Re: continious operator in first-countable spaces

On 11-04-2006 17:12, 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.

Best regards,

Jose Carlos Santos
.

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