[FOM] "Mathematician in the street" on AC

W.Taylor@math.canterbury.ac.nz W.Taylor at math.canterbury.ac.nz
Fri Aug 28 23:22:14 EDT 2009

Quoting Vaughan Pratt <pratt at cs.stanford.edu>:

>> No, because the latter is a countable union of null sets and hence
>> itself null.
> So where exactly must Choice be used here?

The main idea is that one has a countable collection of sets which
are each coverable by (necessarily countable) collections of intervals
of arbitrarily small total finite length.  Then when one comes to try
to cover the union set with a collection of intervals, one must pick
a countable (and small-lengthed) bunch of intervals from the countable
union of countable sets.  This requires making a countable number of
choices to do so, hence AC.

-- Bill Taylor

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