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