2
votes
0answers
143 views

Hahn-Banach via Hamel Basis

my question for tonight: Is there a proof for Hahn-Banach using a Hamel Basis? I know, the proof for existence of Hamel Bases uses already Axiom of Choice, but I'd like to apply this without refering ...
5
votes
1answer
425 views

On every infinite-dimensional Banach space there exists a discontinuous linear functional.

On every infinite-dimensional Banach space there exists a discontinuous linear functional. Assuming the axiom of choice, every vector space has a basis. With an infinite basis, I can define on a ...
10
votes
1answer
444 views

A Hamel basis for $l^{\,p}$?

I am looking for an explicit example for a Hamel basis for $l^{\,p}$?. As we know that for a Banach space a Hamel basis has either finite or uncountably infinite cardinality and for such a basis one ...
90
votes
1answer
3k views

Does the open mapping theorem imply the Baire category theorem?

A nice observation by C.E. Blair1, 2, 3 shows that the Baire category theorem for complete metric spaces is equivalent to the axiom of (countable) dependent choice. On the other hand, the three ...
32
votes
2answers
1k views

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$?

Is there an explicit isomorphism between $L^\infty[0,1]$ and $\ell^\infty$? In some sense, this is a follow-up to my answer to this question where the non-isomorphism between the spaces $L^r$ ...
9
votes
1answer
803 views

Nonnegative linear functionals over $l^\infty$

My purpose is a clarification of the role of the axiom of choice in constructing limits for bounded sequences. Namely, we want a linear functional of norm 1 defined on the space of all bounded complex ...
11
votes
1answer
506 views

Are the coordinate functions of a Hamel basis for an infinite dimensional Banach space discontinuous?

The question is in the title really, but I suppose I could at least fix some notation here. Let $X$ be an infinite-dimensional Banach space - over the reals for the sake of concreteness. Use choice ...