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

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

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.

Thank you for all kind help and comments.

• The title of your question does not have much to do with your actual question. The answer is "yes" if the axiom of choice is assumed. However, from your use of the term "Dedekind-infinite" instead of just plain "infinite", I surmise that the axiom of choice is not assumed?
– bof
May 9, 2017 at 22:42
• No. When you assume the axiom of choice you have that Dedekind-infinite and Infinite are equivalent, but my question just consider the definition I wrote. In other words, from this definition, is it possible deduce what I ask?
– Byag
May 9, 2017 at 23:12
• @Edonbrayn bof's point is that it is possible to deduce that if you assume the axiom of choice - you need to specify what axioms you're using. Of course it's clear from context that you mean ZF, but that information is technically part of the question. May 10, 2017 at 0:03
• Ok, I didn't understand you before. So, assuming the axiom of choice, how I can prove this?
– Byag
May 10, 2017 at 0:10

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.

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.

• It seems to me you don't need an amorphous set, just a common garden variety D-finite infinite set. Have I erred?
– bof
May 9, 2017 at 23:06
• @bof Nope, I just really like amorphous sets! May 9, 2017 at 23:08
• But this does not yet answer the question. You need to argue that there are not many subsets of size $X$, but they don't need to be disjoint. May 9, 2017 at 23:59
• Oh, I see the title is a different question. Anyway... this does not yet answer the title. May 10, 2017 at 0:01