Re: compact are bounded?

"Antonio" <anton.cali@xxxxxxxx> writes:

For what value of "have an example"? The algebraic dual of an
infinite-dimensional topological vectorspace is always strictly
larger than its topological dual, isn't it? And any member of
the first which isn't in the second is a discontinuous rank-1
linear map, no?

What do you mean with "rank-1" linear map ? More precisely, what's the
role of " -1 " ?

Typographical convention: I mean "of rank 1" (that "-" is a hyphen,
not a minus sign). --However, I see that my comparative unfamiliarity
with the area led me into error; G. Edgar's version of the construction
got it right.

Lee Rudolph
.

Relevant Pages

  • Re: compact are bounded?
    ... infinite-dimensional topological vectorspace is always strictly ... And any member of ... the first which isn't in the second is a discontinuous rank-1 ... What do you mean with "rank-1" linear map? ...
    (sci.math)
  • Re: compact are bounded?
    ... is it true that a finite rank linear map T: ... discontinuous linear maps are not compact. ... infinite-dimensional topological vectorspace is always strictly ...
    (sci.math)