Equipotence when removing a subpotent subset

Question

I think that the tree following statements are equivalent. I'd like to have a proof of one of them.

Definition. I say that the set $B$ is strictly subpotent to $A$ if there exists an injective map $B\to A$, but no bijections.

1.

Let $A$ be an infinite set and $B$ be a subset of $A$ which is strictly subpotent to $A$. Then $A\setminus B$ is equipotent to $A$.

2.

Let $A$ be an infinite set and $B$ be a subset which is strictly subpotent to $A$. There exists a subste $U\subset A$ disjoint to $B$ and equipotent to $B$.

3. Let $A$ be infinite and $A'$ be equipotent to $A$. Then $A$ is equipotent to $A\cup A'$.

The common point of these tree statements is that "add or remove a smaller cardinal does not change the cardinality".

I'd like a proof of that without explicit reference to cardinals.

Comptements :

• I'm pretty sure that the proof will need the Zorn lemma on the set of parts of $B$ that can be substracted from $A$ without changing the cardinality (for my first statement).
• With cardianls, there is and answer here
• My purpose is to understand a step in the proof that $A\times A$ is equipotent to $A$, given here

Answer 1

I think I've managed to prove that $A\setminus B$ is equipotent to $A$.

I've written a full proof here, in the French part; lemma 1.45 for the moment. Look for the string "LEMooIVCBooHWQiZB".

The main lines are as follow.

Prerequisites

• If $A$ is infinite, there exist a bijection $\{1,2\}\times A\to A$ (this one requires Zorn lemma). See here.
• If $A$ is surpotent to $B$, then $A\cup B$ is equipotent to $A$.
• If $B_1$ and $B_2$ are equipotent subsets of $A$, then $A\setminus B_1$ is equipotent to $A\setminus B_2$.

Main lines

• There exists disjoint $A_1$ and $A_2$ inside $A$, both equipotent to $A$.
• Consider the copies $B_1$ and $B_2$ of $B$ inside $A_1$ and $A_2$.
• The set $A\setminus B_1$ contains $B_2$ and is surpotent to $B_2$
• We deduce that $A\setminus B$ is surpotent to $B$.
• Thus $(A\setminus B)\cup B$ is equipotent to $A\setminus B$.