TITLE: On the finiteness of an Auslander-Reiten component
QUESTION [6 upvotes]: I am reading a paper called A NOTE ON THE RADICAL OF A MODULE CATEGORY by
CLAUDIA CHAIO AND SHIPING LIU. This is Theorem 2.7: 
And this is part of it's proof, in which the direction (2) $\Rightarrow $ (1) is shown.

$\iota_S: S \rightarrow I(S)$ denotes the injective envelope and $\pi_S: P(S) \rightarrow S$ denotes the projective cover of a simple module $S$. Also dp$(f)$ denotes the depth of $f$.
As far as I know a quiver is locally finite means iff between each two vertices there is only a finite number of arrows between them
I really don't understand why it follows that the Auslander-Reiten component $\Gamma$ is finite. Also why does $\Gamma$ contain at most finitely many indecomposable injective modules?
Can anybody help with his? Any help is highly appreciated!

REPLY [2 votes]: Since $A$ is an Artin algebra, it has only finitely many indecomposable injective modules in total (up to isomorphism), so there are finitely many in $\Gamma$.
In a locally finite quiver, given any $d\geq0$ and vertex $v$, there are finitely many paths ending at $v$ and having length at most $d$, because there are finitely many choices for each arrow. In particular, $v$ is reachable from only finitely many vertices by such paths.
This means that there only finitely many vertices of $\Gamma$ from which one of the finitely many indecomposable injectives may be reached via a path of length at most $d$, for any fixed $d>0$. Since the proof exhibits an $r$ such that every vertex of $\Gamma$ has a path of length at most $r$ to some indecomposable injective, it follows that $\Gamma$ has finitely many vertices.