I am given that $F$ is a definite unary operation on the ordinals which is normal. That is, $x
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
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.