Measurable cardinal
A measurable cardinal $\kappa$ is an uncountable cardinal such that it is possible to "measure" the subsets of $\kappa$ using a 2-valued measure on the powerset of $\kappa$, $\mathcal{P}(\kappa)$. There exists several other equivalent definitions: For example, $\kappa$ can also be the critical point of a nontrivial elementary embedding $j:V\to M$.
Measurable cardinals were introduced by Stanislaw Ulam in 1930.
Contents
Definitions
There are essentially two ways to "measure" a cardinal $\kappa$, that's to say we can require the measure to be $\sigma$-additive (a "classical" measure) or to be $\kappa$-additive (for every cardinal $\lambda$ such that $\lambda < \kappa$, the union of $\lambda$ null sets still has measure zero).
Let $\kappa$ be an uncountable cardinal.
Theorem 1 : The following are equivalent :
- There exists a 2-valued ($\sigma$-additive) measure on $\kappa$.
- There exists a $\sigma$-complete nonprincipal ultrafilter on $\kappa$.
The equivalence is due to the fact that if $\mu$ is a 2-valued measure on $\kappa$, then $U=\{X\subset\kappa|\mu(X)=1\}$ is a nonprincipal ultrafilter (since $\mu$ is 2-valued) and is also $\sigma$-complete because of $\mu$'s $\sigma$-additivity. Similarly, if $U$ is a $\sigma$-complete nonprincipal ultrafilter on $\kappa$, then $\mu:\mathcal{P}(\kappa)\to[0,1]$ defined by $\mu(X)=1$ whenever $X\in U$, $\mu(X)=0$ otherwise is a 2-valued measure on $\kappa$. [1]
An uncountable cardinal which satisfies the equivalent conditions of theorem 1 is sometimes called a 2-measurable cardinal (because "2-valued"). This is not a traditional notation, but it was used in an article of Gustave Choquet : "Cardinaux 2-mesurables et cônes faiblement compacts", Annales de l'Institut Fourier, tome 17, n°2 (1967), P.383-393.
Note : It is clear that, if $\kappa$ is 2-measurable, then every cardinal $\lambda$ such that $\lambda > \kappa$ is also 2-measurable. Thus, the notion of 2-measurability separates the class $C$ of all cardinals in two subclasses : the "moderated" cardinals and the 2-measurable cardinals, the first one being an initial segment of $C$, and therefore this notion is of weak interest for the study of the hierarchy of large cardinals.
Embedding Characterization
Theorem 2 : The following are equivalent :
- There exists a $\kappa$-complete nonprincipal ultrafilter on $\kappa$.
- There exists a nontrivial elementary embedding $j:V\to M$ with $M$ a transitive class and such that $\kappa$ is the least ordinal moved (the critical point).
- There exists a nonprincipal ultrafilter $U$ on $\kappa$ such that the ultrapower $(\text{Ult}_U(V),\in_U)$ of the universe is well-founded.
To see that the second condition implies the first one, one can show that if $j:V\to M$ is a nontrivial elementary embedding, then the set $\mathcal{U}=\{x\subset\kappa|\kappa\in j(x)\})$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and in fact a normal fine measure. To show the converse, one needs to use ultrapower embeddings: if $U$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$, then the canonical ultrapower embedding $j:V\to\text{Ult}_U(V)$ is a nontrivial elementary embedding of the universe. [1]
An uncountable cardinal $\kappa$ is called measurable if the equivalent conditions of theorem 2 are satisfied.
The two theorems are related by the fact (easy to prove) that the least cardinal $\kappa$ (if it exists) which carries a $\sigma$-complete nonprincipal ultrafilter is measurable, and in this case every $\sigma$-complete nonprincipal ultrafilter on $\kappa$ is $\kappa$-complete (see for example Patrick Dehornoy : "La théorie des ensembles", Calvage et Mounet, 2017).
In other words, the first 2-measurable cardinal is measurable.
Therefore, the two notions are equiconsistent, but in the general case they differ : every measurable cardinal is 2-measurable, and the converse is false.
