I'm currently reading Hrbacek's Introduction To Set Theory and exercise 4.1.11 goes like this:
Where Lemma 1.7 is $$\text{If } A_1\subseteq{B}\subseteq{A} \text{ and } |A1| = |A| , \text{ then }|B| = |A|, $$
$f$ is a bijection from $A$ to $A_1$ such that $A_1\subseteq{B}\subseteq{A}$ and $g$ is defined as:$$g(x)= \begin{cases} f(x) & \text{if $x\in{C}$}\\ x & \text{if $x\in{D}$}\\ \end{cases} $$
I did the exercise, but it seems to me that the proof still works if we simply let $F(X) = f[X]$.
Is this true or am I missing something?
If you define $F(X)=f[X]$, then the fixed-point of $F$ will fail to contain $A-B$ (for example, $\varnothing$ is a fixed-point!). Then $g$ will fail to map the elements of $A-B$ into $B$.