# Proving that two subsets of a set of bijections are either equal or disjoint

$$X$$ is a set and $$X!$$ is the set of bijections from $$X$$ to $$X$$. Let $$A(m)$$ be the set defined as $$A(m)=\{f^x(m)\mid x\in\mathbb Z\}$$ for some $$f$$ in $$X!$$ and some $$m$$ in $$X$$. Now, how do I prove that for $$m, n\in X$$, $$A(m)\cap A(n)=\emptyset$$ or $$A(m) = A(n)$$?

I have tried the following strategy. Assume $$A(m) \cap A(n) \ne \emptyset$$. Then, there exists $$x \in A(m) \cap A(n)$$, which implies $$x \in A(m)$$ and $$x \in A(n)$$. Then, I think we have to use the set membership criteria. What I am not sure is how does that help us use the double containment argument to prove that $$A(m) = A(n)$$? To me, it seems very trivially right. I am not sure how to proceed in these kinds of questions.

• You perhaps need that $X$ is finite? – enedil Feb 10 at 20:54
• I do not think so. – Ufomammut Feb 10 at 20:55
• Ah, $x$ in the definition of $A$ can be negative, right? – enedil Feb 10 at 20:56
• No, that is false. Consider $X=\mathbb Z$ and $f(x)=x+1$. This is a bijection, but doesn't have this property. – enedil Feb 10 at 20:58
• Just a word on notation. Using $X$ for a set, and then $x$ for an element of $\Bbb Z$ is awful. Even worse, $m$ is a common letter for denoting an element of $\Bbb Z$, and you use it to denote an element of $X$ (I suppose?) – Asaf Karagila Feb 10 at 21:10

I don't think I'm understanding the definition as written. It appears that you've written $$A(m)=\{f^x(m): x\in\Bbb{Z}, f\in X!\}$$. Do you perhaps mean to instead say: Fix $$f\in X!$$, then for each $$m\in X$$, we define $$A(m)=\{f^x(m):x\in\Bbb{Z}\}$$? It must be the latter I think.
Suppose $$A(m)\cap A(n)\ne \varnothing$$. Let $$f^x(m)=f^y(n)$$. Then $$m=f^{y-x}(n)$$, so $$f^k(m)=f^{y-x+k}(n)\in A(n)$$. Thus $$A(m)\subseteq A(n)$$, and by symmetry, we also have $$A(n)\subseteq A(m)$$. Thus $$A(m)=A(n)$$.
• Yes, the $f$ is fixed. What I do not understand is in your solution, why did you say "Let $f^x(m) = f^y(n)$." I mean I got the same, but in following way: If $A(m)$ $\cap$ $A(n)$ $\neq$ $\emptyset$, then there exists an $a$ $\in$ $X$ such that a $\in$ $A(m)$ $\cap$ $A(n)$, and then I concluded that $f^x(m) = f^y(n)$. Is that your way as well? – Ufomammut Feb 10 at 21:08
• @Ufomamut, yes, that's the same. Since we know that $A(m)\cap A(n)$ is nonempty, then there is an element in $A(m)$ that equals an element of $A(n)$, and those elements have the form $f^x(m)$ and $f^y(n)$ for some $x$ and $y$ in $\Bbb{Z}$. – jgon Feb 10 at 21:11