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Full Idea
Ante rem structuralism, eliminative structuralism formulated over a sufficiently large domain of abstract objects, and modal eliminative structuralism are all definitionally equivalent. Neither is to be ontologically preferred, but the first is clearer.
Gist of Idea
The main versions of structuralism are all definitionally equivalent
Source
Stewart Shapiro (Philosophy of Mathematics [1997], 7.5)
Book Ref
Shapiro,Stewart: 'Philosophy of Mathematics:structure and ontology' [OUP 1997], p.242
A Reaction
Since Shapiro's ontology is platonist, I would have thought there were pretty obvious grounds for making a choice between that and eliminativm, even if the grounds are intuitive rather than formal.
9153 | Dedekindian abstraction talks of 'positions', where Cantorian abstraction talks of similar objects [Dedekind, by Fine,K] |
10270 | The main versions of structuralism are all definitionally equivalent [Shapiro] |
10169 | Relativist Structuralism just stipulates one successful model as its arithmetic [Reck/Price] |
10179 | There are 'particular' structures, and 'universal' structures (what the former have in common) [Reck/Price] |
10181 | Pattern Structuralism studies what isomorphic arithmetic models have in common [Reck/Price] |
10182 | There are Formalist, Relativist, Universalist and Pattern structuralism [Reck/Price] |
14085 | 'Deductivist' structuralism is just theories, with no commitment to objects, or modality [Linnebo] |
14084 | Non-eliminative structuralism treats mathematical objects as positions in real abstract structures [Linnebo] |
14086 | 'Modal' structuralism studies all possible concrete models for various mathematical theories [Linnebo] |
14087 | 'Set-theoretic' structuralism treats mathematics as various structures realised among the sets [Linnebo] |
8699 | Are structures 'ante rem' (before reality), or are they 'in re' (grounded in physics)? [Friend] |