TITLE: Stability condition for vector bundles
QUESTION [5 upvotes]: Let $X$ be a smooth projective variety over $\mathbb{C}$, and fix $A$ an ample divisor as the polarization. We say a vector bundle $E$ to be (semi-)stable, if for any proper subsheaf $F$ of $E$, $\mu(F)<\mu(E)$ (resp. $\leq$). 
I guess it is not sufficient to check just subbundles $F$ of $E$. But is there a counterexample? Or more precisely, is there an example of $(X, A, E, F)$ satisfying the following conditions?
(1)$X,A,E$ are as above, and $F$ is a proper subsheaf of $E$ breaking the stability condition, i.e., $\mu(F)\geq \mu(E)$.
(2)There exists no vector bundle which break the stability condition, i.e., $\mu(F')<\mu(E)$ for any subbundle $F'$ of $E$.

REPLY [8 votes]: There are many examples of unstable bundles on a projective surface that have no non-trivial subbundles. For example, if $k$ is an integer with $k < 3$ and $I$ is the sheaf of ideal of $m$ distinct points in $\mathbb P^2$, with $m > 0$, there exists an extension
$$
0 \longrightarrow \mathcal O \longrightarrow E \longrightarrow I(k) \longrightarrow 0
$$
on $\mathbb P^2$ in which $E$ is locally free. Furthermore, the Chern classes of $E$ are $c_1(E) = k$ and $c_2(E) = m$ (for this, see page 103 of "Vector bundles on complex projective spaces", by Okonek, Schneider and Spindler). If $k < 0$, this vector bundle is clearly unstable; but for most values of $k$ and $m$ it can not split as a direct sum of line bundles, hence it cannot contain a line subbundle (since every extension of line bundles on $\mathbb P^2$ splits).