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Yog79

A field with one element.

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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?

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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.

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