Hayut Property
There is also another quite interesting model-theoretic characterization of measurability. Let a theory $T$ be $\kappa$-unboundedly satisfiable iff for every $\lambda<\kappa$, there is a model $\mathcal{M}\models T$ with $\lambda\leq|M|<\kappa$. In other words, the sizes of models of $T$ are unbounded in $\kappa$.
A class of formulae $Q$ is $\kappa$-Hayut iff for any $\kappa$-unboundedly satisfiable theory $T\subseteq Q$, there is a model of $T$ of size at least $\kappa$. More intuitively, $\kappa$-many small models of size less than $\kappa$ can combine to make one big $\kappa$-sized model.
An abstract logic $\mathcal{L}$ is called almost $\kappa$-favorable iff there is some way to represent every sentence of $\mathcal{L}$ with vocabulary $\tau$ as a sequence of length below $\kappa$ of symbols of $\tau$ and ordinals in $\kappa$ in such a way that the satisfaction relation is upward absolute for inner models $M$ of ZFC elementarily equivalent to $V$ with $M^{<\kappa}\subset M$. If $\kappa$ is an uncountable regular cardinal, the following are almost $\kappa$-favorable:
- $\mathcal{L}_{\lambda,\mu}$ for any $\lambda,\mu\leq\kappa$
- $\mathcal{L}_{\kappa,\omega}(q_{<\kappa})$, which is $\mathcal{L}_{\kappa,\omega}$ with universal cardinality quantifiers $q_\lambda$ for every $\lambda<\kappa$ (where $M\models q_\lambda$ iff $|M|\geq q_\lambda$)
- $\mathcal{L}_{\kappa,\kappa}$ with the addition of a single existential 2nd-order quantifier, where negation on the resulting sentences is not allowed
Assuming $V=L$, every $\mathcal{L}$ where sentences are represented as sequences of length below $\kappa$ of symbols of $\tau$ and ordinals in $\kappa$ ($\kappa$-sequential logic) that has an extension with an $\mathcal{L}_{\omega,\omega}$-definable satisfaction relation is almost $\kappa$-favorable. For example: if $V=L$, then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is almost $\kappa$-favorable, but if a measurable exists then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is not almost $\kappa$-favorable, and in fact if $\kappa$ is the least measurable then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is not $\kappa$-Hayut; however, if $\kappa$ is extendible, then $\mathcal{L}_{\kappa,\kappa}^{<\omega}$ is $\kappa$-Hayut, though it still isn't almost $\kappa$-favorable.
An uncountable regular cardinal $\kappa$ is measurable if and only if $\mathcal{L}_{\kappa,\kappa}$ is $\kappa$-Hayut, if and only if $\mathcal{L}_{\kappa,\omega}(q_{<\kappa})$ is $\kappa$-Hayut up to $2^\kappa$. Furthermore, an uncountable regular cardinal $\kappa$ is measurable if and only if every almost $\kappa$-favorable logic is $\kappa$-Hayut.
For more information, see this post.
Other Embedding Characterizations
There are also other embedding characterizations of measurable cardinals. Namely (under NBG or ZFC + $j$) the following are equivalent for any cardinal $\kappa$:
- $\kappa$ is measurable.
- $\kappa$ is the critical point of some $j:V\prec_{\Delta_0}V$.
- $\kappa$ is the critical point of some $j:V\prec_{\Delta_1}V$.
- $\kappa$ is the critical point of some $j:V\rightarrow V$ such that for any $\Sigma_1$-formula $\varphi$, $\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$.
