Hi I was reading topology without tears appendix 1 :
This made sense at first as you can always just take the power-set. But can't you really construct a largest cardinal number? The operation of taking power-set is monotonic on an infinite set. Couldn't you encounter an aleph fixpoint (isn't one assured to exist because of Knaster-Tarski theorem) which would be the largest cardinal number? Or am I plain wrong. Thanks in advance.
On
One trivial reason which prevents Knaster-Tarski: your proposed operation "add one to a cardinal" is not a function, and hence is not a monotonic function. Indeed, there is no set of all cardinals, so your "function" doesn't have a domain. (The class of all sets is not a set, by Russell's paradox; the class of all ordinals is not a set, by Burali-Forti, and hence neither is the class of cardinals, which can be indexed by the ordinals.) In general this sort of quibble doesn't necessarily mean a theorem about sets becomes false when we naively translate it to the context of classes, but it does mean you have to prove the new theorem for function-classes rather than just hope that it's true because it was true for genuine functions.
So what happens when we try and adjust Knaster-Tarski to apply to the class of all cardinals?
The Knaster-Tarski theorem applies to complete lattices. The class of all cardinals is of course not a complete lattice, because it's not a set. So we need to adjust the definition of a "complete lattice" to try and make the proof go through.
The proof of Knaster-Tarski goes like "take $\sup \{x : x \le f(x)\}$ and show that this is a fixed point of $f$". So the right notion of "complete" will have to encompass "has sups of all sub-collections". But the cardinals certainly don't have this property, as you pointed out: let $f(x) = x + 1$, and we find ourselves trying to take the sup of the class of cardinals, which (by your quoted corollary) doesn't exist.
So no analogue of Knaster-Tarski (at least with its usual proof) will help you show that there's an aleph fixpoint.
There are Aleph fixed points, but they're not a largest cardinal. By Hartogs' theorem, for a every cardinal, there's a larger cardinal. This dates back to 1915. Of course, as mentioned, by the power-set axiom and Cantor's earlier theorem it was already known that for any cardinal there's a larger cardinal. Hartogs obtained the result using well-orders.
The lowest Aleph fixed point $\lambda$ is $$\alpha_0 = \omega_0\quad \alpha_{n+1}=\omega_{\alpha_n}\quad \lambda = \lim_{n\rightarrow \omega_0} \alpha_n$$ Then $\lambda = \omega_\lambda = \aleph_\lambda$, but $\lambda$ is not very big. It has cofinality $\operatorname{cf} \lambda =\aleph_0$ because it's the limit of a $\omega_0$-indexed sequence.
A cardinal $\kappa$ is called weakly inaccessible if $\kappa = \operatorname{cf}\kappa = \aleph_\kappa$. These are already big enough that their existence isn't provable in ZFC.
A cardinal $\kappa$ is called strongly inaccessible if $\kappa = \operatorname{cf}\kappa = \aleph_\kappa = \beth_\kappa$. Again, their existence isn't provable in ZFC, but if they exist they're impressive. Because then $V_\kappa$ is a model of ZFC, where $V_\kappa$ is from the von-Neumann hierarchy. In this model built from a set inside the model, $\kappa$ serves as the proper class of ordinals of $V_\kappa$. So thinking about it externally, you can think of the proper class $\text{Ord}$ of all ordinals of a model as a kind of 'largest cardinal', but it's not a set, not an ordinal of the model in the regular sense.