Separability of Banach spaces

Thursday, March 30. 2006

Separability of Banach spaces

I mulled repeatedly for some weeks over the fact, why a Banach space is separable if any element can be approximated by a sequence of finite dimensional elements. It has been formulated as $S_nx\to x$ , where $S_n\in F(X)$ is a continuous linear operator with finite rank, and this conclusion was never obvious to me. For another time, there was a fact which should be clear and not difficult to show.

Finally, I remembered: Every (finite dimensional) vector space has a (finite) basis. I.e., if $\dim(S_my)=k(m)$ , then there is a basis $\{b_1,\ldots,b_{k(m)}\}$ which can be extended as soon as $\dim(S_{m+\ell}\,y)>k(m)$ for any $\ell\in\mathbb{N}$ . Any basis is countable, and a theorem says that a space $X$ is separable iff there is a countable set $A$ with $X=\overline{\operatorname{lin}\,A}$ , where $\overline{\operatorname{lin}\,A}$ is the closure of the set of all (finite) linear combinations of elements in $A$ .

Posted by Stephan Paukner in Master's Thesis at 10:22 | Comments (0) | Trackbacks (0)

Defined tags for this entry: mathematics

Trackbacks

Trackback specific URI for this entry

No Trackbacks

Comments

Display comments as (Linear | Threaded)

No comments

Add Comment

Name
Email
Homepage
In reply to
Comment	Phone* What is five plus zero? Enclosing asterisks marks text as bold (word), underscore are made via _word_. Standard emoticons like :-) and ;-) are converted to images. To prevent automated Bots from commentspamming, please enter the string you see in the image below in the appropriate input box. Your comment will only be submitted if the strings match. Please ensure that your browser supports and accepts cookies, or your comment cannot be verified correctly. Enter the string from the spam-prevention image above: BBCode format allowed
	Remember Information? Subscribe to this entry
Submitted comments will be subject to moderation before being displayed.