elementary set theory - Suppose that $A_1subseteq A$ and $B_1subseteq B$, and that $Asim B$ and $A_1sim B_1$. Then $Asetminus A_1 sim Bsetminus B_1$.

For $A,B,A_1,B_1$ are infinite sets. Suppose that $A_1\subseteq A$ and $B_1\subseteq B$, and that $A\sim B$ and $A_1\sim B_1$. Then $A\setminus A_1 \sim B\setminus B_1$.

---

My attempt:

We denote $A$ and $B$ are equinumerous by $A\sim B$.

Since $A\sim B$, there is a bijection $f:A\to B$. Similarly, there is a bijection $h:A_1\to B_1$.

From here, I don't know how to proceed to define a bijection from $A\setminus A_1$ to $B\setminus B_1$. I think the proof may require Axiom of Choice, but don't know how. Please shed some lights!

Answer

A counterexample. Let $\mathbb N = \{1,2,3,4,\dots\}$ be the natural numbers, and let $\mathbb E = \{2,4,6,\cdots\}$ be the even natural numbers. Then our counterexample is: $$A_1 = \mathbb N, \quad A=\mathbb N,\quad B_1 = \mathbb E,\quad B = \mathbb N .$$ We have $A_1 \sim B_1$ and $A \sim B$ but $A \setminus A_1$ is empty and $B\setminus B_1$ is the set of all odd numbers.