Talk:Measurable cardinal

"Subsets" of a cardinal number?

The current Wikipedia article on cardinal number, https://en.wikipedia.org/wiki/Cardinal_number, defines cardinality in terms of an equivalence relation that is defined on a collection of sets. By that definition, if the set S "has cardinality K", this indicates that S is a member of some ordered pair P of sets and that P is an member of the set K (since a "relation" like K consists of a set of ordered pairs). So "subsets of S" are not the same collection of sets as "the subsets of K".

However, the current article on "Measurable cardinal" seems to treat "the subsets of K" as if they were the same sets as "the subsets of a set S that has cardinality K". This terminology may be a cultural tradition in discussing measurable sets. If so, it should be mentioned. Tashiro~enwiki (talk) 17:43, 28 March 2017 (UTC)

Please see my response to a related question here. JRSpriggs (talk) 20:24, 28 March 2017 (UTC)

Proof of inaccessibility of real-valued measurable cardinals

I added a proof for two reasons:

• The proof is in itself notable in my opinion.
• The proof demonstrates that the mathematics of large cardinals need not always be totally inaccessible.

YohanN7 (talk) 11:11, 10 July 2017 (UTC)

Gobbledygoop

The Wikipedia article on Measurable cardinals will be all but gobbledygoop to anyone who knows only basic Set Theory and Real Analysis or General Topology. No beginner can possibly grasp what the Hell a measurable cardinal is either from the standard definition using a two valued K-additive on an ultrafilter etc., or from the still more obscure definition in terms of embeddings into the Von Neuman universe. Since such folks, -- and I would to some extent include myself here -- have no initial way of grasping what such a cardinal might be composed of, or better, what in the world it is supposed to enumerate, and also why it is so very large compared to lesser cardinals, the rest of the article will be worthless gibberish to them. That doesn't mean that it won't remain a useful reference for more advanced Set Theorists, but I have always assumed it to be the purpose of the Wikipedia to explain things clearly and simply enough, and with sufficient illustrations where needed, that anyone with a basic grasp of the larger subject, in this case Set Theory, should be able to understand the more advanced topics discussed in the Wikipedia fairly quickly. This article, however, does not do that. Indeed, here the reader is given no good intuitive way to grasp what in the world a measurable cardinal is, and where it comes from etc., before being spirited away into proofs about its many known qualities. — Preceding unsigned comment added by Boscolovich (talk • contribs) 22:42, 20 September 2019 (UTC)

(Assuming ZFC is consistent.) One cannot prove the existence of a Measurable cardinal nor provide any examples of one. So talking about it is kind of like saying "suppose there is a magical fire-breathing flying dragon, what can we infer from this?". In other words, the point of defining it is not to point out actual examples, but to give one a way of deducing things about it, if it exists. JRSpriggs (talk) 02:09, 21 September 2019 (UTC)

I think workers from a fair range of foundational views will at least take the position that, given the current state of knowledge, it is more natural to assume that measurables do exist than that they don't. An "example" of a measurable cardinal could then be "the least measurable cardinal", which is of course not very helpful vis a vis Boscolovich's concerns.

The general complaint that the article is not as useful as it could be, even to readers who have sufficient background to understand the topic, is always worth considering and seeing if we can do better. At a quick look, though, no clear plan for improvement presents itself to my mind. Possibly we could add motivation by adding some history, fleshing out the real-valued measurable section and starting with Lebesgue measure, move to 2-valued measures on cardinals, talk about what the additivity of such a measure can be (if it's uncountable, it must be measurable). Would that help? If done right, maybe it could help a little, but I'm kind of unsure. In any case it's unlikely to appear in the lead. --Trovatore (talk) 00:31, 22 September 2019 (UTC)

See List of large cardinal properties where it says "Existence of a cardinal number κ of a given type implies the existence of cardinals of most of the types listed above that type, and for most listed cardinal descriptions φ of lesser consistency strength, Vκ satisfies 'there is an unbounded class of cardinals satisfying φ'.". So one is dancing on the edge of contradiction, coming as close as possible without falling in.

