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

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
  • 39.8k
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
  • 39.8k
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
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