Proof:
(1 implies 4). If $\kappa$ is measurable, then $\kappa$ is the critical point of a $j:V\prec M$ for some inner model $M$. Therefore $\kappa$ is the critical point of a $j:V\prec_{\Sigma_1}M$ and so for any $\Sigma_1$-formula $\varphi$, $\varphi[x,y,z...]\rightarrow M\models\varphi[j(x),j(y),j(z)...]$. Then, let $\varphi$ be a $\Sigma_1$-formula. If $\varphi[x,y,z...]$ then $M\models\varphi[j(x),j(y),j(z)...]$ and because $\Sigma_1$-formulae are upward absolute for inner models, $\varphi[j(x),j(y),j(z)...]$. Therefore: $$\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$$
(4 implies 3). Let $j:V\rightarrow V$ be as described. Then any $\Delta_1$ formula $\varphi$ is $\Sigma_1$ and $\neg\varphi$ is also $\Sigma_1$. So: $$\varphi[x,y,z...]\rightarrow\varphi[j(x),j(y),j(z)...]$$ $$\neg\varphi[x,y,z...]\rightarrow\neg\varphi[j(x),j(y),j(z)...]$$ $$\therefore\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]$$
(3 implies 2). If $j:V\prec_{\Delta_1}V$ then $j:V\prec_{\Delta_0}V$ by definition.
(2 implies 1). If $j:V\prec_{\Delta_0}V$ has critical point $\kappa$, then $\{X\subseteq\kappa:\kappa\in j(X)\}$ is a $\kappa$-complete measure on $\kappa$.
This gives further characterizations (which are weakenings of the $j:V\prec M$ characterizations). Namely, $\kappa$ is measurable iff it is the critical point of some $j:V\prec_{\Delta_0} M$ for some inner model $M$.
Proof:
(1 implies 2). Let $\kappa$ be measurable. Then $\kappa$ is the critical point of $j:V\prec M$ for some inner model $M$, and so $j:V\prec_{\Delta_0} M$.
(2 implies 1). Let $\kappa$ be the critical point of a $j:V\prec_{\Delta_0}M$. Then, $j:V\rightarrow V$ and for any $\Delta_0$ formula $\varphi$, $\varphi[x,y,z...]\Leftrightarrow M\models\varphi[j(x),j(y),j(z)...]$. Of course, since $\varphi$ is $\Delta_0$, $M\models\varphi[j(x),j(y),j(z)...]$ iff $\varphi[j(x),j(y),j(z)...]$. So: $$\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]$$ $$j:V\prec_{\Delta_0}V$$ $$\therefore\kappa\text{ is measurable.}$$
Finally, this gives much more general (and astonishing) characterizations of measurability in terms of nontrivial elementary embeddings of classes into themselves. The following are equivalent:
- $\kappa$ is measurable.
- There is some transitive class (or set) $M\models\text{ZFC}$ with $\mathcal{P}^2(\kappa)\subseteq M$ and some $j:M\prec_{\Delta_0}M$ with critical point $\kappa$.
- For every transitive class (or set) $M\models\text{ZFC}$ with $\kappa\subseteq M$ there is a $j:M\prec_{\Delta_1}M$ with critical point $\kappa$.
Proof:
(1 implies 3). Let $\kappa$ be measurable and let $j:V\prec_{\Delta_1}V$. Then, let $M\models\text{ZFC}$ be a transitive class with $\kappa\subseteq M$. Consider $j\upharpoonright M:M\rightarrow M$. Let $\varphi$ be a $\Delta_1$ formula. Then, $\varphi[x,y,z...]\Leftrightarrow M\models\varphi[x,y,z...]$ (because $\Delta_1$ formulae are upward and downward absolute for transitive classes). Since $j$ is $\Delta_1$-elementary, for any $x,y,z...\in M$:$$\varphi[x,y,z...]\Leftrightarrow\varphi[j(x),j(y),j(z)...]\Leftrightarrow\varphi[j\upharpoonright M(x),j\upharpoonright M(y),j\upharpoonright M(z)...]$$ $$\Leftrightarrow M\models\varphi[x,y,z...]\Leftrightarrow M\models\varphi[j\upharpoonright M(x),j\upharpoonright M(y),j\upharpoonright M(z)...]$$ Therefore $j\upharpoonright M:M\prec_{\Delta_1}$. Now note that $j$ has critical point $\kappa$ and so $j\upharpoonright M$ does too.
