Date: Nov 12, 2012 2:53 PM
Author: Achimota
Subject: Re: countable set of closed subspaces in separable Hilbert space question
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.