On Thursday, January 10, 2013 10:48:46 AM UTC, William Elliot wrote: > On Thu, 10 Jan 2013, firstname.lastname@example.org wrote: > > > > > Let A and B be sets. Assume ZF without assuming choice. Then, for all > > > non negative integers n, I believe (correct me if I'm wrong) that, if n > > > x A is equipotent to n x B, then A is equipotent to B. > > > > It's false for n = 0. > > > > If A is infinite, the so is B and > > A equipotent n x A equipotent n x B equipotent B > > > > If A finite, then so is B and for n > 0, if n|A| = n|B| in N. > > then |A| = |B|. > > > > > There's a famous Conway/Doyle paper which proves this for n = 2 and n = > > > 3. > > > > What do you mean this for n = 2 and n = 3, when this is about all n >= 0? > > > > > However, it doesn't seem rigorous or clear and I have trouble > > > understanding it. > > > > I don't uunderstand what you're talking about. > > > > > Does anyone know a more axiomatic treatment? (I don't have access to a > > > university, and I'm not in the market for maths purchases, so only free > > > references would be helpful.) > > > > > How about an explicit statement of "this for n = 2 and n = 3"?
You are correct that my statement is false for n = 0 -- apologies for the error.
The requested explicit statement is:
Let n be an integer such that either n = 2 or n = 3. Let A and B be sets such that a bijection exists between n x A and n x B. Prove without using the axiom of choice that a bijection exists between A and B.
As for your comment about A being equipotent to n x A when A is infinite, there are 3 (exhaustive but not mutually exclusive) possibilities. Possibility 1 is that you are assuming the axiom of choice. Possibility 2 is that your asssertion is not even true without the axiom of choice. Possibility 3 is that your assertion is true without assuming the axiom of choice, but is highly non-trivial.
I am interested in extending the above to the case where n > 3.