# Axiom of choice: Replacing a strong condition with a weaker condition

• A
• jose diez
jose diez
This set-theory theorem is very easy to prove:
(*) if A≈B & C≈D & A∩C=∅ & B∩D=∅ then A∪C≈B∪D
It seems intuitive that if one replaces the strong
A∩C=∅ & B∩D=∅
condition by the weaker
A∩C≈B∩D
the implication
(**) if A≈B & C≈D & A∩C≈B∪∩D then A∪C≈BD
still holds.
(**) does not seem to be much stronger than (*), nevertheless I have been able to prove (**) only using Ax of Choice (ACh). This suggested to me that (**) might be other equivalent to ACh, but I have not found it in the standard lists, nor I have been able to prove that (**) implies ACh.
Does anybody have any clue on this?

Sorry there was a typo in (**)

Corrected:
(**) if A≈B & C≈D & A∩C≈B∩D then A∪C≈B∪D

What do the squiggly equal signs stand for? Equal cardinality?

yes

jose diez said:
Does anybody have any clue on this?
I haven't worked through it, but isn't it possible to prove both (*) or (**) by induction without invoking AC? In this case, they would be independent of AC so cannot be equivalent.

jose diez said:
Corrected:
(**) if A≈B & C≈D & A∩C≈B∩D then A∪C≈B∪D
Shooting from hip, I'd say choice is not necessary for this to occur. To prove choice, we would have an arbitrary family of nonempty sets and we must find a choice function. The condition (**) is formulated in only finite terms. I don't see how it provides an angle to tackle with infinite families.

That said, I don't have a counterexample.

If you subtract ##A\cap C## from the left side of everything, and ##B\cap D## on the right side, then you have reduced to the case of no intersection as long as you can show the following result:

If ##B\subset A##, ##D\subset C##, ##D\approx B##, ##A\approx C##, then ##A-B \approx C- D##

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