(3 implies 2). Simply choose $M=V$ and note that any $j:V\prec_{\Delta_1}V$ with critical point $\kappa$ is already a $j:M\prec_{\Delta_0}M$ with critical point $\kappa$.
(2 implies 1). If $j:M\prec_{\Delta_0}M$ is as described, then simply choose $U=\{X\subseteq\kappa:\kappa\in j(X)\}$ as usual. The proof that $U$ is a $\kappa$-complete nonprincipal ultrafilter is a little more difficult than usual:
- If $X\in U$ and $X\subseteq Y\subseteq\kappa$, then $M\models (j(X)\subseteq j(Y)\subseteq j(\kappa))$ so $j(X)\subseteq j(Y)\subseteq j(\kappa)$ and therefore $\kappa\in j(Y)$ and $Y\in U$ ($U$ is closed upwards under subset).
- If $X,Y\in U$ then $\kappa\in j(X)\cap j(Y)$ and by $\Delta_0$-ness of $\cap$, $j(X\cap Y)=j(X)\cap j(Y)$ so $\kappa\in j(X\cap Y)$ and therefore $X\cap Y\in U$ ($U$ is a filter).
- Assume $X\in U$ is a finite set. Then it is easily verified that $j(X)=j"X$ and so $\kappa\in j"X$. This is a contradiction as $\kappa\neq j(x)$ for any set $x$; if it were then it would be an ordinal below $\kappa$ and so $\kappa$ would not be the critical point. ($U$ is nonprincipal).
- If $X\notin U$ then $\kappa\notin j(X)$ so $\kappa\in (j(\kappa)\setminus j(X))$ and by $\Delta_0$-ness of $\setminus$, $\kappa\in j(\kappa\setminus X)$ meaning $\kappa\setminus X\in U$. ($U$ is an ultrafilter).
- Finally, if $F\subseteq U$ is a family of size $\lambda<\kappa$, then $j(F)=j"F$ and since $\kappa\in j(X)$ for every $X\in F$, $\kappa\in X$ for every $X\in j"F=j(F)$ and so $\kappa\in\cap j(F)$. By $\Delta_0$-ness of $\cap$, $\kappa\in j(\cap F)$ and so $\cap F\in U$. ($U$ is $\kappa$-complete).
Category Theoretic Characterization
Interestingly, measurable cardinals have certain category theoretic properties about them. This connection is likely at heart due to the fact that certain embeddings $j:V\rightarrow V$ share connections with functors $F:\text{Set}\rightarrow\text{Set}$, and measurable cardinals can be characterized in terms of those embeddings.
Specifically a measurable cardinal exists if and only if there is a nonidentity exact functor $F:\text{Set}\rightarrow\text{Set}$ [2].
For more information, read [2].
Properties
Every measurable is a large cardinal, i.e. $V_\kappa$ satisfies $\text{ZFC}$, therefore $\text{ZFC}$ cannot prove the existence of a measurable cardinal.
$\kappa$ is inaccessible, the $\kappa$th inacessible, the $\kappa$th weakly compact cardinal, the $\kappa$th Ramsey, and similarly bears most of the large cardinal properties under Ramsey-ness.
Measurable cardinals are ineffable and stationary limits of ineffable cardinals.
It is notable that every measurable has the mentioned properties in $\text{ZFC}$, but in $\text{ZF}$ they may not (but their existence remains consistency-wise much stronger than existence of cardinals with those properties), in fact under the axiom of determinacy, the first two uncountable cardinals, $\aleph_1$ and $\aleph_2$, are both measurable.
Any measurable cardinal $κ$ has super Ramsey M-rank $κ^+$[3], is super completely Ramsey and a stationary limit of super completely Ramsey cardinals[4] and is a limit of regular cardinals $α$ which are $α$-Ramsey.[5]
If $\kappa$ is measurable, then it has a measure that take every value in $[0,1]$. Also there must be a normal fine measure on $\mathcal{P}_\kappa(\kappa)$.
