1. Bruno, I fixed my proof. Hopefully convincing now

So we assumed $\displaystyle f$ had no fixpoint and $\displaystyle S=\left\{x:f(x)>x\right\}$, and $\displaystyle T= \left\{x:f(x)< x\right\}$.

As shown before we know that $\displaystyle S,T$ are closed under $\displaystyle f$

Now let $\displaystyle u= \sup(S)$. Then we have $\displaystyle u\in T$. Suppose that $\displaystyle u\in S$ then $\displaystyle f(u)>u$ but since $\displaystyle f(u)\in S$ this contradicts the fact that $\displaystyle u= \sup(S)$.

Now since $\displaystyle u\in T$ we have $\displaystyle f(u)< u$ and we can find $\displaystyle x_n\in (f(u), u)\cap S$ ( this is clearly true since $\displaystyle u = \sup(S)$

But then we have $\displaystyle f(u)<x_n<f(x_n)<u$ and this is a contradiction.

2. Originally Posted by Dinkydoe
Bruno, I fixed my proof. Hopefully convincing now

So we assumed $\displaystyle f$ had no fixpoint and $\displaystyle S=\left\{x:f(x)>x\right\}$, and $\displaystyle T= \left\{x:f(x)< x\right\}$.

As shown before we know that $\displaystyle S,T$ are closed under $\displaystyle f$

Now let $\displaystyle u= \sup(S)$. Then we have $\displaystyle u\in T$. Suppose that $\displaystyle u\in S$ then $\displaystyle f(u)>u$ but since $\displaystyle f(u)\in S$ this contradicts the fact that $\displaystyle u= \sup(S)$.

Now since $\displaystyle u\in T$ we have $\displaystyle f(u)< u$ and we can find $\displaystyle x_n\in (f(u), u)\cap S$ ( this is clearly true since $\displaystyle u = \sup(S)$

But then we have $\displaystyle f(u)<x_n<f(x_n)<u$ and this is a contradiction.
That works!

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