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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 4:14 PM
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> On Sun, 11 Nov 2012 03:51:24 -0800 (PST), "Daniel J. Greenhoe" wrote:
> >There is no such thing as "algebraic closure" in mathematics.

On Monday, November 12, 2012 12:26:06 AM UTC+8, David C. Ullrich wrote:
> There most certainly is.

Thank you for your correction of my mistake. The term does seem to come up in literature about algebra, of all places. For example here:
http://books.google.com.tw/books?id=jdx5K7CdL_4C&pg=PA517
Obviously this is yet one more topic that I don't know too much about (but yet does not stop me from talking about it), but it seems "algebraic closure" and "topological closure" have very little in common except for having similar names. Topological closure relates to sequences in a topological space while algebraic closure seems to relate to the binary operators of a field.

> ... William likes to misunderstand things for no good reason.
I appreciate William's responses to my post. They have been helpful to me.

> Otoh you do need to explain the difference between limit and union here...

I don't know if there is a difference and hence my original post. The countable union of nested closed subspaces comes up in the "multiresolution analysis" (MRA) axioms of wavelet analysis. The MRA was apparently first published by Stéphane Mallat in 1989:
http://blanche.polytechnique.fr/~mallat/papiers/math_multiresolution.pdf
The concept of the MRA has become extremely common in wavelet literature.

I have noted that the "countable union of closed sets" is called F_sigma in the literature while G_delta is the "countable intersection of open sets":
http://books.google.com/books?vid=ISBN0521497566&pg=PA130
Would there be some connection or fundamental result involving F_sigma sets? Could the Baire Category Theorem be useful here?

Dan

On Monday, November 12, 2012 12:26:06 AM UTC+8, David C. Ullrich wrote:


> On Sun, 11 Nov 2012 03:51:24 -0800 (PST), "Daniel J. Greenhoe" 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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