Every measurable cardinal is regular, and (under AC) bears most large cardinal properties weaker than it. It is in particular $\Pi^2_1$-indescribable. However the least measurable cardinal is not $\Sigma^2_1$-indescribable. [1] Independently of the truth of AC, the existence of a measurable cardinal implies the consistency of the existence of large cardinals with the said properties, even if that measurable is merely $\omega_1$.
If $\kappa$ is measurable and $\lambda<\kappa$ then it cannot be true that $\kappa<2^\lambda$. Under AC this means that $\kappa$ is a strong limit (and since it is regular, it must be strongly inaccessible, hence it cannot be $\omega_1$).
If there exists a measurable cardinal then $0^\#$ exists, and therefore $V\neq L$. In fact, the sharp of every real number exists, and therefore $\mathbf{\Pi}^1_1$-determinacy holds. Furthermore, assuming the axiom of determinacy, the cardinals $\omega_1$, $\omega_2$, $\omega_{\omega+1}$ and $\omega_{\omega+2}$ are measurable, also in $L(\mathbb{R})$ every regular cardinal smaller than $\Theta$ is measurable.
Every measurable has the following reflection property: let $j:V\to M$ be a nontrivial elementary embedding with critical point $\kappa$. If $x\in V_\kappa$ and $M\models\varphi(\kappa,x)$ for some first-order formula $\varphi$, then the set of all ordinals $\alpha<\kappa$ such that $V\models\varphi(\alpha,x)$ is stationary in $\kappa$ and has the same measure as $\kappa$ itself by any 2-valued measure on $\kappa$.
Measurability of $\kappa$ is equivalent with $\kappa$-strong compactness of $\kappa$, and also with $\kappa$-supercompactness of $\kappa$ (fragments of strong compactness and supercompactness respectively.) It is also consistent with $\text{ZFC}$ that the first measurable cardinal and the first strongly compact cardinal are equal.
If a measurable $\kappa$ is such that there is $\kappa$ strongly compact cardinals below it, then it is strongly compact. If it is a limit of strongly compact cardinals, then it is strongly compact yet not supercompact. If a measurable $\kappa$ has infinitely many Woodin cardinals below it, then the axiom of determinacy holds in $L(\mathbb{R})$, also the axiom of projective determinacy holds.
A cardinal $\kappa$ is measurable if and only if it is $\kappa^+$-hypermeasurable.
Measurable cardinals are not necessarily tall, but it is relatively consistent that the least measurable cardinal is tall.
The least Woodin cardinal is not measurable (and not even weakly compact), but Shelah cardinals are measurable.
If $\kappa$ is measurable in a ground model, then it is measurable in any forcing extension of that ground model whose notion of forcing has cardinality strictly smaller than $\kappa$. Prikry showed however that every measurable can be collapsed to a cardinal of cofinality $\omega$ and no other cardinal is collapsed.
If $κ$ is uncountable, $κ = κ^{<κ}$ and $2^κ = κ^+$, then the following are equivalent:
- $κ$ is measurable.
- $κ$ satisfies the $κ^+$-filter property.
- $κ$ satisfies the strategic $κ^+$-filter property.
On the other hand, starting from a $κ^{++}$-tall cardinal $κ$, it is consistent that there is a cardinal $κ$ with the strategic $κ+$-filter property, however $κ$ is not measurable.[5]
If $κ$ is a measurable cardinal, then $κ$ is $κ$-very Ramsey. If a cardinal is $ω_1$-very Ramsey (=strategic $ω_1$-Ramsey cardinal), then it is measurable in the core model unless $0^\P$ exists and an inner model with a Woodin cardinal exists.[6, 7]
The existence of a strategic $ω+1$-Ramsey cardinal (and of a strategic fully Ramsey cardinal) is equiconsistent with the existence of a measurable cardinal.[6]
A cardinal $κ$ is $C^{(n)}$-measurable iff there is a transitive class $M$ and an elementary embedding $j : V → M$ with critical point $crit(j) = κ$ and with $j(κ) ∈ C^{(n)}$. Every measurable cardinal is $C^{(n)}$-measurable for all $n$.[8]
If $\mathrm{I}_4^0(\kappa)$, then $\kappa$ is measurable and $\{\alpha\lt\kappa|\alpha\text{ is measurable}\}$ has measure 1. If $i$ witnesses $\mathrm{I}_4^n(\kappa)$, then $i^n(\kappa)$ is measurable.[9]
Measurable cardinals are equiconsistent with weakly measurable cardinals. If GCH holds, then all weakly measurable cardinals are measurable.
