Is there a metric space on $\omega^\omega$ such that $\alpha+n\to\alpha+\omega$ as $n\to\infty$?
Let $\omega^\omega$ be the set of all ordinals less than $\omega^\omega$ then I seek a function:
$d:\omega^\omega\times\omega^\omega\to\Bbb R$ such that $\omega^\omega,d$ is a metric space
and for all $\alpha\in\omega^\omega$, adding further integers converges to $\alpha+\omega$
I'm aware of the Order Topology but this looks to be far from a metric space.
On
$\omega^\omega$ is a countable ordinal. One can embed every countable totally ordered set in an order-preserving way into $\Bbb Q$ and so into $\Bbb R$. Let's do this with $\omega^\omega$. But this need not be a homeomorphism of $\omega^\omega$ onto its image $X$ say. One can get round that by adding the upper bound of the image of every bounded increasing sequence in $\omega^\omega$. You'll then then a new subset $X'$ of $\Bbb R$ with the same order-type as $X$, and homeomorphic to $\omega^\omega$ in the order topology. Of course, $\Bbb R$ is a metric space, and so is $X'$.
You can do this for any countable ordinal.
Let $\alpha$ be any countably infinite ordinal. Fix a bijection $\varphi:\alpha\to\omega$ and define
$$f:\alpha\to\Bbb R:\eta\mapsto\sum_{\xi<\eta}2^{-\varphi(\xi)}\;;$$
then $f$ is an order-embedding of $\alpha$ in $\Bbb R$. If $\alpha$ has the order topology, $f$ is a homeomorphism of $\alpha$ onto $f[\alpha]$, and you can use it to define a metric on $\alpha$.