## locall compactness and compact operators

I need help understanding an argument.

Here is a brief
exposition

>>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.