Suppose that $\phi: Y \to Y$ be a bijection. Suppose that $B\subseteq A\subseteq Y$ and $\phi(A)\subseteq A$ and $\phi(B)\subseteq B$. Is it true that $\phi(A\setminus B)\subseteq A\setminus B$? I took $x\in A\setminus B$. So $x\in A$ but $x\notin B$. So $\phi(x)\in A$. But I coulnt conclude that $\phi(x) \notin B$. when does $\phi$ preserves set difference? What more assumption do we need?

  • $\begingroup$ You can also use math.stackexchange.com/questions/359693/… as a starting point for searching for the "more assumptions". $\endgroup$
    – Asaf Karagila
    Aug 24 at 9:47
  • 2
    $\begingroup$ @AsafKaragila - I don't think this question is a duplicate of that one. Note that here the question is about proving that $\phi(A \smallsetminus B) \subseteq A \smallsetminus B$ (which does not hold), there the question is about proving $\phi(A \smallsetminus B) \subseteq \phi(A) \smallsetminus \phi(B)$ (which holds). $\endgroup$ Aug 24 at 10:04

Clearly the inclusion

$$\phi(A\setminus B)\subseteq A\setminus B$$

doesn’t hold. Just consider the counterexample

$Y=A=\mathbb N$, $B=2 \mathbb N$ with $$\begin{cases} \phi(n)=n +2 & \text{ if } n \in 2\mathbb N\\ \phi(n)=n-2 & \text{ if } n \in \mathbb N \setminus \left(\{1\} \cup 2\mathbb N\right)\\ \phi(1)=2 \end{cases}$$

  • $\begingroup$ It seems to me that your counterexample works fine if you set $\phi(0) = 0$. Otherwise, $\phi(0) = 2 = \phi(1)$ and so $\phi$ would not be injective. $\endgroup$ Aug 24 at 9:59
  • $\begingroup$ Here $\mathbb N=\{1,2,3,\dots\}$. $\endgroup$ Aug 24 at 10:00
  • $\begingroup$ So if $\phi|B$ is surjective then the inclusion will work right? ie $\phi(B)=B$ $\endgroup$
    – budi
    Aug 24 at 10:02
  • $\begingroup$ @budi - For $B = 2\mathbb{N} = \{2,4,6,\dots\}$, you have $\phi(B) = B \smallsetminus \{2\} \subseteq B$. The fact that $2 \notin \phi(B)$ is crucial for the counterexample. $\endgroup$ Aug 24 at 10:12
  • $\begingroup$ Yes. This is coherent with your requirement $\phi(B) \subseteq B$. $\endgroup$ Aug 24 at 10:13

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