Does every strictly increasing, unbounded sequence of positive real numbers contain arbitrarily long, finite subsequences which are "sort of increasing" or "sort of decreasing" (as defined below)? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T08:07:45Zhttp://mathoverflow.net/feeds/question/95244http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/95244/does-every-strictly-increasing-unbounded-sequence-of-positive-real-numbers-contaDoes every strictly increasing, unbounded sequence of positive real numbers contain arbitrarily long, finite subsequences which are "sort of increasing" or "sort of decreasing" (as defined below)?Yann Peresse2012-04-26T10:50:50Z2012-04-26T10:50:50Z
<h3>Is the following true?</h3>
<p>If $(x_0, x_1, \dots)$ is a strictly increasing, unbounded sequence of positive real numbers, then there exist fixed $M,N \geq 1$ such that the sequence $(x_0, x_1, \dots)$ contains an ($M,N$)-expander of length $k$ for every $k\in \mathbb{N}$.</p>
<h3>Definition of an $(M,N)$-expander:</h3>
<p>If $M,N \geq 1$ are integers, then
an $(M, N)$-expander of length $k$ of $(x_0, x_1, \dots)$ is a subsequence
$(x_{i[1]},x_{i[2]},\cdots,x_{i[k]})$ of $\mathfrak{X}$ such that
$i[j+1]-i[j]\leq M$ for all $1\leq j\leq k-1$ and either
\begin{equation}
\frac{x_{i[n+1]}-x_{i[n]}}{x_{i[m+1]}-x_{i[m]}}\leq N
\textrm{ for all }1\leq m\leq n\leq k-1
\end{equation}
or
\begin{equation}
\frac{x_{i[m+1]}-x_{i[m]}}{x_{i[n+1]}-x_{i[n]}}\leq N
\textrm{ for all }1\leq m\leq n\leq k-1
\end{equation}</p>
<h3> Is this an open question?</h3>
<p>This is a question that was asked (formulated a little differently) in the following paper, of which I am one of the authors:</p>
<p>‘Relative ranks of Lipschitz mappings on countable discrete metric spaces’, Topology and its Applications 158 (2011) 412-423;</p>
<p>In that sense, it is an open problem. However, as far as I know, this question has been not been widely considered, and so it is not a well-known open problem that is known to be difficult. If, nevertheless, this question is inappropriate for this forum, then I appologise.</p>
<h3>Motivation </h3>
<p>If the answer is "yes, it is true", then the results in the the paper mentioned above prove the following conjecture about the semigroup $\mathfrak{L}_{\mathfrak{X}}$ of all Lipschitz functions from a countable subset $\mathfrak{X}$ of $\mathbb{R}$ to itself (where the semigroup operation is composition of functions):</p>
Conjecture:
<p>If $\mathfrak{X}$ is any countable subset of the real numbers, then</p>
<p>either $\mathfrak{X}$ contains a Cauchy sequence and there exists a single function from $\mathfrak{X}$ to $\mathfrak{X}$ that together with $\mathfrak{L}_{\mathfrak{X}}$ generates all functions from $\mathfrak{X}$ to $\mathfrak{X}$;</p>
<p>or $\mathfrak{X}$ contains no Cauchy sequences and the least number of functions from $\mathfrak{X}$ to $\mathfrak{X}$ that together with $\mathfrak{L}_{\mathfrak{X}}$ generate all functions from $\mathfrak{X}$ to $\mathfrak{X}$ is the uncountable cardinal $\mathfrak{d}$ (the dominating number).</p>