The axiom of choice has many counterintuitive consequences like the Banach-Tarski paradox. The Hahn-Banach theorem is a consequence of the axiom of choice, but it is weaker.

I would like to know some counterintuitive consequences of the Hahn-Banach theorem. Can the Banach-Tarski paradox be derived from the Hahn-Banach theorem?

  • $\begingroup$ Asaf Karagila answers affirmatively here: math.stackexchange.com/a/156216/43208 I found this through a quick Google search. $\endgroup$
    – Todd Trimble
    Nov 5, 2014 at 16:27
  • $\begingroup$ This question appears to be off-topic because it is about consequences of the axiom of choice and the OP did not search the standard reference. $\endgroup$ Nov 5, 2014 at 16:30
  • $\begingroup$ Also see here: matwbn.icm.edu.pl/ksiazki/fm/fm138/fm13813.pdf $\endgroup$
    – Todd Trimble
    Nov 5, 2014 at 16:31
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    $\begingroup$ Let me elaborate a bit on my vote to close above. Every time you have a question about whether two weak forms of axiom of choice imply each other, the very first thing you do is to search the consequences of AoC project. Using the string search you find that Banach-Tarski is form 309, and Hahn-Banach is form 52. Then you type into the first text box the two numbers "52, 309", hit enter, and you get a nice little mutual implication box. $\endgroup$ Nov 5, 2014 at 16:32
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    $\begingroup$ No, I think it's a fine question; my only objection is that it's so easy to find the answer. But, not everybody knows what Willie Wong asserts is "standard", and he provides information well worth knowing. Let it stand. $\endgroup$
    – Todd Trimble
    Nov 5, 2014 at 16:33

1 Answer 1


To aid future inquiries, let us record here the go-to reference kindly provided by Willie Wong in a comment above: http://consequences.emich.edu/conseq.htm. This provides a data base with a utility to search for known implications between weak forms of the axiom of choice.

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    $\begingroup$ This type of answer is why you got my vote for moderator :) $\endgroup$ Nov 5, 2014 at 21:37
  • $\begingroup$ @BenjaminSteinberg Thanks. This particular case wasn't a difficult one. :-) $\endgroup$
    – Todd Trimble
    Nov 5, 2014 at 21:53

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