About the existence of Hamel basis for vector spaces

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Discussion Overview

The discussion centers on the existence of Hamel bases for vector spaces, particularly in the context of the Axiom of Choice (AC). Participants explore whether vector spaces can exist without a Hamel basis when AC is not assumed, examining implications for specific examples like ##\mathbb R_\mathbb Q##.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that every vector space has a Hamel basis, contingent on the Axiom of Choice being true.
  • Others argue that if the Axiom of Choice is false, there can exist vector spaces without a Hamel basis, citing ##\mathbb R_\mathbb Q## as a potential example.
  • There are claims that the exclusion of AC only affects the proof of the existence of Hamel bases, not the existence itself.
  • Some participants express uncertainty about whether the absence of AC guarantees the existence of vector spaces without a basis.
  • Discussion includes the equivalence of the Axiom of Choice and the existence of Hamel bases, with some questioning the implications of this equivalence.
  • Participants note that while it is consistent with ZF set theory that ##\mathbb R_\mathbb Q## does not have a basis, it remains unclear whether it actually does or does not have one.
  • There is a suggestion that the discussion may relate to Gödel's incompleteness theorems, though some participants believe it does not.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether vector spaces can exist without a Hamel basis when the Axiom of Choice is excluded. Multiple competing views remain regarding the implications of AC on the existence of bases in vector spaces.

Contextual Notes

Participants highlight that without the Axiom of Choice, it is challenging to define dimensions for arbitrary vector spaces and that the existence of bases may depend on the cardinality of generating subsets.

  • #31
A well-formed formula (wff) is (syntactically) provable if and only if it is valid. This defines the relation between (Semantic) truth and (Syntactic) provability. I'm not aware of how the term 'hold' is used, by there are wff's in Sentence Logic that are contingent. And, yes, every provable statement in FOL os Sentence Logic is a Tautology; true in every model.
 
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  • #32
WWGD said:
A well-formed formula (wff) is (syntactically) provable if and only if it is valid. This defines the relation between (Semantic) truth and (Syntactic) provability. I'm not aware of how the term 'hold' is used, by there are wff's in Sentence Logic that are contingent. And, yes, every provable statement in FOL or Sentence Logic is a Tautology; true in every model.
Ok, a well-formed formula (wff) is (syntactically) provable if and only if it is valid since we're assuming a sound and (semantically) complete logic system (like FOL or Sentential (or propositional) logic are).

What does it mean that in Sentential logic there are contingent wffs ?
 
  • #33
cianfa72 said:
Ok, a well-formed formula (wff) is (syntactically) provable if and only if it is valid since we're assuming a sound and (semantically) complete logic system (like FOL or Sentential (or propositional) logic are).

What does it mean that in Sentential logic there are contingent wffs ?
Statements that aren't tautologies, like ##A \rightarrow B ##, which is not true when A is true and B is false.
 
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