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Let X be a set and let $f: X\rightarrow X$ be a projection.

We define a relation on X such that $x \sim y$, means that there exist a natural numbers m and n such that $f^m(x)=f^n(y)$

($f^m$ means $f \circ f\circ ... \circ f$ $m$ times)

I´m suppose to show that this is equivalence relation. And I started doing this

  1. $x\sim x$ $\space$ $f^m(x)=f^n(x)$ for all $x\in X$
  2. If $x\sim y$ then $y\sim x$ $\space$ $f^m(x)=f^n(y) \Rightarrow f^m(y)=f^n(x)$
  3. if $x\sim y$ and $y\sim z$ then $x\sim z$ $\space$ $f^m(x)=f^n(y) $ and $f^m(y)=f^n(z)$ than $f^m(x)=f^n(z)$

But than I though this might not be true if m does not equal n. And im not really sure how to show this if m and n are not equal, so if anyone has some idea about it, it would help alot

Yanko
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Abc123
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1 Answers1

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The fact that $f$ is a projection doesn't seem to be useful.

For reflexity: $$ \forall x \in X, \quad x = x \iff f^0(x) = f^0(x) \implies x \sim x $$

For symmetry if $x, y \in X$ are such that $x \sim y$ then there exists $n, m$ such that $f^n(x) =f^m(y)$. Then $f^m(y) = f^n(x)$ thus $y \sim x$.

For transitivity if $x \sim y$ and $y \sim z$ then there exists $n, m, k, l$ such that $f^n(x) = f^m(y)$ and $f^k(y) = f^l(z)$. Then $f^{kn}(x) = f^{km}(y) = f^{mk}(y) = f^{ml}(z)$ therefore $x \sim z$.

I think you're missing the fact that the $m$ and $n$ in the definition of $\sim$ are not fixed.

bitesizebo
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