I've been thinking about Grassmannians recently. Think of $\mathbb{R}^k$ as a $k$-dimensional vector space. Let $\text{Gr}_n(\mathbb{R}^k)$ denote the Grassmannian of all $n$-dimensional vector subspaces of $\mathbb{R}^k.$ (This is a compact, Hausdorf topological manifold of dimension $n(k-n)$.) Let

$ \Gamma^n(\mathbb{R}^k) := \{ (X,v) : X \in \text{Gr}_n(\mathbb{R}^k) \text{ and } v \in X \} . $

There's a standard idea of a vector bundle $\pi : \Gamma^n(\mathbb{R}^k) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^k)$ given by $\pi(X,v) := X.$ This bundle has the nice property that lots of other bundles can be realised as sub-bundles of it. There is a more general definition, where we use $\mathbb{R}^{\infty}$ in place of $\mathbb{R}^k$. My question is about why we define $\mathbb{R}^{\infty}$ the way we do.

We define $\mathbb{R}^{\infty}$ as the set of infinite sequences $(x_1,x_2,x_3,\ldots)$ where each $x_i \in \mathbb{R}$ and only finitely many of the $x_i$ are non-zero. We identify $\mathbb{R}^k$ with the sequences of the form $(x_1,\ldots,x_k,0,0,\ldots),$ and then topologize $\mathbb{R}^{\infty}$ as the direct limit of the sequence $ \mathbb{R}^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3 \subset \ldots$ Then we get the universal bundle $\pi : \Gamma^n(\mathbb{R}^{\infty}) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^{\infty}).$

My question is why do we insist that only finitely many of the $x_i$ are non-zero for each $(x_1,x_2,x_3,\ldots) \in \mathbb{R}^{\infty}$? I understand that it gives a countably infinite dimensional vector space, but that's a result of the definition; it doesn't explain why we define it the way we do. I suspect that it's related to the topology, but I don't really know.

Edit: The context is the OP is reading Milnor and Stasheff.

I did not vote to close your question. The fact that the answer you wanted is "It's a CW complex" is only obvious with hindsight. People work with spaces that are not CW complexes, so how is one meant to have known that this was the desired answer? I reiterate that while your original question may have been clear to you, it was not clear what kind of answer you were expecting or desiring. $\endgroup$ – Yemon Choi Aug 21 '11 at 4:33