Algebraic isomorphism classes of vector bundles of rank $r$ on $\mathbb P^1_\mathbb C$ are in bijective correspondence with $r$-tuples of integers $a=(a_1,\dots,a_r)$ such that $a_1\geq a_2\geq \dots a_r$. Two distinct such "classes" $a,b$ define the same topological vector bundle if and only if the corresponding vector bundles have the same degree (i.e. $\sum a_i=\sum b_i$).
This fact says that the algebraic classification produces "more" isomorphism classes than the topological one (in other words, if $F\cong_{alg}E$, then $F\cong_{top}E$).
But in A. Langer's notes on moduli of sheaves I read the phrase:
"Later, vector bundles on $\mathbb P^n_\mathbb C$ were classified for $n \leq 6$, but it is not known which of these vector bundles can be realized as algebraic vector bundles."
This suggests that there should be an inclusion
$$\{\textrm{Algebraic isomorphism classes}\}\subset\{\textrm{Topological isomorphism classes}\},$$ which is contradicting the fact that the algebraic classification is "finer".
Question. What am I misunderstanding? Which one is actually finer, and in which sense, between the two classifications?
As suggested by @Asal Beag Dubh (thanks again!), the point is that in the case of vector bundles on $\mathbb P^1$ there is a surjective map
$$\{\textrm{Algebraic iso classes}\}\twoheadrightarrow \{\textrm{Topological iso classes}\},$$ as the right hand side "equals" $\{\textrm{Algebraic iso classes}\}/$degree.
General situation (on $\mathbb P^n$, say): we know that there can be several holomorphic structures on the same topological bundle; in other words, $[E]_{alg}\subseteq [E]_{top}$, but no one knows whether $[E]_{alg}$ is empty for a fixed topological bundle $E$. In other words, the map
$$\{\textrm{Algebraic iso classes}\}\to \{\textrm{Topological iso classes}\}$$ might not be surjective, i.e. there might exists a topological bundle $E$ such that no algebraic vector bundle is isomorphic to $E$.
Still, I would be glad to see: