# Cantor-Bernstein with "weakly injective" functions

Let us call a map $$f: X \to Y$$ between non-empty sets a "weak injection" if $$f^{-1}(\{y\})\subseteq X$$ is finite for every $$y \in Y$$.

Recall that the (Schroeder-)Cantor-Bernstein-Theorem (sometimes abbreviated by (CB)) states that if there are injections between two sets $$X, Y$$, then there is also a bijection between $$X$$ and $$Y$$. (CB) is a theorem of $${\sf (ZF)}$$.

Consider the following statement:

(wiCB) If $$X, Y$$ are infinite sets such that there are weak injections between $$X$$ and $$Y$$, then there is a bijection between $$X$$ and $$Y$$.

Obviously, (wiCB) implies (CB). Can (wiCB) also be proved in $${\sf (ZF)}$$?

• I have also heard "finite-to-one" and, more cutely, "finjective" used for weakly injective. Commented Feb 5 at 14:42

No, it is not provable in $$\mathsf{ZF}$$.

It is consistent with $$\mathsf{ZF}$$ that there is a sequence of disjoint two-element sets whose union is not countable, i.e. $$\vert A_i\vert=2$$ but there is no bijection between $$A:=\bigcup_{i\in\mathbb{N}}A_i$$ and $$\mathbb{N}$$.

WLOG, $$A\cap\mathbb{N}=\emptyset$$. Now let $$B_i=A_i\sqcup\{i\}$$ and let $$B=\bigcup_{i\in\mathbb{N}}B_i$$. We have obvious weak injections $$B\rightarrow\mathbb{N}$$ and $$\mathbb{N}\rightarrow B$$, the former sending $$x\in B_i$$ to $$i$$ and the latter sending $$i$$ to $$i$$, but $$B$$ is not countable.

• And more generally (wiCB) implies that the union of $\aleph_\alpha$ finite sets has cardinality at most $\aleph_\alpha$.
– bof
Commented Feb 4 at 22:21
• Hmm, this raises an interesting separate question of if wiCB is equivalent to Choice. My guess is that it is weaker. Commented Feb 5 at 15:33
• @JoshuaZ Probably decently weaker. E.g. if every set can be linearly ordered and $|X\times\omega|=|X|$ for all infinite $X$ then we get wiCB. That sounds doable in the "$\omega_1$ Cohen model", but I'm not certain. Commented Feb 5 at 17:24
• @CalliopeRyan-Smith: It's not clear to me which is the "$\omega_1$ Cohen model". Note that $|X\times\omega|=|X|$ if and only if $|X|=|X|+|X|$, so we really just need that every set is linearly orderable along with that. Luckily, Sageev's model (which is really the only model we really know about where $|X|+|X|=|X|$ holds for every infinite set $X$ and choice fails) satisfies that every set can be linearly ordered. (This is Lemma 9.30 in his massive paper.) Commented Feb 8 at 21:11
• @AsafKaragila $\operatorname{Add}(\omega_1,\omega_1)$, countable-support permutations of $\omega_1$, etc. My naive hope is (was?) that $\mathsf{SVC}(A)$ for a linearly orderable $A$ with $\aleph(A)>\aleph_0$ would be enough. Commented Feb 9 at 13:36