given $B=\left\{ [c_{n}]_{R}:n\in\mathbb{N}\right\} $ as $c_{n}:\mathbb{N\longrightarrow N} $ is the fixed function $c_{n}(m)=n$ for every $m$.
let $R$ be an Equivalence relation on $\mathbb{N^{N}}$ defined to be:
$\mathrm{f}R\mathrm{g}\Longleftrightarrow\exists n\in\mathbb{N}:\forall m\geq n\quad f(m)=g(m)$ .
we'll define a linear order on the quotient set $\mathbb{N^{N}}/R $ for which:
$\left[f\right]_{R}\leq\left[g\right]_{R}\Longleftrightarrow f\leq_{*}g$
as $\leq_{*}$ is a relation on $\mathbb{N^{N}}$ fixed to be:
$f\leq_{*}g\Longleftrightarrow\exists n\in\mathbb{N}\; s.t \;\;\negthickspace\forall m\geq n\;f(m) \le g(m) $
I had to prove that $[Id]_{R}$ is a upper bound for $B$ ,and than to explain why it's not the supremum of $B$, which I did.
than I need to prove that every countable set $B$ , as $B\subseteq\mathbb{N^{\mathrm{\mathbb{N}}}}/R$ - has a supremum.
I defined $g$:
$\forall n\in\mathbb{N}:\negmedspace g(n)=max\{f_{j}(n):|0\leq j\leq n\}$ as $B=\{f_{j}(n)\in\mathbb{N^{N}}|\forall n,j\in\mathbb{N}\}$.
how do I prove that the set of all values of $g$ is the upper bound for $B$?
To show [g] is an upper bound for
B = { $[f_j] \in N^N/R$ : j in N },
notice that from j onward $f_j \le g$.
Now if hR$f_j,$ eventually $h \le g$. Thus [$f_j$] $\le$ [g].