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I am given that $F$ is a definite unary operation on the ordinals which is normal. That is, $$x and furthermore $$F(y)=\sup\{F(x): x 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 then if $y\geq a$ for all $a\in A$ then we would get a contradiction (how?) and hence we must have that $y 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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