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Let $Y\subseteq X$ be nonempty sets and $f:X\to Y$ be a one-to-one function. Denote $C_0=X\setminus Y$ and $C_n=f(C_{n-1})$, prove all sets are pairwise disjoint.

This came to me while looking at Michael Greinecker's solution to Intuition behind Cantor-Bernstein-Schröder

I understand the point of his proof, but at the start he claims that the sets $f^n(C)$ are disjoint.

Now if one of the sets is $C_0$ then it's obvious, otherwise I expect it should regress to the $C_0$ case somehow, but wasn't able to see exactly how. Any proofs of this?

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Nescio
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1 Answers1

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Let $B=\{\langle m,n\rangle\in\Bbb N\times\Bbb N:m<n\text{ and }C_m=C_n\}$. If $B=\varnothing$, we’re done, so assume that $B\ne\varnothing$. Let $\langle m,n\rangle\in B$ be such that $m$ is minimal. If $m>0$, then $C_m=f[C_{m-1}]$, and $C_n=f[C_{n-1}]$, so $C_{m-1}=C_{n-1}$, contradicting the choice of $m$. Thus, $m=0$, and you’ve already disposed of that case. Hence $B=\varnothing$, and the sets $C_n$ are pairwise disjoint.

Brian M. Scott
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