Hi I was reading topology without tears appendix 1 :

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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.

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    $\begingroup$ What makes you think that there might be a fixed point in the first place? The operation "+1" doesn't have any fixed points but is monotonic on the ordinals - or are you not convinced of that either? $\endgroup$ Jul 8, 2023 at 9:23
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    $\begingroup$ (Or, for example, do you think there might be a fixed point for the operation "wrap as a singleton", i.e. $x \mapsto \{x\}$?) $\endgroup$ Jul 8, 2023 at 9:25
  • $\begingroup$ en.wikipedia.org/wiki/Knaster%E2%80%93Tarski_theorem if any function is monotonic it must have at least two fixed points a largest and a smallest fixed point. See Tarski. Sorry I reaaly should have mentioned this in my question I ll edit. $\endgroup$
    – Jip Helsen
    Jul 8, 2023 at 9:25
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    $\begingroup$ 1) your statement is importantly false (you missed out the assumption of a complete lattice), and 2) your statement doesn't apply to your question ("take the power set" is not a function on the class of sets, because it doesn't have a set which is a domain). $\endgroup$ Jul 8, 2023 at 9:26
  • $\begingroup$ ah okay thx could you write answer (you can just copy paste) and I ll approve. $\endgroup$
    – Jip Helsen
    Jul 8, 2023 at 9:27

2 Answers 2


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.


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.

  • $\begingroup$ thx I indeed got it mixed up. I ll approve asap (4 minutes) $\endgroup$
    – Jip Helsen
    Jul 8, 2023 at 9:31
  • $\begingroup$ Your point (1) is on the money — the problem being the largeness of the lattice in question — but your point (2), that these “functions” on cardinals are not set-size, is not so pertinent, and I think confuses the real point. Point (2) is a fairly inessential point of terminology: these are certainly “class functions”, and many mathematicians (including set theorists) often take “functions” to include these, and little if anything is lost by doing so. [cont’d] $\endgroup$ Jul 8, 2023 at 19:42
  • $\begingroup$ Point (1) however is a genuinely mathematical point that can’t be evaded by terminology: cardinals are certainly a complete class-lattice, and one may reasonably choose to allow “lattice” to cover large lattices, but in that case, the statment of K-T and similar fixed point theorems has to explicitly restrict to “small lattices”. The core point is that these theorems need the lattice to have suprema of sets/families as large as the lattice itself — so a class-size lattice, with suprema for sets, doesn’t suffice. $\endgroup$ Jul 8, 2023 at 19:42
  • $\begingroup$ @PeterLeFanuLumsdaine Fair enough, thanks; I've moved bits around and rejigged to discuss why Knaster-Tarski can't work here in general. $\endgroup$ Jul 8, 2023 at 22:38

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