Let $B$ denote a fixed topological space, which will be called the base space. A real vector bundle $\xi$ over $B$ consists of a total space $E=E(\xi)$, a projective map $\pi: E\to B$ and for each $b\in B$ the structure of a vector space over the real numbers in the set $\pi^{-1}(b)$. It also satisfy the condition of local triviality.
The vector space $\pi^{-1}(b)$ is called the fiber over $b$. It may be denoted by $F_b$ or $F_b (\xi)$. The dimension $n$ of $F_b$ is allowed to be a locally constant function of $b$. However, my question is why it is also a constant function on the connected component of the topological space $B$.
ps: In most case of interest this function is constant. One then speaks of $n$-plane bundle. Any book about vector bundle are dealing with the $n$-plane bundle, particularly, the material above are from John milnor. I was wondering is there any vector bundle is not an $n$-plane bundle? vector bundle On Wikipedia, someone said it is constant on connected component. I learn from this and ask myself what could happenes if we put together two vector bundles of different (constant) dimensional of connected base space respectively? It do make sense but it is useless. Although I solved my original problem, i.e, to give a non $n$-plane bundle, I have a new one, which I ask here.
Not to this specific problem, my friends have no interest on something old stuff such as point set topology since we have to learn something new, say, characteristic classes. I have no idea about learn math but just want to solve this uselessness problem.
edit: Connected topological space is a topological space that cannot be divided into two disjoint open set which union is the original space.
Show that the following two definitions of "locally constant" for a function $f : X \to Y$ are equivalent (Edit: As Eric Wofsey says in the comments, under the hypothesis that $X$ is locally connected):
You can show that the dimension of the fiber of a vector bundle satisfies the second definition using local triviality.