[FOM] Algebraic closure of Q

[Thomas Forster]

The way in which this question involves AC calls to mind a
related problem.   It is known that if AC fails badly enough,
then the reals can be a union of countably many countable sets.
Can there be - again if AC fails badly enough - a countable 
subgroup of the full symmetric group on the naturals - of 
countable index?   The point is that altho' the cosets of a 
subgroup are always "pairwise equipollent", they are not 
*uniformly* equipollent....

      tf


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