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I read that Alexander subbase theorem is equivalent to the Boolean prime ideal theorem which is weaker than AC. But AC implies Alexander subbase theorem and subbase theorem implies Tychonoff, see here: Help needed with last step in proof of Tychonoff theorem

Also, Tychonoff implies AC. Therefore subbase theorem is equivalent to AC? Which is wrong: this or that subbase theorem is equivalent to Boolean prime ideal theorem? Thank you for correcting me.

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dolan
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    If you carefully go through the proof of Tychonoff’s Theorem that uses the Alexander Subbase Theorem as one of its steps, you will realize that the Axiom of Choice is used at some other step. – Haskell Curry Feb 16 '13 at 08:05

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The act of choosing the $ x_{i} $’s in the linked thread requires the Axiom of Choice, and this step is separate from the application of the Alexander Subbase Theorem. The Wikipedia article on Subbase states this quite explicitly.

The role of the Alexander Subbase Theorem in the proof of Tychonoff’s Theorem is simply one of simplification; it is not powerful enough on its own to imply Tychonoff’s Theorem.

Haskell Curry
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