Assume (iii). Take and let be the two coordinate projections, continuous by the definition of the product topology. Their equality set is the diagonal, so it is closed. For distinct , the point lies in its open complement. Choose a basic rectangle containing it and contained in that complement. If were nonempty, a point in their intersection would give in the rectangle, impossible. Thus are disjoint open neighborhoods of , proving that is Hausdorff. Hence (iii) implies (i), completing all three equivalences.

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