# Tag Info

### Axiom of choice, Banach-Tarski and reality

There are two ingredients in the Banach-Tarski decomposition theorem: The notion of space, together with derived notions of part and decomposition. The axiom of choice. Most discussion about the ...
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### Zorn's lemma: old friend or historical relic?

I agree with almost everything in your post. But still, I believe I know why people use Zorn's lemma. My answer. Zorn's lemma encapsulates succinctly many of the consequences of AC via transfinite ...
Accepted

Accepted

### A Krull-like Theorem and its possible equivalence to AC

Nice question ! I believe the homework exercise implies AC. Indeed, assume its conclusion holds, and let $R$ be a ring with no maximal ideal. I'm going to prove that $R$ is zero, thus proving Krull's ...
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### Do vector spaces without choice satisfy Cantor-Schroeder-Bernstein?

There are models of ZF+DC in which every subset of every Polish space has the property of Baire (I can try to add references later, I think to Solovay and Shelah, but these are pretty well known). ...
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### Bishop quote stating that axiom of choice is constructively valid

According to the BHK interpretation of intuitionistic logic we have that: A proof of $\exists x \in A . \phi(x)$ consists of a pair $(a, p)$ where $a \in A$ and $p$ is a proof of $\phi(a)$. A proof ...
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### Applications of Zorn’s lemma that aren’t chain-complete/directed-complete?

On the one hand, one might expect that there can be no fully satisfactory example of the phenomenon, in light of the observations mentioned in the comments, namely, that every partial order fulfilling ...

### Does Con(ZF + Reinhardt) really imply Con(ZFC + I0)?

The answer to your question is (almost) yes (almost is because of the addition of DC to the statement). Recently Gabriel Goldberg has proved ''Con(NBG+DC+Reinhardt)$\implies$ Con(ZFC+I0)''. ...
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### Axiom of choice, Banach-Tarski and reality

Are there reasons why it is plausible (for physicists, philosophers, mathematicians) to believe that not all sets should be measurable? Yes. If every set of reals is Lebesgue measurable, then you ...
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