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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.

14 votes

Continuous linear functionals and the Axiom of Choice

No, this is equivalent to the Hahn–Banach theorem already in the case of Banach spaces. See in my write up: https://arxiv.org/abs/2010.15632
Asaf Karagila's user avatar
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2 votes

Cardinality of $C^*([0,1])$

Recall that $[0,1]$ is compact and therefore $C([0,1])$ is a Polish space. By standard arguments, there are only $2^{\aleph_0}$ continuous functions between two Polish spaces. Therefore there are at …
Asaf Karagila's user avatar
  • 39.8k
18 votes
Accepted

Non continuous Linear form on $E=C([0,1],\mathbb{R})$ without AC

No. It's impossible. In certain models of $\sf ZF+DC$ there is a property known as "automatic continuity" for Banach spaces, that means that every linear operator to a normed space is continuous. Su …
Asaf Karagila's user avatar
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4 votes

Can we choose an element from a class?

It is often tempting to think that all our existential instantiations happen "in advance", a sort of proof theoretic "mise en place" if you will. But they don't, and they don't have to be. For a given …
Asaf Karagila's user avatar
  • 39.8k