While reading through some issues of Baez's (wonderful) "This Week's Finds in Mathematical Physics," I came across this statement (from week 149):
$K(\mathbb{Z},2)$ is a bit more complicated: it's infinite-dimensional complex projective space, $\mathbb{C}P^{\infty}$! [...] you can define $\mathbb{C}P^{\infty}$ as a "direct limit" of these [$\mathbb{C}P^n$] as $n$ approaches infinity, using the fact that $\mathbb{C}P^n$ sits inside $\mathbb{C}P^{n+1}$ as a subspace. Alternatively, you can take your favorite complex Hilbert space $\mathcal{H}$ with countably infinite dimension and form the space of all 1-dimensional subspaces in $\mathcal{H}$. This gives a slightly fatter version of $\mathbb{C}P^{\infty}$, but it's homotopy equivalent, and it's a very natural thing to study if you're a physicist: it's just the space of all "pure states" of the quantum system whose Hilbert space is $\mathcal{H}$.
This is fairly standard (I've seen similar statements in numerous other places, I just haven't thought much about it), but I'm just wondering: are these two realizations $\mathbb{C}P^{\infty}$ merely homotopy equivalent? Or can more be said (e.g. homeomorphic, diffeomorphic, etc.)?
The $CP^n$'s constructed as the direct limit are a $CW$ complex. Hence $CP^\infty$ is the countable union of closed subspaces with empty interior (in $CP^\infty$). However, a Hilbert space (and also Hilbert manifolds) are Baire. The countable union of closed subspaces with empty interior have empty interior hence cannot be the whole space. So these are not homeomorphic.
Another way of seeing the "fatness" is yet another model. We can also see $CP^\infty$ as the space of lines in $\mathbb{C}^\infty$. This is the vector space of sequences $(a_1,\ldots,)$ such that all but a finite number of $a_1$ are zero. This $\mathbb{C}^\infty$ can be seen to sit inside the Hilbert space (after choosing a basis). However, in the Hilbert space there are "more points" namely ones such that $\sum_{i=1}^\infty |a_i|^2<\infty$. One can argue that $C^\infty$ is "thin" inside the Hilbert space, hence the space of lines in $C^\infty$ is thin inside the space of lines in $H$.