
Re: countable set of closed subspaces in separable Hilbert space question
Posted:
Nov 11, 2012 6:51 AM


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  xx_n  < e for all n>N
Dan

