# Equipotence when removing a subpotent subset

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 $$B$$.

1. 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
• Why do you want to avoid ordinals? This is exactly where they simplify a lot of the technical part and provide you a whole new kind of understanding (not to mention that ordinals are great). Feb 5, 2020 at 6:23
• @Asaf Karagila. Same reason as people saying "square matrix times column matrix" instead of "linear map applied to the vector". It's more my target public than me. (in fact my aim is to prove that every field has an algebraic closure) Feb 5, 2020 at 11:13
• I think that avoiding ordinals "at all costs" is a huge disservice to mathematicians. Why do you need all of that to prove that every field has an algebraic closure? This follows from weaker principles, e.g. the ultrafilter lemma (equivalently the compactness theorem). No need to bother them with Zorn's lemma and cardinal arithmetic. Feb 5, 2020 at 11:18
• (You can find the proofs of my previous suggestion in mathoverflow.net/questions/46566/…) Feb 5, 2020 at 11:20

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 $$A\setminus B$$.