Is there a proof that $BO(k)$ is not of the homotopy type of a finite dimensional complex?


The Grassmannian $BO(k) := \{ k\text{-dim subspaces of } \mathbb{R}^\infty \}$ classifies the $k$-dimensional vectorbundles on a $CW$ complex $X$ as:
$Vect^k(X) \cong [X,BO(k)]$.


$BO(k)$ can be constructed as the direct limit of finite dimensional Grassmann manifolds $G\_k(\mathbb{R}^n)$, but it is not finite dimensional itself.
Is there a proof that there cannot be a better (=finite-dimensional) construction ?


My idea was showing that the cohomology of $BO(k)$ is not bounded in dimension. This works for $BO$ as follows:
There are bundles with non-vanishing $k$th Stiefel-Whitney $\omega\_k$ class for all $k$. (Take for example the $k$th product $\gamma\_1 \times \dots \times \gamma\_1$ of the tautological line bundle $\gamma\_1$ over the real projective space $\mathbb{R}P^1$.)
By naturality $\omega\_k$ of the classifying bundle over $BO(k)$ cannot vanish
and so $H^k(BO(k),\mathbb{Z}/2)$ is non-trivial. Then it should follow that $BO= colim\_\to BO(k)$ cannot be finite-dimensional.


However, this idea does not work for $BO(k)$ because all characteristic classes I know "live" in the cohomology of dimension atmost rank of the bundle.
Are there unstable classes in higher degrees, or is there some other way to proof the result?



Infinite Grassmannian does not have the homotopy type of a finite-dimensional complex
We have $H^\*(BO(k); \mathbb{Z}\_2) \cong \mathbb{Z}\_2[w\_1, \dots, w\_k]$ where $\deg w\_i = i$. In particular, $H^n(BO(k); \mathbb{Z}\_2) \neq 0$ for every $n$ as $w\_1^n$ is a non-zero element. Therefore $BO(k)$ cannot be homotopy equivalent to a finite-dimensional CW complex.


That is, the degrees of the usual choice of generators of the $\mathbb{Z}\_2$ cohomology ring of $BO(k)$ are bounded by $k$, but they generate a graded ring which is non-zero in each degree.


