skip to primary navigationskip to content
 

Set Theory and its Philosophy

 Oxford University Press, 2004STAIP image

Hardback £78 ISBN 0-19-926973-4

Paperback £25 ISBN 0-19-927041-4

A comprehensive philosophical introduction to set theory. Anyone wishing to work on the logical foundations of mathematics must understand set theory, which lies at its heart. I offer a thorough account of cardinal and ordinal arithmetic, and the various axiom candidates. I discuss in detail the project of set-theoretic reduction, which aims to interpret the rest of mathematics in terms of set theory. The key question here is how to deal with the paradoxes that bedevil set theory. I offer a strikingly simple version of the most widely accepted response to the paradoxes, which classifies sets by means of a hierarchy of levels. What makes the book unusual is that it interweaves a careful presentation of the technical material with a detailed philosophical critique: I do not merely expound the theory dogmatically but at every stage discuss in detail the reasons that can be offered for believing it to be true.

Contents

Part I: Sets

  1. Logic
  2. Collections
  3. The hierarchy
  4. The theory of sets

Part II: Numbers

  1. Arithmetic
  2. Counting
  3. Lines
  4. Real numbers

Part III: Cardinals and Ordinals

  1. Cardinals
  2. Basic cardinal arithmetic
  3. Ordinals
  4. Ordinal arithmetic

Part IV: Further Axioms

  1. Orders of infinity
  2. The axiom of choice
  3. Further cardinal arithmetic

Appendices

  1. Traditional axiomatizations
  2. Classes
  3. Sets and classes

Reviews

(Links may not work from your web address.)

Errata

p.44 l.14 For V V' read V V'.

p. 46 For the proof of proposition 3.6.13 substitute the following (thanks to Franz Fritsche):

Suppose that the proposition is false and V is the lowest level for which it fails, so that there is a history W of V such that W ≠ {V' : V'V}. Certainly V'WV'V. So suppose now that V'W. Then for every V''W we have V''V' and V'V'' (since if V'V''W then V' ∈ W), and so V''V' [proposition 3.6.11]. So W ⊆ {V'' : V''V'}, whence V = acc(W) ⊆ acc{V'' : V''V'} = V' and therefore V'V (since otherwise V'V'). Contraposition gives V'VV'W. Hence W = {V' : V'V}. Contradiction. ☐