Cardinals of set operations without AC

Question

Given info: $|A|=\mathfrak{c}$ , $|B|=\aleph_0$ in ZF (no axiom of choice).

Prove: $|A\cup B|=\mathfrak{c}$

If $B \subset A\implies|A \backslash B|=\mathfrak{c}$?

I have found several places proving that for $|\mathbb{R} \backslash \mathbb{Q}|,$ but none of the solutions appears to work for arbitrary sets. Maybe one should prove it for $|\mathbb{R} \backslash \mathbb{Q}|$ and then show that it will work for arbitrary $A$ and $B$ as well?

Here Showing that $\mathbb{R}$ and $\mathbb{R}\backslash\mathbb{Q}$ are equinumerous using Cantor-Bernstein David has a very nice idea (constructing bijection), but it requires that infinite set has countably infinite subset, which again!? needs some choice axiom.

In some sources I even saw statements that these can't be proved in ZF.

From what my teacher said I percieved that solution has something to do with Cantor–Bernstein theorem and that knowing how to prove $\mathfrak{c}+\mathfrak{c}=\mathfrak{c}$ would help as well.

Thanks!

Answer 1

Prove it first for disjoint $A$ and $B$, relaxing the condition on $B$ to $|B|\le \aleph_0$. You then recover the full statement by considering $A\cup B = A\cup(B\setminus A)$.

You can restrict your attention even further to, say $A=(0,1)$ and $B$ being a subset of the integers. Once you have proved it for that case, the definition of "same cardinality" guarantees that it will be true for every other choice of disjoint $A$ and $B$ of the appropriate cardinalities.

It is true without any choice axiom that a set of size continuum has a countably infinite subset. By definition, because it has size continuum, there's a bijection from $\mathbb R$, and the image of $\mathbb N$ under that bijection is a countably infinite subset.

Answer 2

Note that if $A$ has cardinality $\frak c$ then $A$ has a countably infinite subset. This does not require the axiom of choice. $|\mathcal P(\Bbb N)|=\frak c$ and $|X|<|\mathcal P(X)|$.

As for the suggestion to use the fact $\frak c+\frak c=\frak c$, this is again helpful because you can replace $B$ with its power set and you have: $$\mathfrak c=|A|\leq |A|+|B|\leq |A|+|\mathcal P(B)|=\frak c+c=c.$$

Answer 3

By assumption, there exist bijections $a\colon A\to \mathbb R $ and $b\colon B\to\mathbb N_0$. We can define the injective map $b'\colon B\setminus A\to\mathbb N_0$, $x\mapsto\lvert\{y\in B\setminus A\mid b(y)<b(x)\}\rvert$. If $b'$ is a bijection, we can define $c\colon A\cup B\to\mathbb R$ by $$c(x)=\begin{cases}a(x)&\text{if }x\in A\text{ and }a(x)\notin\mathbb N_0,\\ 2a(x)&\text{if }x\in A\text{ and }a(x)\in\mathbb N_0,\\ 2b'(x)+1&\text{if }x\in B\setminus A.\end{cases} $$ If on the other hand $b'$ is not a bijection, let $N=\min(\mathbb N\setminus b'(B\setminus A))$. Then clearly $b'(A\setminus B)= \{k\in\mathbb N_0\mid k<N\}$ and $b'$ is a bijection with this set so that we can define $c\colon A\cup B\to\mathbb R$ by $$c(x)=\begin{cases}a(x)&\text{if }x\in A\text{ and }a(x)\notin\mathbb N_0,\\ a(x)+N&\text{if }x\in A\text{ and }a(x)\in\mathbb N_0,\\ b'(x)&\text{if }x\in B\setminus A.\end{cases} $$ In both cases, $c$ is readily seen to be a bijection.