  Let $\Lambda$ be a finite-dimensional $k$-algebra with $k$ algebraically
closed. Bongartz has recently shown that the existence of an indecomposable
$\Lambda$-module of length $n > 1$ implies that also indecomposable
$\Lambda$-modules of length $n-1$ exist. Using a slight modification of his
arguments, we strengthen the assertion as follows: If there is an
indecomposable module of length $n$, then there is also an accessible one.
Here, the accessible modules are defined inductively, as follows: First, the
simple modules are accessible. Second, a module of length $n \ge 2$ is
accessible provided it is indecomposable and there is a submodule or a factor
module of length $n-1$ which is accessible.
