another question here to see if I'm on the right track. Here is the problem that I have been proposed.
Determine if the set $H$ of all functions $f:Z_+\rightarrow Z_+$ that are eventually constant is countable.
Here is my solution.
I claim the set $H$ is countable. By definition, we say that $f$ is eventually constant if there is a positive integer $N$ such that $f(n)=c,\textrm{c a constant,}\ \forall\ n\ge N.$ Let $F$ be a set containing some $f:Z_+\rightarrow Z_+.$ Let $$H=\bigcup_{n\in\mathbb{Z_+}}F_n=\{f:Z_+ \rightarrow Z_+ |f\in F_n\ \textrm{for at least one}\ n\in Z_+\}.$$ Since $H$ is an arbitary union of sets, we can define a function $g:Z_+\rightarrow H$, where $g$ is an indexing function of $H$ and $Z_+$ is the index set given. Since $g$ is an indexing function, this implies that $g$ is surjective. By Theorem 7.1, which states if $H$ is a nonempty set and there is a surjective function $f:Z_+ \rightarrow H$ then $H$ is countable, we have that $H$ is countable as claimed.
Best Answer
Alternatively, consider the following:
For a function $f:Z_+\rightarrow Z_+ $ such that $\exists N \in Z_+ : \forall n > N, f(n) = f(N) $
Let $M = min\{ N : \forall n > N, f(n) = f(N) \}$
and $S(f) = \sum_{i=1}^{M} f(i) + M$
Then for any value $k$, there are only finitely many such functions with $S(f) = k$. List all $f$ with $S(f)=1$ then all $f$ with $S(f)=2$ etc.