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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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Achimota

Posts: 254
Registered: 4/30/07
Re: countable set of closed subspaces in separable Hilbert space question
Posted: Nov 11, 2012 2:54 AM
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> On Sat, 10 Nov 2012, Daniel J. Greenhoe wrote:
> > Let H be a separable Hilbert space.

On Sunday, November 11, 2012 1:56:02 PM UTC+8, William Elliot wrote:
> Vectorial subspaces or topological subspaces?

A Hilbert space is a special case of a linear space. The elements of the sets are called vectors, so in that sense one could call the subspaces "vector subspaces".

A Hilbert space is a complete inner product space. The inner product induces a norm which in turn induces a metric which in turn induces a topology on the space. So, in that sense one could also call the subspaces "topological subspaces".

The conclusion is this: In a Hilbert space, there is no distinction between a "vector subspace" and a "topological subspace". All subspaces in a Hilbert space are both "vectorial" (in the sense they contain vectors) and "topological" (in the sense that the inner product induces a topology).

Dan

On Sunday, November 11, 2012 1:56:02 PM UTC+8, William Elliot wrote:
> On Sat, 10 Nov 2012, Daniel J. Greenhoe wrote:
>
>
>

> > Let H be a separable Hilbert space.
>
> > Let (X_n) be a sequence of nested subspaces in H such that
>
> > X_n subset X_{n+1}, X_n not= X_{n+1}.
>
>
>
> Vectorial subspaces or topological subspaces?
>
>
>

> > What is the relationship between the following two conditions in H?
>
> > 1. closure{ Union X_n } = H
>
> > 2. closure{ lim_{n->infty} X_n } = H
>
>
>

> > Does one imply the other? Are they equivalent?
>
>
>
> Is there a difference between lim(n->oo) X_n and \/_n X_n?
>
>
>
> Closure in the topological sense?





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