
Re: countable set of closed subspaces in separable Hilbert space question
Posted:
Nov 12, 2012 2:53 PM


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 illdefined 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 scalarvector multiplication and so a in A ==> a in A which implies (under the above definition) A = A and AB = 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 illdefined. In fact, I don't even > > > have a definition for XX_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.

