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Assume that $A\sim A\cup \{ A\}$, does that imply that $P(P(A))\sim P(P(A))\times P(P(A))$ without using the axiom of choice?

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This is a matter of cardinal arithmetic. Let $a$ denote $|A|$.

Then $2^{2^a}\times 2^{2^a}=2^{2^a+2^a}$. So it is enough to prove that $2^a+2^a=2^a$. But $2^a+2^a=2^{a+1}$. Finally, using the assumption, the conclusion follows.

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    Could it be done without cardinal arithmetic and using functions and well known fact about cardinality? – 2017-02-17
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    Just "unroll" the cardinal arithmetic to produce bijections. – 2017-02-17
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    Thank you so much for your time, but can you show me the proof explicitly in order to have it if I give up, because I am completely stuck. – 2017-02-17
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    Open any book, and work the proofs about cardinal arithmetic. They are often constructive. From one bijection to the next. And just work this sort of argument here. – 2017-02-17
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    But the first property works only if they are disjoint – 2017-02-17
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    I only have proof when they are disjoint. Can you give me more information about where I can see proof when they are not disjoint? – 2017-02-17