You should try to see how each of these definitions of large cardinals seeks to establish the existence of similar large cardinals smaller than it. JRSpriggs (talk) 08:05, 22 September 2019 (UTC)

JR, I have a decent background in the subject, and I would argue with the claim that the definitions "seek" to show the similar large cardinals; it seems to me more to flow from a natural big picture. But that's neither here nor there; if you want we can talk about it on our personal talk pages. The question at hand is how to make the article more useful to more people (it being understood, of course, that there is only so much we can realistically do). --Trovatore (talk) 21:33, 22 September 2019 (UTC)

Gobbledygoop. The emperor has no clothes. These guys are smoking weed in the math department of some university or another, there are certain basic facts and logical consequences of accepted axioms they are not willing to accept, and they keep contradicting themselves. The fact of the matter is, that Lebesgue measure is not defined for all subsets of the reals, and no amount of expostulating and smoking ultra slim filtered cigarettes can change that. 24.237.158.249 (talk) 13:28, 17 January 2022 (UTC)

Regarding the "Gobbledygoop" comment above

The above comments section is damning, so lets work this out, and see where things first go wrong. The informal definition in the lede is just fine, and entirely understandable (to naive readers). The formal definition currently reads like so:

Here, κ-additive means: For every λ < κ and every λ-sized set {A_β}_β<λ of pairwise disjoint subsets A_β ⊆ κ, we have

μ(⋃_β<λ A_β) = Σ_β<λ μ(A_β).

The left-hand side looks fine: any naive understanding of set union yields acceptable intuitive results. The right-hand side is, however, confusing. What kind of addition is this? This is a confusion for the naive reader: if there are two sets A_β with μ(A_β)=1 then 1+1=2 so this contradicts the idea that the measure is 0-1-valued. So maybe Σ_β<λ μ(A_β) actually means MAX_β<λ μ(A_β)? Wildly guessing that some form of cardinal arithmetic is intended for the summation, and then clicking into that article, does not resolve the confusion. Perhaps the intent is that one and only one of the A_β has measure 1, and all the others have measure zero? Perhaps the sum is meant to be "logical-OR" (So that 0=F and 1=T is an indicator function indicating when A_β is small or not?) The mental juggling needed to explore this interpretation of this sentence seems pointless. So this is the first "gobbledygoop" issue to be fixed.

Subsidiary to this, it might be useful to state that ZFC is 'equivalent'(?) to first-order logic, and so indicator functions can be cleanly defined in ZFC for all cardinalities. i.e. there is no danger or risk of working with indicator functions (they are just predicates in FOL). This, at least, avoids potential confusion for a different class of naive readers, who might be trying to map concepts of classifying functors (see Yoneda lemma) onto this article. 84.15.187.169 (talk) 14:25, 10 July 2024 (UTC)

The point is that there are NOT "two [disjoint] sets A_β with μ(A_β)=1". JRSpriggs (talk) 14:24, 10 June 2026 (UTC)

More "Gobbledygoop"

The next point of confusion appears in the "properties" section, which currently reads as follows:

It is trivial to note that if κ admits a non-trivial κ-additive measure, then κ must be regular. (By non-triviality and κ-additivity, any subset of cardinality less than κ must have measure 0, and then by κ-additivity again, this means that the entire set must not be a union of fewer than κ sets of cardinality less than κ.) Finally, if λ < κ, then it can't be the case that κ ≤ 2^λ. If this were the case, we could identify κ with some collection of 0-1 sequences of length λ. For each position in the sequence, either the subset of sequences with 1 in that position or the subset with 0 in that position would have to have measure 1. The intersection of these λ-many measure 1 subsets would thus also have to have measure 1, but it would contain exactly one sequence, which would contradict the non-triviality of the measure. Thus, assuming the Axiom of Choice, we can infer that κ is a strong limit cardinal, which completes the proof of its inaccessibility.

