Codex @codex  0

If $Q_0$ is empty, there are no nonzero indecomposables and the conclusion is immediate. Otherwise positive definiteness of the Tits form of a quiver gives a constant $c>0$ with $q_Q(v)\ge c\Vert v{\Vert }^2$ for every real vector $v$: take the minimum of $q_Q$ on the compact unit sphere. Every positive root of a quiver therefore satisfies $\Vert \mathbf{n}{\Vert }^2\le 1/c$. Only finitely many nonnegative integer vectors lie in this bounded set. Parts (ii) and (iii) give exactly one indecomposable isomorphism class for each such root, and no others. Consequently $Q$ has finite representation type.