Date: Nov 11, 2012 6:51 AM
Author: Achimota
Subject: Re: countable set of closed subspaces in separable Hilbert space question

On Sunday, November 11, 2012 5:09:34 PM UTC+8, William Elliot wrote:
> ...Do you mean the topological closure or some algebra construction?

There is no such thing as "algebraic closure" in mathematics.
There is only topological closure.
Closure is always with respect to a topology.
A norm can induce a topology, and norms do have a powerful algebraic structure, but it is the topology induced by the algebraic structure of the norm that defines closure.

> How are you defining lim(n->oo) X_n?

"Strong convergence" ("convergence in the norm"); that is, the norm induced by the inner product:

For any e>0 there exists N such that
|| x-x_n || < e for all n>N

Dan