Inmathematics, specifically inset theory, ameasurable cardinal is a certain kind oflarge cardinal number. In order to define the concept, one introduces a two-valuedmeasure on a cardinal, or more generally on any set. For a cardinal, it can be described as a subdivision of all of itssubsets into large and small sets such that itself is large, theempty set and allsingletons with are small,complements of small sets are large and vice versa. Theintersection of fewer than large sets is again large.[1]
It turns out thatuncountable cardinals endowed with a two-valued measure are large cardinals whose existence cannot be proved fromZFC.[2]
The concept of a measurable cardinal was introduced byStanisław Ulam in 1930.[3]
Formally, a measurable cardinal is an uncountablecardinal number such that there exists a-additive, non-trivial, 0-1-valuedmeasure on thepower set of.
Here,-additive means that for every and every-sized collection of pairwise disjoint subsets, we have
Equivalently, is a measurable cardinal if and only if it is an uncountable cardinal with a-complete, non-principalultrafilter. This means that the intersection of anystrictly less than-many sets in the ultrafilter is also in the ultrafilter.
Equivalently, is measurable if it is thecritical point of a non-trivialelementary embedding of theuniverse into atransitive class. This equivalence is due toJerome Keisler andDana Scott, and uses theultrapower construction frommodel theory. Since is aproper class, a technical problem that is not usually present when considering ultrapowers needs to be addressed, by what is now calledScott's trick.
It is trivial to note that if admits a non-trivial-additive measure, then must beregular: by non-triviality and-additivity, any subset of cardinality less than must have measure 0, and then by-additivity again, this means that the entire set must not be a union of fewer than sets of cardinality less than. Finally, if then it can't be the case that. If this were the case, we could identify with some collection of 0-1 sequences of length. 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-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 theaxiom of choice, we can infer that is astrong limit cardinal, which completes the proof of itsinaccessibility.
Although it follows fromZFC that every measurable cardinal isinaccessible (and isineffable,Ramsey, etc.), it is consistent withZF that a measurable cardinal can be asuccessor cardinal. It follows from ZF + AD that is measurable,[4] and that every subset of contains or is disjoint from aclosed and unbounded subset.
Ulam showed that the smallest cardinal that admits a non-trivial countably-additive two-valued measure must in fact admit a-additive measure. (If there were some collection of fewer than measure-0 subsets whose union was, then the induced measure on this collection would be a counterexample to the minimality of.) From there, one can prove (with the axiom of choice) that the least such cardinal must be inaccessible.
If is measurable and and (the ultrapower of) satisfies, then the set of such that satisfies is stationary in (actually a set of measure 1). In particular, if is a formula and satisfies, then satisfies it and thus satisfies for a stationary set of. This property can be used to show that is a limit of most types of large cardinals that are weaker than measurable. Notice that the ultrafilter or measure witnessing that is measurable cannot be in since the smallest such measurable cardinal would have to have another such below it, which is impossible.
If one starts with an elementary embedding of into withcritical point, then one can define an ultrafilter on as. Then, taking an ultrapower of over, we can get another elementary embedding of into. However, it is important to remember that. Thus, other types of large cardinals such asstrong cardinals may also be measurable, but not using the same embedding. It can be shown that a strong cardinal is measurable and also has-many measurable cardinals below it.
Every measurable cardinal is a 0-huge cardinal because, that is, every function from to, is in. Consequently,.
If a measurable cardinal exists, every (with respect to theanalytical hierarchy) set of reals has aLebesgue measure.[4] In particular, anynon-measurable set of reals must not be.
A cardinal is calledreal-valued measurable if there is a-additiveprobability measure on the power set of that vanishes on singletons. Real-valued measurable cardinals were introduced byStefan Banach (1930).Banach & Kuratowski (1929) showed that thecontinuum hypothesis implies that is not real-valued measurable.Stanislaw Ulam (1930) showed (see below for parts of Ulam's proof) that real valued measurable cardinals are weakly inaccessible (they are in factweakly Mahlo). All measurable cardinals are real-valued measurable, and a real-valued measurable cardinal is measurable if and only if is greater than. Thus a cardinal is measurable if and only if it is real-valued measurable and strongly inaccessible. A real valued measurable cardinal less than or equal to exists if and only if there is acountably additive extension of theLebesgue measure to all sets of real numbers if and only if there is anatomless probability measure on the power set of some non-empty set.
Solovay (1971) showed that existence of measurable cardinals in ZFC, real-valued measurable cardinals in ZFC, and measurable cardinals in ZF, areequiconsistent.
Let be an outer measure on a set. Say that a cardinal number is anUlam number if whenever[5][nb 1]
| , | 1 |
| for every, | 2 |
| all areμ-measurable, | 3 |
then implies.
Equivalently, if is a set of pairwise disjoint subsets of, is an Ulam number if whenever
then implies.
The smallest infinite cardinalℵ0 is an Ulam number. The class of Ulam numbers is closed under thecardinal successor operation.[6] If an infinite cardinal has an immediate predecessor that is an Ulam number, assume satisfies properties (1)–(3) with. In thevon Neumann model of ordinals and cardinals, for each, choose aninjective function and define the sets
Since the functions are injective, the sets
and
are pairwise disjoint. By property (1) of, the set
iscountable, and hence
Thus, there is a such that
which implies, since is an Ulam number and using the second definition, that
If and then. Thus
By property (1),, and since, by (3), (1) and (2),. It follows that. The conclusion is that is an Ulam number.
There is a similar proof[7] that the supremum of a set of Ulam numbers with an Ulam number is again a Ulam number. Together with the previous result, this implies that a cardinal that is not an Ulam number isweakly inaccessible.