Let be $(a_n)_{n\in\Bbb N}$ and $(b_n)_{n\in\Bbb N}$ and increasing and a decreasing sequence of reals numbers such that $$ a_0<\dots <a_n<b_n<\dots<b_0 $$ for any $n\in\Bbb N$. So by the last inequality it is evident that $b_0$ is an upper bound for $(a_n)_{n\in\Bbb N}$ and $a_0$ is a lower bound for $(b_n)_{n\in\Bbb N}$ so that we can claim that $(a_n)_{n\in\Bbb N}$ has a supremum $M$ and $(b_n)_{n\in\Bbb N}$ has a infimum $m$ but unfortunately I was not able to prove or disprove if $M\le m$ so that I thought to put here a specific question where I ask to prove or disporve it with a counterexample.
So could someone help me, please?
So if there existed $h,k\in\Bbb N$ such that $b_k<a_h$ then these would be such that $$ a_k<b_k<a_h<b_h $$ So if $(b_n)_{n\in\Bbb N}$ is decreasing then the inequality $b_k<b_h$ implies that $k>h$ but if $(a_n)_{n\in\Bbb N}$ is increasing then the inequality $a_k<a_h$ implies that $k<h$ so that the inequality $b_k<a_h$ would imply that $k>h$ and $k<h$ but this is clearly impossible: so we conclude that $$ a_h<b_k $$ for any $h,k\in\Bbb N$ and this implies that $b_k$ is an upper bound for $(a_h)_{h\in\Bbb N}$ so that $M\le b_k$ but this means that $M$ is a lower bound for $(b_k)_{k\in\Bbb N}$ and so finally $M\le m$.
OBSERVATION
Alternatively as TheSilverDoe showed in the comments above the inequality $a_h<b_k$ can follow immediately observing that $$ a_h<a_{h+k}<b_{h+k}<b_k $$