¿A Dedekind-infinite set $A$ has as many proper subsets, equipotent to $A$, as its own cardinality?

Question

Definition: A set $A$ is said to be Dedekind-infinite if it has a proper subset $X$ equipotent (similar/in bijection) to $A$.

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My question is as follows, how to prove that if $A$ is a Dedekind-infinite set there exists $\{A_i\}_{i\in I}$, a collection of proper subsets of $A$, such that $A_i\subsetneq A$ and $A_i\sim A$ for each $i\in I$, where $I\sim A$, and $A_i\cap A_j=\emptyset$ if $i\not=j$?

Here $X\sim Y$ means "exists $f:X\rightarrow Y$ a bijective function".

The question comes to me in the middle of a proof and I think is intuitive correct. In addition, in the case of $\mathbb{N}$ we have that the sets of powers of prime numbers is a collection with that property.

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Thank you for all kind help and comments.

Answer 1

Suppose there is a Dedekind-finite infinite set $S.$ Then $A=\mathbb N\cup S$ is a Dedekind-infinite set.

Claim. There do not exist two disjoint subsets $A_1,A_2$ of $A$ such that $A_1\sim A_2\sim A.$ In fact, there do not exist two disjoint subsets $S_1,S_2$ of $A$ such that $S_1\sim S_2\sim S.$

Proof. Assume for a contradiction that $S_1,S_1\subseteq A$ and $S_1\cap S_2=\emptyset$ and $S_1\sim S_2\sim S.$ Since $S_1$ and $S_2$ are Dedekind-finite infinite sets, $S_1\cap\mathbb N$ and $S_2\cap\mathbb N$ are finite, and $S_2\setminus\mathbb N$ is infinite. Thus we have $S_1\cap\mathbb N\sim F$ for some finite set $F\subset S_2\setminus\mathbb N,$ and then $S'=(S_1\setminus\mathbb N)\cup F\sim S$ and $S'$ is a proper subset of $S,$ contrary to the assumption that $S$ is Dedekind-finite.

Answer 2

The statement you are trying to prove is not true.

Note that if $A$ is Dedekind-infinite, then so is any set containing $A$. So weird Dedekind-infinite sets can be built by taking some nice Dedekind-infinite set - say, $\mathbb{N}$ - and looking at the union of that with some really weird set $A$. For example, we could take $A$ to be amorphous - that is, infinite but not partitionable into disjoint infinite sets! (Such sets can exist in ZF, but this is not easy to prove.)

Now look at $X=\mathbb{N}\cup A$ for $A$ amorphous. EDIT: As bof points out below, we actually don't need $A$ to be amorphous, it's enough for $A$ to be any Dedekind-finite infinite set; I just like amorphous sets. I claim that $X$ can't be split into $X$-many sets of size $X$. In fact, I claim:

There are no disjoint $Y, Z\subseteq X$ with $Y\sim Z\sim X$.

That is, we can't even find two sets that we want!

The culprit is $A$, via the following fact: $$\mbox{There is no injection from any Dedekind-finite infinite set into any well-ordered set}.$$ This is a good exercise (hint: think about preimages), and in this context means that if $Y\subseteq X$ with $f: X\rightarrow Y$ a bijection, then $f(a)\in Y\cap A$ for all but finitely many $a\in A$; that is, when we "shrink" $X$ down, "most" of the $A$-part of $X$ stays in the $A$-part.

So suppose $Y, Z$ were disjoint subsets of $X$ of size the same as $X$. Fix bijections $f, g: X\rightarrow Y, Z$, and let

• $U=\{b\in A: f(a)=b\mbox{ for some $a\in A$}\}$,

• $V=\{b\in A: g(a)=b\mbox{ for some $a\in A$}\}$.

Clearly $U$ and $V$ are disjoint, and are infinite by the above observation; but this contradicts the fact that $A$ is amorphous.