Errata

The purpose of this page is to host any and all errata for any work posted on this site.

21-268 Recitation Notes – Errata:

• Exercise 2.4: Property P needs to be stable under intersections.

• Section 4.3:  I claimed that κ : C^1([0,1]) -> R via κ(f) = f'(0) is a discontinuous linear map. Of course, this is false if you give C^1 the natural normed structure 

||f||_C^1 = ||f||_∞ + ||f'||_∞. 

However, it is indeed discontinuous if you just view C^1 as a subspace of C([0,1]) with the uniform norm, and this is what I meant.

Model theory pocket guide – Errata:

• "M models T" means T is a subset of Th(M), not equal to Th(M).

Realcompactness and fundamental groups – Errata:

• Definition 6: It is true that V ≠  L is independent of ZFC, and that the existence of a measurable cardinal implies V ≠  L. However, this does not immediately imply that the existence of measurable cardinals is independent of ZFC. In fact, one can show measurable cardinals are (strongly) inaccessible: they are uncountable regular cardinals λ such that κ < λ implies 2^κ < λ. Let I  be the axiom "there exists an inaccessible cardinal." The consistency strength of I  and related axioms is more complicated than one might expect:

  • It is true (in ZFC) that Con( ZFC ) implies Con( ZFC + ¬I ). Indeed, towards a contradiction, suppose that ZFC ⊢ I . Let λ be the least such inaccessible cardinal; one can show (in ZFC) that if λ is inaccessible, then the von Neumann universe V_λ of rank λ is a model of ZFC. Moreover, if λ is the least such, then this model has no inaccessible cardinals. So ZFC, if it proves I , would prove its own consistency and hence be inconsistent by Gödel's second incompleteness theorem. Hence, if ZFC is consistent, it cannot prove I  and thus ZFC + ¬I  is consistent. 

  • The same reasoning shows that ZFC + I  can prove Con( ZFC ). Thus even ZFC + I ,  let alone ZFC, cannot prove "Con( ZFC ) implies Con( ZFC + I )." At best, we have a conditional independence result:

IF ZFC + I  is consistent, THEN I  is independent of ZFC. 

• If we set J  = "there are two inaccessible cardinals," then ZFC + J  proves that Con( ZFC ) implies Con( ZFC + I  ), so ZFC + J  proves that I  is independent of ZFC but similarly cannot prove that J  is independent of ZFC. 

• Let M  = "there is a measurable cardinal." If ZFC + M  is consistent and κ is the first measurable cardinal, then |{λ < κ: λ is inaccessible}| = κ. Suffice it to say, the existence of a measurable cardinal is a rather strong assumption (which has seen incredible use in e.g. inner model theory), though not as strong as some other large cardinal axioms.

The point of this erratum was to clarify that the existence of a measurable cardinal is of strictly stronger consistency strength than the consistency of ZFC – it's not like the continuum hypothesis CH, where if ZFC is consistent then we can prove ZFC+CH and ZFC+¬CH both consistent.