# Tagged Questions

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### Why is adopting Russell's Axiom of Reducibility as strong as eliminating the Ramified Hierarchy?

In order to respond to concerns of impredicativity, Bertrand Russell developed a system of ramified second-order logic, which is like regular second-order logic except the comprehension schema is ...

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### Has the Ramified Theory of Types been applied to Predicative Set Theories?

Questions of predicativity are well-studied in the context of arithmetic. We have a base theory, first-order Peano arithmetic. Some people, like Edward Nelson (in chapter 1 of his book) and Charles ...

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### Did Gödel prove that the Ramified Theory of Types collapses at $\omega_1$?

Second-Order Arithmetic is considered impredicative, because the comprehension scheme allows formulas with bound second-order variables that range over all sets of natural numbers, including the set ...

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343 views

### Is there a notion of “predicative given the real numbers”?

A definition is called impredicative if it involves quantification over a domain that contains the thing being defined. For instance, if you define hereditary property to be a property which applies ...

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### Does the Feferman-Schutte analysis give a precise characterization of Predicative Second-Order Arithmetic?

A definition is called impredicative if it involves quantification over a domain that contains the thing being defined. For instance, if you define hereditary property to be a property which applies ...

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### Illustrating Edward Nelson's Worldview with Nonstandard Models of Arithmetic

Mathematician Edward Nelson is known for his extreme views on the foundations of mathematics, variously described as "ultrafintism" or "strict finitism" (Nelson's preferred term), which came into the ...

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349 views

### Cardinal Arithmetic, foundations and constructive math

This is not my area but a question occurred to me that I can not find the answer to. There is a very strong axiom of constructibility which ironically gives us highly non-constructive math (GCH is one ...

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### Is there a finite-dimensional vector space whose dimension cannot be found? [closed]

Is there a finite-dimensional vector space whose dimension cannot be found? Assume, we have somehow constructed a vector space whose dimension is finite, but yet unknown. Is there always an algorithm ...