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We know that if $S$ is an amorphous set then $\mathcal{P}(S)$ is Dedekind-finite. I was wondering if the implication was reversible and decided to generalize a bit though I think the question in the title is perhaps a bit too open-ended.

If $\mathcal{P}(S)$ is an infinite Dedekind-finite set, is $S$ amorphous? Can we say much else about $S$?

Asaf Karagila
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P-addict
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    If $X$ and $Y$ are disjoint infinite sets such that $\mathcal P(X)$ and $\mathcal P(Y)$ are Dedekind-finite, then $\mathcal P(X\cup Y)\cong\mathcal P(X)\times\mathcal P(Y)$ is an infinite Dedekind-finite set, but $X\cup Y$ is not amorphous. – bof Aug 24 '19 at 05:32
  • Ah, that was pretty quick. Ok, thanks. – P-addict Aug 24 '19 at 05:33

1 Answers1

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There is one essential theorem here, due to Kuratowski.

$\mathcal P(X)$ is Dedekind-infinite if and only if $X$ can be mapped onto $\omega$.

It follows that $\mathcal P(X)$ being Dedekind-finite means that $X$ cannot be mapped onto $\omega$. From this it is easy to see that if $A$ and $B$ both cannot be mapped onto $\omega$, then $A\cup B$ cannot either. But if $A$ and $B$ are infinite, then $A\cup B$ is definitely not amorphous. This was pointed out by bof in the comments as well.

But now you might ask, perhaps it is true that every infinite set which cannot be mapped onto $\omega$ contains an amorphous subset? Maybe amorphous sets are the root of the problem.

The answer to that is also negative. It is possible that there is a linearly ordered space which is densely ordered, but has a Dedekind-finite power set. In particular, such set cannot have an amorphous subset. Since it's linearly ordered.

(See more here.)

Asaf Karagila
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