Re: continious operator in first-countable spaces

On 11-04-2006 21:01, eugene wrote:

Prove that in a vector topological space with
first-countability axiom
every bounded linear operator is continious.
What's the meaning of "bounded linear operator" in
this context? I'm
aware of the meaning in the context of linear
operators between normed
vector spaces; if T is such an operator, we then
say
that it is bounded
if there's some K > 0 such that, for every vector
_v_
on the domain of
T, ||T(v)|| <= K.||v||. Then, yes, bounded =>
continuous.

Yes, if you are dealing with normed vector spaces,
it is to
prove this fact and i know how to do it. But here
is more
general consept: vector topological space, here we
have a
topological structure and vector space structure on
a set,
such that (x,y)->x+y and (k,x)->kx are
continious(is a sense
of topology)-here is in vector topological
space(you can
search for surely better definition).

I know what a topological space. That was not my
question.

So, in vector topological spaces the notion of
boundedness is
slighly different: the set E is bounded is for
every
neighborhood of 0 there is a number s such that E
\in tV
for all t>s.

And boundedness here in this sense.

This makes no sense or, at least, it is no an answer
to my question.
What you did was to define the meaning of the word
"bounded" for subsets
of a topological vector space. My question was about
the meaning of that
word when it is applied to linear maps between
topological vector
spaces. My guess is that it means that t maps bounded
sets into bounded
sets.
Ah, obviously i misunderstood you. Yes, you guess is right, i meant that a linear map called bounded if it maps bounded sets to bounded sets.
Sorry if my poor explanation of vector space and boundedness may have offended you a little, it was not my purpose.
.

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