n\geq N$, we have $|a_n-b_n| \leq \epsilon$, meaning that taking $a=a_N$ and $b=b_N$ will yield $$b-a=b_N-a_n=|a_N-b_N|\leq \epsilon$$ thus proving that $A$ and $B$ are adjacent.
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  • $\begingroup$ for the first part you want to prove $\forall\ a \in A,\ \forall\ b \in B,\ a \le b$ so you used Contraposition in ur proof ? $\endgroup$
    – Educ
    Jan 7 '15 at 9:40
  • $\begingroup$ @Educ I used proof by contradiction, i.e. I proved that the negation of $\forall a\in A, \forall b\in B: a\geq b$ leads to a contradiciton. $\endgroup$
    – 5xum
    Jan 7 '15 at 9:41

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