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Topic: countable set of closed subspaces in separable Hilbert space question
Replies: 14   Last Post: Nov 13, 2012 10:29 PM

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David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: countable set of closed subspaces in separable Hilbert space question
Posted: Nov 11, 2012 11:26 AM
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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
> || x-x_n || < e for all n>N
>
>Dan





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