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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 12, 2012 2:53 PM
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On Monday, November 12, 2012 3:52:24 PM UTC+8, William Elliot wrote:
> In general, if A is a set and f a function f(A)
> is defined as the set { f(A) | a in A }. Accordingly,
> A + B = { a + b | a in A, b in B }
> -A = { -a | a in A }


"A+B" and "-A" are here ill-defined because + and - is not in general defined on sets. The definition would be valid if the definition specified a group (A,+) or the equivalent rather than simply that "A is a set".

The case I am working with is linear subspaces, and in this case the definition
> -A = { -a | a in A }
is trivial because linear subspaces are closed under scalar-vector multiplication and so
a in A ==> -a in A
which implies (under the above definition)
A = -A
and
A-B = A+B

On Monday, November 12, 2012 3:52:24 PM UTC+8, William Elliot wrote:
> On Sun, 11 Nov 2012, Daniel J. Greenhoe wrote:
>

> > On Monday, November 12, 2012 11:21:30 AM UTC+8, William Elliot wrote:
>
>
>

> > > How for example, are you defining
>
> > > ||X - X_n|| for sets X and X_n?
>
> >
>
> > William, I think you bring up a very good point. My "definition" of the
>
> > limit of a sequence of subspaces is ill-defined. In fact, I don't even
>
> > have a definition for X-X_n, and am not sure a good way to define the
>
> > norm ||Y|| of a subspace Y.
>
>
>
> In general, if A is a set and f a function f(A)
>
> is defined as the set { f(A) | a in A }. Accordingly,
>
>
>
> A + B = { a + b | a in A, b in B }
>
> -A = { -a | a in A }
>
>
>
> and perhaps, if ever used,
>
> ||A|| = { ||f|| : f in A }
>
>
>
> would be a collection of numbers and as such
>
> one can't write ||A|| < r without the special
>
> definition A < r when for all a in A, a < r.
>
>
>
> By special, I mean I use A <= r when for all a in A, a <= r;
>
> a definition not used by others. I don't use A < r.





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