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In article <Bx81Ho.9HL@cantua.canterbury.ac.nz> wft@math.canterbury.ac.nz 
(Bill Taylor) writes:

   >A = { 1 , { 2 , { 3 , { 4 , {.....}}}}}
   >B = { 1 , { 1 , { 1 , { 1 , {.....}}}}}
   >Can B above, be obtained from A above, by ZF methods.

  The question is vague in several respects, but the most likely
answer is no. A, to put this into formal ZF terms, is, let us say, the
range of a function f defined on the natural numbers, where
f(n)={n,f(n+1)} for all n.  Assuming the existence of such a function
in ZF (without foundation) does not in any obvious way (and hence not
at all, by the argument from experience) imply the existence of a set
x such that x={1,x}.  In particular, Replacement by itself is no help
since to get x using replacement, we need to define a function h by
h(n)=G(f(n)), where G({a,b})={1,G(b)}, and there is no way of defining
G in ZF (without foundation) so that this is provable. In other words:
Replacement allows us to make infinite "substitutions" in sets, but to
make infinite substitutions in non-well-founded sets we need a
corresponding non-well-founded recursion principle.

  Of course, even if this is correct it doesn't follow that accepting
A and rejecting B makes any sense. Or in other words: it doesn't help
explain what mathematical objects, if any, you have in mind.