Show that in the Schroeder-Bernstein Theorem proof the set $X_0$ is the smallest set such that the function h in the proof is bijective I think basically the question means that if there is a set $X_1$ that works then $X_0 \subset X_1.$ How do I go about doing this? 
This is the proof we are using.

 A: Although it is not quite clear from how you wrote the question, you want $X_0$ to be a fixed point of the map $F$, so the question is how to determine $X_0$ so that it is the smallest fixed point. To see this, define sets $C_0,C_1,\dots$ by


*

*$C_0=\emptyset$,

*$C_1=F(C_0)$,

*$C_2=F(C_1)$,

*etc., that is, $C_{n+1}=F(C_n)$ for all $n$.


Suppose $X$ is a fixed point of $F$. Since $X\supseteq \emptyset=C_0$, monotonicity of $F$ gives us that $C_1=F(C_0)\subseteq F(X)=X$. Similarly, we have that if $C_n\subseteq X$, then $C_{n+1}=F(C_n)\subseteq F(X)=X$. It follows that $X\supseteq \bigcup_n C_n$.
It is now a simple matter to check directly that $\bigcup_n C_n$ is a fixed point of $F$, and therefore $X_0=\bigcup_n C_n$ is the smallest fixed point of $F$.

This argument also illustrates a common theme: When looking for fixed points of maps, typically iterations of the map give clues of what these fixed points are. The paper where the fixed point lemma is proved, 

Alfred Tarski, A lattice-theoretical ﬁxpoint theorem and its applications, Pacific J. Math., 5 (1955), 285–309. MR0074376 (17,574d),

contains several applications and variants where fixed points are explicitly identified via iterating (perhaps transfinitely) the relevant maps.
