# The set of fixed points is closed (ordinals)

I am given that $F$ is a definite unary operation on the ordinals which is normal. That is, $$x<y\Rightarrow F(x)<F(y)$$ and furthermore $$F(y)=\sup\{F(x): x<y\}$$ if $y$ is a limit ordinal. A set $M$ is called closed if for all non empty subset $A$ we have that $\sup A\in M$. We want to show that the set of fixed points $M=\{\alpha\in ON: F(\alpha)=\alpha\}$ is first of all non-empty (which I already proved) and closed.

This is the way I was trying to do the problem. Let $A$ be a non-empty subset of $M$. Then let $x=\sup A$. I want to show that $x\in M$, which means that $F(x)=x$, as supremums are unique, I want to prove that $F(x)$ is the supremum of $A$. Note that since $$x\geq a\Rightarrow F(x)\geq F(a)=a$$Then we have that $F(x)$ is an upper bound for $A$. Thus I only remain to show that it is the least upper bound. My approach was to let $y<F(x)$ then if $y\geq a$ for all $a\in A$ then we would get a contradiction (how?) and hence we must have that $y<a$ for some $a\in A$ which means that $F(x)$ is indeed the supremum of $A$.

Working out the "how?": I have that $y\geq a$, then $F(y)\geq F(a)=a$, and by definition of $x$, I must have that $x\leq y$, and thus, $F(x)\leq F(y)$, after this I have been going around in circles. Hints are greatly appreciated. Thanks.

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$\bf Hint:$ If $\sup A=x$ is a successor ordinal then $x\in A$ and there is nothing to prove. On the other hand, if $x$ is a limit ordinal then by normality $F(x)=\sup\{F(y): y<x\}$.
ok and the way to prove $\sup\{F(y):y<x\}=\sup\{F(a):a\in A\}$ is by letting the left hand side be $\beta$, then $\beta\geq F(a)$ and so it is an upper bound, and if $\gamma<\beta$ then we have that there is a $z<x$ such that $F(z)>\gamma$, (by definition of $\beta$), and since $x$ is a sup for $A$, we can find an $a\in A$ such that $z<a$ and by normality: $F(z)<F(a)$ meaning that $\gamma<F(a)$ so $\gamma$ is not an upper bound. –  Daniel Montealegre Feb 27 '12 at 6:31
HINT: If $x\in A$, there is nothing to prove, so assume that $x\notin A$. Show that in this case $F(y)<x$ for every $y<x$ and conclude that $F(x)=\sup\{F(y):y<x\}=x$.
Hey so I read online that normal functions commute with supremums. That is, $F(\sup M)=\sup\{F(\alpha):\alpha\in M\}$, then this problem would follow since $F(x)=\sup\{F(a):a\in A\}=\sup\{a:a\in A\}=x$, and I think this last statement is the one proved below? –  Daniel Montealegre Feb 27 '12 at 8:33