A couple questions about a countable set I encountered Last week, we worked on a proof in class, but I'm having a hard time following several passages in the proof.  ${}^\omega \omega$ denotes the set of functions from $\omega$ to $\omega$.
At one point in the proof, we assumed that for every $f \in {}^\omega \omega$, there exists a $g \in {}^\omega \omega$ such that $f(n) \leq g(n)$ for all but finitely many $n$. From here, we defined the set $K$ as 
$$K = \{ k \in {}^\omega \omega : k(n) = g(n) \mbox{ for all but finitely many } n \}.$$
We then said that $K$ was countable. However, I'm not seeing why it would be countable. 
From here, we then said that we could always find a $k \in K$ such that $f \leq k$, that is, $f(n) \leq k(n)$ for all $n \in \omega$. This is the second part of the proof that I'm not understanding. Why can we assume a stronger inequality?
I would really appreciate if anyone would be able to help me out with this one. Thanks!!!
 A: I couldn't find a nice duplicate (which doesn't amount to the hint you already said to not understand). So let me give you a summary answer.
Let us think about a function in ${}^\omega\omega$ as a sequence of integers, now given $f,g$ which are equivalent, there is some finite initial segment of $f$ that when removed, and replaced with the finite initial segment from $g$ (of the same length) we have $g$ itself.
For example, if $f(n)=n$ and $g(n)=n$ for $n>0$, and $g(0)=1$, we replace the initial segment of length $1$ from $f$, by the same initial segment of $g$.
Recall that the set of finite sequences from $\omega$ is countable. This shows that every equivalence class is countable.
The second fact follows from the fact is trivial if you consider $k=f$. But the point is that you can also modify any initial segment to be as large as you would like, so if you are given several $f_i\in K$ you can consider some initial segment after which all the $f_i$ agree with one another, and take $k(n)=\max f_i(n)+1$ as values over the initial segments where the $f_i$'s differ.
