Is this true?

Every ordered field is Archimedean

No — it is false.

The counterexample

The object that kills this claim comes with Trap Analysis.

86 counterexamples are free — including most of the ones Part C leans on hardest.

The kind of mistake this is

Finite-dimensional intuition

Something true in ℝⁿ, or in a nice space, assumed in general.

Drill statements like this

Others that fail the same way

An arbitrary intersection of open sets is open

Explained in full in The real line: supremum, Archimedes and density — free to read

From The Real Line › Completeness, sup/inf, Archimedean property

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