I was looking at the Bourbaki-Witt Fixed Point Theorem which states that
If $X$ is a non-empty, chain complete poset and $f: X \to X$ s.t. $f(x) \geq x$ for all $x$, then $f$ has a fixed point.
I was wondering if one could modify the proof of this theorem to prove a version of the Tarski-Knaster Fixed point theorem. Suppose also that $X$ has a minimum element $a$ and every subset of $X$ has a supremum. Let $f: X \to X$ be a monotone function i.e. $x \leq y \implies f(x) \leq f(y)$. Then $f$ must have a fixed point.
Proof: Define the functional:
$$ \begin{align} g(0) &= a \\ g(\alpha^+) &= f(g(\alpha)) \\ g(\lambda) &= \text{sup}\{g(\alpha) : \alpha < \lambda\} \end{align} $$
where $\lambda$ is a limit ordinal. If there is no ordinal $\alpha$ s.t. $g(\alpha) = g(\alpha^+)$ (which would be a fixed point), then $g$ must be a monotonically increasing function and is thus an injection from the ordinals into $X$ which is a contradiction.
The reasoning seems a little dubious to me so I would appreciate any thoughts!
Edit:
We can see $g$ is weakly increasing by noting that, since $g(0)$ is the minimum element, we have $g(0) \leq g(1)$. Then by monotonicity of $f$, we have $g(1) = f(g(0)) \leq f(g(1)) = g(2)$ and so on. This notion could be formalized via induction.
Yes, this proof works. When $\lambda$ is a limit ordinal, you can prove $g(\lambda)\leq g(\lambda^+)$ as follows. For any $\alpha<\lambda$, $g(\alpha)\leq g(\lambda)$ and hence $g(\alpha)\leq g(\alpha^+)=f(g(\alpha))\leq f(g(\lambda))=g(\lambda^+)$. Since $g(\lambda)$ is the least upper bound of all these $g(\alpha)$, we must have $g(\lambda)\leq g(\lambda^+)$.