
Re: countable set of closed subspaces in separable Hilbert space question
Posted:
Nov 11, 2012 11:26 AM


On Sun, 11 Nov 2012 03:51:24 0800 (PST), "Daniel J. Greenhoe" <dgreenhoe@yahoo.com> wrote:
>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 most certainly is.
Which is not to say your meaning was unclear  William likes to misunderstand things for no good reason.
Otoh you do need to explain the difference between limit and union here...
>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

