Let $\omega^\omega$ denote the set of all functions $f:\omega\to\omega$ ordered by $f\leq g$ iff $f(n) \leq g(n)$ for all $n\in \omega$.


Set $K = \{f\in \omega^\omega: m<n\in \omega \implies f(m)<f(n)\}$. If $\textbf{DM}(\cdot)$ denotes the [Dedekind-MacNeille completion](http://en.wikipedia.org/wiki/Dedekind%E2%80%93MacNeille_completion), do we have $\textbf{DM}(\omega^\omega) \cong \textbf{DM}(K)$?



Dedekind-MacNeille completion of the strictly increasing members of $\omega^\omega$
The answer is no.


To see this, consider the bottoms of $K$ and $\omega^\omega$ under
the pointwise $\leq$ order you have described. Both structures
have a least element:


* The constant $0$ function is least in $\omega^\omega$.
* The diagonal function $d(n)=n$ is least in $K$.


Notice further that $\omega^\omega$ has infinitely many atoms,
that is, minimal elements above the least element. For example,
the characteristic function of a singleton is strictly above $0$
but there are no functions in between. The existence of atoms is
preserved by the Dedekind-MacNeille completion.


But $K$ has no atoms. Rather, there is a coinitial
$\omega$-sequence converging down to the least element $d$. For each natural number $k$, consider
the function $d\_k$ defined by $d\_k(i)=i$ for $i<k$ and
$d\_k(i)=i+1$ for $i\geq k$. So $d\_0>d\_1>d\_2>\cdots$ and so on, with respect to your order (note that $f<g$ means $f\leq g$ and $g\not\leq f$, which is not the same as pointwise $f(n)<g(n)$), and
furthermore, any function in $K$ that is above $d$ is above some
$d\_k$, since it must jump up at least one at some point, and at
that point, it is forced to jump up at all later points in order
to be strictly increasing. So $K$ has no atoms. I claim that the
Dedekind-MacNeille completion will not add any atoms, because the
elements of the completion correspond to subsets $A\subset K$ such
that $(A^u)^l=A$, that is, for which $A$ is equal to the set of lower bounds of its set of upper bounds; but if such an $A$ has a function above $d$,
then it will also contain infinitely many $d\_k$, and hence not
correspond to an atom in the completion.


