Topology on the set of linear subspaces

Hello,

let $X$ be a seperable Hilbert space. Let $(e_i)_i$ be a Hilbert basis, and for each index let $E_i = \langle e_1,\dots,e_i \rangle \subset X$ the span of the first $i$ basis vectors. For any $x \in X$, let $x_i$ be the best-approximation of $x$ in $E_i$, and it is clear that $x_i \rightarrow x$.

It seems intuitive to say, that the $(E_i)$ approximate $X$ in a certain sense. Nevertheless, I am not aware of a topology on the set of linear subspaces, which would give such a result rigorously.

A first attempt might be to identify each linear subspace with its projection-onto, and inspect these projections as a topological (no more linear) space. A next step might be to take into account the order of the basis vectors for each linear subspace (which might be crucial for stability in numerical analysis). I am not known a theory Grassmannian manifolds in infitinte-dimensional vector spaces, nor how to relate non-equidimensional Grassmannian manifolds.

Can you give me hints where to find theory into this direction?

• The projection operators from X onto E_i converge to the identity operator from X to X in, say, the strong operator topology. – Qiaochu Yuan Jan 9 '11 at 17:41

Denote the intersection of a closed linear subspace $A$ with the unit sphere by $S_A$, say. You can define the distance of $A$ and $B$ to be the Hausdorff distance between $S_A$ and $S_B$. This will give you a metric topology on the set of closed linear subspaces but it seems that it does not quite do what you want: the distance between a proper closed subspace and the ambient space is always equal to $1$.