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

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

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

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

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

But then we have $f(u) and this is a contradiction.

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

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

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

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

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

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

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