# Tagged Questions

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### Does existence of a non-continuous linear functional depend on Axiom of Choice?

Well, it is easy to construct a non-continuous linear functional on an arbitrary infinite-dimensional vector space (assuming Choice, and taking a basis etc.). I think it is intuitive to say that: ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 ...
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### 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 $L^r$ ...
### Nonnegative linear functionals over $l^\infty$
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 ...