Yog79 0 Posted May 3 Since $\mathbb{R}$ is not first order definable, the statement $\forall x \in \mathbb{R}(x = x)$ is not a first order statement and thus not provable in ZFC. Does that mean we can assume $\exists x \in \mathbb{R}(x \neq x)$? If so, would this provide us with the basis for a field with one element? 0 Share this post Link to post Share on other sites

John Cuthber 3703 Posted May 3 Am I the only one who thought of this? 0 Share this post Link to post Share on other sites

taeto 92 Posted May 3 The truth of the predicate \(x=x\) follows from logic, i.e., it is how equality is defined. It has nothing to do with \(\mathbb{R}\). The field \( ( \{0\}, +, \cdot ) \) with one element is kind of okay as a field, as it satisfies the important axioms. Yet most authors exclude it explicitly, because it is a cumbersome exception for some theorems. In this 'field' you can always divide by zero, for example.so it is a little too exceptional. 0 Share this post Link to post Share on other sites