# Assume ZFC is consistent

1. Apr 21, 2009

### sairalouise

If you assume that ZFC is consistent, then by the main theorem of model theory ZFC has a model, let the model be countable.
Since ZFC proves: "there is a set consisting of all real numbers" there is a point a belonging to M such that:
M satisfies " a is the set of all real numbers"
But since M is countable there are only countably many points b belonging to M such that:
M satisfies b belonging to a, so a contains only countably many elements. But the real numbers is uncountable, what has happened? Shouldnt a be uncountable?

2. Apr 21, 2009

### Hurkyl

Staff Emeritus
This is called Skolem's paradox. It arises by (accidentally) equivocating the word "countable" used in two different contexts.

e.g. you'll find that your countable model of ZFC does not contain any bijection between the model's version of N and the model's version of R. Any bijections that do exist between them are "external", meaning they do not correspond to something in the model.