For Infinite $B\subseteq A$, do $|A\times A|=|A|$, $B\subset A$, and $|B|<|A|$ Imply $|A\setminus B|=|A|$ without the axiom of choice?

Question

Inspired by a proof that if $S\subset\mathbb{R}$ is countable then $|\mathbb{R}\setminus S|=2^{\aleph_{0}}$ using the fact that $|\mathbb{R}\times\mathbb{R}|=|\mathbb{R}|$, I tried to generalize to all infinite sets $A$ with the property $|A\times A|=|A|$.

Conjecture: Assuming $\sf{ZF}$ (that is, without $\sf{AC}$), if $|A\times A|=|A|$, and $B\subset A$ satisfies $|B|<|A|$, then $|A\setminus B|=|A|$.

Attempt: Let the image of $B$ in a bijection from $A$ to $A\times A$ be $C$. Consider the function $f:C\rightarrow A$ satisfying $f((x,y))=x$. Since $|C|=|B|<|A|$, the range of this function is a proper subset of $A$. Thus there exists $a\in A$ such that $\{(a,x):x\in A\}\subseteq(A\times A)\setminus C$, and thus $|A|\leq|(A\times A)\setminus C|\leq|A|$. Reversing the bijection gives $|A\setminus B|=|A|$.

The issue is that I am not sure if the claim that the range of the function is a proper subset of $A$ is provable in $\sf{ZF}$. I know it isn't true in general: For example, it is consistent that there is a surjection from (i.e. a partition of) $\mathbb{R}$ to $2^{\aleph_{0}}+\aleph_{1}>|\mathbb{R}|$ where $2^{\aleph_{0}}$ and $\aleph_{1}$ are incomparable.

If $B$ (and hence $C$) is well-orderable, the claim seems to be true. This is because if we consider $f^{-1}:f(A)\rightarrow\mathcal{P}(C)$, defined to map elements $a$ of $f(A)$ to the set of elements in $C$ which map to $a$ under $f$, then since $f^{-1}(a)\subset C$ we may choose a least element $c_{a}$ (without $\sf{AC}$). If $f(A)=A$, this defines an injection $g:A\rightarrow C$ ($g(a)=c_{a}$), a contradiction.

If the claim is provable: How do I justify that step? Or is there an alternative approach?

If the claim is not provable: What is an example of sets $A$ and $B$ not satisfying the claim? Obviously such sets cannot be "explicitly" constructed, but what are some properties they must have (e.g. some description like "$A$ must be $\aleph_{\alpha}$-amorphous")?

Answer 1

Unfortunately, this is not provable in $\sf ZF$.

Monro, G. P., Decomposable cardinals, Fundam. Math. 80, 101-104 (1973). ZBL0272.02085.

In particular, it is consistent that every set which cannot be well-ordered is the union of two strictly smaller sets, and indeed that situation can include $\Bbb R$.

In other words, in some models the property $B\subsetneq A\to|A\setminus B|=|A|$ implies well-orderability.

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On the other hand, assume that $|A\times 2|=|A|$ for every infinite set, then in that case we have that:

$$|A|+|B|=|A\setminus B|+|B|+|B|=|A\setminus B|+|B|=|A|.$$

Now fix a bijection from $f\colon A\sqcup B\to A$, and since sets are homogeneous, we can ensure that the copy of $B$ is mapped to the actual $B$. This means that $f\restriction A$ is a bijection between $A$ and $A\setminus B$.

Sageev, Gershon, An independence result concerning the axiom of choice, Ann. Math. Logic 8, 1-184 (1975). ZBL0306.02060.

There Sageev shows that $|A\times 2|=|A|$ for every infinite set $A$ does not imply the axiom of choice. Therefore even requiring that the statement holds for all infinite sets is not enough.

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Finally, note that all we needed is that $B$ satisfies $|B\times 2|=|B|$. So really what we want to do is require that if $|B|<|A|$, then $|B\times 2|=|B|$. And this holds for any $\kappa$-amorphous set where $\kappa>\omega$, for example. And so this shows that the property does not even characterises sets of the form $|A\times 2|=|A|$, since $\kappa$-amorphous sets are the classical counterexamples.

Answer 2

Such problems are quite sensitive to the explicit axiomatic system one uses to formalise mathematics, but unless one works with a rather particularly restrictive (should we even dare say handicapped..) system, the usual variations of either ZFC or Gödel-Bernays systems will exhibit the following result as a theorem:

for any infinite cardinal number $\mathbf{x}$ one has $\mathbf{x}^2=\mathbf{x}$

having the consequence that:

for any infinite cardinal $\mathbf{x}$ the relation $2\mathbf{x}=\mathbf{x}$ also holds.

From these one easily infers the property that:

if $\mathbf{x}, \mathbf{y}$ are cardinals at least one of which infinite then $\mathbf{x+y}=\mathrm{max}\{\mathbf{x}, \mathbf{y}\}$.

This directly applies to your instance of $A$ infinite and $B \subseteq A$ being such that $|B|<|A|$ (if I understand the setting well; if not, may I apologise for making irrelevant comments): by denoting $|B|=\mathbf{x}, |A\setminus B|=\mathbf{y}$, since their sum is infinite $\mathbf{x}, \mathbf{y}$ cannot be both finite and thus by the result above $|A|=\mathrm{max}\{\mathbf{x}, \mathbf{y}\}$, and it is clear from the hypothesis that the larger cardinal of the two cannot be $\mathbf{x}$, leaving you with $|A|=\mathbf{y}$.