Lets work this out. The first sentence is fine, although perhaps the word "trivial" should be struck. Next, open paren seems to delimit the start of a proof of the first sentence, and the close paren means the proof is concluded. OK, fine. Next it says "Finally". Finally what? Is this part two of the proof encased by the parenthesis? If so, the parenthesis should be moved. Or maybe this is a new paragraph discussing a new topic? Who can say?

Whatever, lets keep going. We read: identify κ with some collection of 0-1 sequences of length λ. Uhhh, is the "collection" of length λ, or are the sequences of length λ? Maybe both? Can I have a collection of five sequences? Why is it possible to "identify" the collection with κ? What kind of "identification" is this?

Without any clear idea of what these sequences are supposed to be, the trouble snowballs. Here: The intersection of these λ-many measure 1 subsets ... would contain exactly one sequence, which would contradict the non-triviality of the measure. Huh? What's the contradiction? So the intersection is one sequence. What's the big deal? Why would this imply that the measure is trivial?

The obvious (counter-)example that a newbie might want to think about is the case where λ=ω countable infinity and κ is the cardinality of the continuum, for which one has κ = 2^λ if CH holds. If CH does not hold, then I guess that the cardinality of the continuum κ is still $\kappa =2^{\omega }=\aleph _{2}$ and, at any rate the "collection" of "sequences" would refer to the Cantor set or possibly some subset of the Cantor set. Each countable-length sequence is a single point in Cantor space (i.e. the fine topology) and there's an uncountable number of points, and so if one tries to plug through this case into the gobbledeygoop paragraph, one gets strung up with no clear idea of how this "proof" results in the desired conclusion that κ must be regular (i.e. that the cardinality of the continuum is not a regular cardinal.) Clean this up, and maybe the rest of the article becomes "accessible". 84.15.187.169 (talk) 15:15, 10 July 2024 (UTC)

The sentence in parentheses is a proof that the cardinal is regular. But that is only a part of the larger proof that the cardinal is a strong inaccessible. A strong inaccessible must also be larger than 2^λ for any λ less than the cardinal. "The sequences [are] of length λ"; while the "collection" is of length κ. A measure being "trivial" (i.e. principal) means that some "one sequence" has measure 1. Remember that any finite set or indeed any set of cardinality less than κ must have measure zero to satisfy the definition of a measurable cardinal. The continuum is much too small to be a measurable cardinal. So your "counter-example" is vacuous. You should not just make up arbitrary facts out of your own prejudices and use them to try to prove something. JRSpriggs (talk) 13:36, 12 June 2026 (UTC)

Why are measurable cardinals so strong?

Merely being the critical point (set theory) of an elementary embedding is not sufficient for an ordinal, κ, to be a measurable cardinal. See for example unfoldable cardinal and worldly cardinal. What more is needed?

Let j be the elementary embedding. Suppose M is its domain and N is its range; and M and N are transitive classes using the usual element relation.

JRSpriggs (talk) 19:03, 12 June 2026 (UTC)

In the case of some kinds of worldly cardinals, we have V_κ is a subset of M but V_κ+1 is not. In the case of unfoldable cardinals, κ is an element of M and V_λ is a subset of N for some λ>κ, but M is not closed under powerset.

If V_κ+1 is a subset of M, then one can easily show that V_κ+1 is a subset of N, but I do not think that that is enough.

If we assume that κ is measurable and that its measure is a normal measure and a subset of the powerset of κ, I believe that we can show that κ is a 0-huge cardinal, that is, every function from κ to N is an element of N. Under these assumptions, if x is an element of V_κ, then

$\{\langle \alpha ,x\rangle :\alpha <\kappa \}$

is mapped to x. For A in V_κ+1

$\{\langle \alpha ,A\cap V_{\alpha }\rangle :\alpha <\kappa \}$

is mapped to A. If f is a function from κ to N and for each α<κ,

$\{\langle \beta ,g_{\beta \alpha }\rangle :\beta <\kappa \}$

maps to f (α), then I think that

$\{\langle \alpha ,\{\langle \beta ,g_{\beta \alpha }\rangle :\beta <\alpha \}\rangle :\alpha <\kappa \}$

maps to f. JRSpriggs (talk) 19:20, 12 June 2026 (UTC)