Without the axiom of choice $\omega_1$ can be measurable. ZFC+“There is a measurable cardinal” is equiconsistent to ZF+DC+“$\omega_1$ is measurable”. (Jech, 1968, Takeuti, 1970, after [1])
See also: Ultrapower
Failure of $\text{GCH}$ at a measurable
Gitik proved that the following statements are equiconsistent:
- The generalized continuum hypothesis fails at a measurable cardinal $\kappa$, i.e. $2^\kappa > \kappa^+$
- The singular cardinal hypothesis fails, i.e. there is a strong limit singular $\kappa$ such that $2^\kappa > \kappa^+$
- There is a measurable cardinal of Mitchell order $\kappa^{++}$, i.e. $o(\kappa)=\kappa^{++}$
Thus violating $\text{GCH}$ at a measurable (or violating the SCH at any strong limit cardinal) is strictly stronger consistency-wise than the existence of a measurable cardinal.
However, if the generalized continuum hypothesis fails at a measurable, then it fails at $\kappa$ many cardinals below it.
Real-valued measurable cardinal
A cardinal $\kappa$ is real-valued measurable if there exists a $\kappa$-additive measure on $\kappa$. The smallest cardinal $\kappa$ carrying a $\sigma$-additive 2-valued measure must also carry a $\kappa$-additive measure, and is therefore real-valued measurable, also it is strongly inaccessible under AC.
If a real-valued measurable cardinal is not measurable, then it must be smaller than (or equal to) $2^{\aleph_0}$. Martin's axiom implies that the continuum is not real-valued measurable. [1]
Solovay showed that the existence of a measurable cardinal is equiconsistent with the existence of a real-valued measurable cardinal. More precisely, he showed that if there is a measurable then there is generic extension in which $\kappa=2^{\aleph_0}$ and $\kappa$ is real-valued measurable, and conversely if there exists a real-valued measurable then it is measurable in some model of $\text{ZFC}$.
Virtually measurable cardinal
(all information from [6])
A cardinal $κ$ is virtually measurable iff for every regular $ν > κ$ there exists a transitive M and a forcing $\mathbb{P}$ such that, in $V^\mathbb{P}$, there is an elementary embedding $j : H_ν^V → M$ with $\mathrm{crit}(j) = κ$.
Every virtually measurable cardinal is strategic $ω$-Ramsey, and every strategic $ω$-Ramsey cardinal is virtually measurable in $L$.
If κ is virtually measurable, then either $κ$ is remarkable in $L$ or $L_κ \models \text{“there is a proper class of virtually measurables”}$.
See also
Read more
- Jech, Thomas - Set theory
- Bering A., Edgar - A brief introduction to measurable cardinals
References
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- Carmody, Erin and Gitman, Victoria and Habič, Miha E. A Mitchell-like order for Ramsey and Ramsey-like cardinals. , 2016. arχiv bibtex
- Feng, Qi. A hierarchy of Ramsey cardinals. Annals of Pure and Applied Logic 49(3):257 - 277, 1990. DOI bibtex
- Holy, Peter and Schlicht, Philipp. A hierarchy of Ramsey-like cardinals. Fundamenta Mathematicae 242:49-74, 2018. www arχiv DOI bibtex
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- Bagaria, Joan. $C^{(n)}$-cardinals. Archive for Mathematical Logic 51(3--4):213--240, 2012. www arχiv DOI bibtex
- Corazza, Paul. The gap between $\mathrm{I}_3$ and the wholeness axiom. Fund Math 179(1):43--60, 2003. www DOI MR bibtex