locall compactness and compact operators

I need help understanding an argument.

Here is a brief

exposition

"In reply to:

>>Let B be the unit ball in l^2. Then

T(B) = B (surjective!). As any open neighbourhood

>>U of 0 contains

some rB for a scalar r, T(U) contains an sB for some scalar s.

As

>>locally compact normed spaces are finite dimensional, T cannot be

compact.

>

>OK, T(B) = B and as any open neighbourhood U of 0

contains some rB for a scalar r, T(U) contains an sB for some scalar

s.

>Now, l2 is not locally compact. How does it follow that T is not

compact?

>

>

T(B) = B is obvious. An operator is compact iff

the image of the unit ball is pre-compact.

The closure of the unit ball is

not compact hense T is not compact. q.e.d.

I'd like to understand your

topological argument though.

Locally compact means that every point of the

space has a compact neighbourhood.

What you are saying is that every

neighbourhood of 0 has an image which contains sB for some real s.

Which

does not mean locally compact. A more detailed explanation would be appreciated.

Please."

For more info check

http://at.yorku.ca/cgi-bin/bbqa?forum=a

... g;msg=4092

and

http://at.yorku.ca/cgi-bin/bbqa?forum=a

... =4087.0001

Apreciate the help

strammer