The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: countable set of closed subspaces in separable Hilbert space question
Replies: 14   Last Post: Nov 13, 2012 10:29 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
David C. Ullrich

Posts: 21,553
Registered: 12/6/04
Re: countable set of closed subspaces in separable Hilbert space question
Posted: Nov 11, 2012 11:26 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

>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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.