display all the ideas for this combination of philosophers
13 ideas
| 13643 | Aristotelian logic is complete [Shapiro] |
| Full Idea: Aristotelian logic is complete. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 2.5) | |
| A reaction: [He cites Corcoran 1972] |
| 10206 | Modal operators are usually treated as quantifiers [Shapiro] |
| Full Idea: It is common now, and throughout the history of philosophy, to interpret modal operators as quantifiers. This is an analysis of modality in terms of ontology. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], Intro) |
| 13651 | A set is 'transitive' if contains every member of each of its members [Shapiro] |
| Full Idea: If, for every b∈d, a∈b entails that a∈d, the d is said to be 'transitive'. In other words, d is transitive if it contains every member of each of its members. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 4.2) | |
| A reaction: The alternative would be that the members of the set are subsets, but the members of those subsets are not themselves members of the higher-level set. |
| 13647 | Choice is essential for proving downward Löwenheim-Skolem [Shapiro] |
| Full Idea: The axiom of choice is essential for proving the downward Löwenheim-Skolem Theorem. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 4.1) |
| 10208 | Axiom of Choice: some function has a value for every set in a given set [Shapiro] |
| Full Idea: One version of the Axiom of Choice says that for every set A of nonempty sets, there is a function whose domain is A and whose value, for every a ∈ A, is a member of a. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], 1) |
| 10252 | The Axiom of Choice seems to license an infinite amount of choosing [Shapiro] |
| Full Idea: If the Axiom of Choice says we can choose one member from each of a set of non-empty sets and put the chosen elements together in a set, this licenses the constructor to do an infinite amount of choosing. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], 6.3) | |
| A reaction: This is one reason why the Axiom was originally controversial, and still is for many philosophers. |
| 10301 | The axiom of choice is controversial, but it could be replaced [Shapiro] |
| Full Idea: The axiom of choice has a troubled history, but is now standard in mathematics. It could be replaced with a principle of comprehension for functions), or one could omit the variables ranging over functions. | |
| From: Stewart Shapiro (Higher-Order Logic [2001], n 3) |
| 13631 | Are sets part of logic, or part of mathematics? [Shapiro] |
| Full Idea: Is there a notion of set in the jurisdiction of logic, or does it belong to mathematics proper? | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], Pref) | |
| A reaction: It immediately strikes me that they might be neither. I don't see that relations between well-defined groups of things must involve number, and I don't see that mapping the relations must intrinsically involve logical consequence or inference. |
| 13654 | It is central to the iterative conception that membership is well-founded, with no infinite descending chains [Shapiro] |
| Full Idea: In set theory it is central to the iterative conception that the membership relation is well-founded, ...which means there are no infinite descending chains from any relation. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 5.1.4) |
| 13640 | Russell's paradox shows that there are classes which are not iterative sets [Shapiro] |
| Full Idea: The argument behind Russell's paradox shows that in set theory there are logical sets (i.e. classes) that are not iterative sets. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 1.3) | |
| A reaction: In his preface, Shapiro expresses doubts about the idea of a 'logical set'. Hence the theorists like the iterative hierarchy because it is well-founded and under control, not because it is comprehensive in scope. See all of pp.19-20. |
| 13666 | Iterative sets are not Boolean; the complement of an iterative set is not an iterative sets [Shapiro] |
| Full Idea: Iterative sets do not exhibit a Boolean structure, because the complement of an iterative set is not itself an iterative set. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 7.1) |
| 13653 | 'Well-ordering' of a set is an irreflexive, transitive, and binary relation with a least element [Shapiro] |
| Full Idea: A 'well-ordering' of a set X is an irreflexive, transitive, and binary relation on X in which every non-empty subset of X has a least element. | |
| From: Stewart Shapiro (Foundations without Foundationalism [1991], 5.1.3) | |
| A reaction: So there is a beginning, an ongoing sequence, and no retracing of steps. |
| 10207 | Anti-realists reject set theory [Shapiro] |
| Full Idea: Anti-realists reject set theory. | |
| From: Stewart Shapiro (Philosophy of Mathematics [1997], Intro) | |
| A reaction: That is, anti-realists about mathematical objects. I would have thought that one could accept an account of sets as (say) fictions, which provided interesting models of mathematics etc. |