What Is a Measurable Cardinal in Set Theory?

[tzimie]

Please help me, I am an idiot )

From here: https://en.wikipedia.org/wiki/Measurable_cardinal

measurable cardinal is a certain kind of large cardinal number. In order to define the concept, one introduces a two-valued measure on a cardinal κ, or more generally on any set. For a cardinal κ, it can be described as a subdivision of all of its subsets into large and small sets such that κ itself is large, ∅ and all singletons {α}, α ∈ κ are small, complements of small sets are large and vice versa. The intersection of fewer than κ large sets is again large​

I don't understand the last part I've put in bold.
It is saying that k can't be split into 2 "large" disjoint sets? For reals (not measurable cardinal of course), I can call a non-countable sets "large". Then x<0 and x>0 are both large with empty intersection. Why it won't work for measurable?

I can believe that measurable cardinal is so big, that there are so many elements, that (countable set of formulas) can't effectively discriminate then to build such sets. But the definition above say nothing about definability, but just about sets in general.And AC is so powerful that can build disjoint sets if we don't require formulas

Please help me understand it on intuitive level.

 

[stevendaryl]

The intersection of fewer than κ large sets is again large

I don't understand the last part I've put in bold.
It is saying that k can't be split into 2 "large" disjoint sets? For reals (not measurable cardinal of course), I can call a non-countable sets "large". Then x<0 and x>0 are both large with empty intersection. Why it won't work for measurable?

I don't know anything about measurable cardinals, but if that property is correct, then yes, you cannot split $\kappa$ into disjoint "large" sets. It does seem weird, but here's an analogy: Let's call a set of natural numbers "large" if its complement is finite. Then the intersection of two "large" sets is again large.

 

[tzimie]

Let's call a set of natural numbers "large" if its complement is finite. Then the intersection of two "large" sets is again large.

Thank you, interesting.
But now I am even more puzzled, because your example satisfies all criteria for being measurable cardinal. Hence I am missing something important.

 

[stevendaryl]

Thank you, interesting.
But now I am even more puzzled, because your example satisfies all criteria for being measurable cardinal. Hence I am missing something important.

Sort-of, but the notion of large and small in my example isn't a measure. A measure has the property that for any omega-sequence $S_1, S_2, ...,$ of disjoint sets, $\mu(\cup_j S_j) = \sum_j \mu(S_j)$. That doesn't work for my example, because you can create a "large" set as a countable union of "small" sets. So that would mean $\mu(S_j) = 0$ but $\mu(\cup_j S_j) = 1$.

 

[Stephen Tashi]

But now I am even more puzzled, because your example satisfies all criteria for being measurable cardinal. .

The definition of the "measurable cardinal" (according the Wikipedia article you cited) says that such a cardinal must be uncoutable. So the natural numbers are are not an example of a measurable cardinal.

For a cardinal κ, it can be described as a subdivision of all of its subsets into large and small sets such that κ itself is large,...

The first logical tangle to straighten out is whether "a cardinal" has subsets.

Is "a cardinal" a "set" ?

If so, is "the cardinality of the set S" equal to the set "S" itself? Or is "the cardinality of the set S", a property of the set "S"?

Let K = "cardinality of the set S". If K is a "property" of the set S then is this property defined as the set of all sets that share the property? Or is K defined as an equivalence relation on sets? - then it would be a set whose elements were ordered pairs of sets. That would make "the subsets of K" is a different collection of sets than "the subsets of S".

 

[stevendaryl]

The first logical tangle to straighten out is whether "a cardinal" has subsets.

Is "a cardinal" a "set"?

I'm not sure if you're actually asking these questions because you want to know, or asking them as Socrates would, to teach. But in set theory with the axiom of choice, you can associate ordinals and cardinals with pure sets. An ordinal is a special set that is defined so that it is equal to the set of all smaller ordinals. A cardinal is an ordinal that is not equal in size to any smaller ordinal (where two sets $A$ and $B$ are equal in size if there is a one-to-one function mapping $A$ to $B$). So by this definition of "cardinal", a cardinal is a set of ordinals.

If so, is "the cardinality of the set S" equal to the set "S" itself?

With the above definition of cardinal, if S is a cardinal, then the cardinality of S is S itself. If S is not a cardinal, then the cardinality of S is the smallest ordinal that is the same size as S.

Let K = "cardinality of the set S". If K is a "property" of the set S then is this property defined as the set of all sets that share the property?

Yes, one approach to defining cardinality is to say that the cardinality of a set S is the collection of all sets S' such that S and S' are the same size. That's an unwieldy definition to work with, though. The more convenient definition is to say that the cardinality of S is the smallest ordinal that is the same size as S. This requires the Axiom of Choice, however (which is equivalent to saying that every set is the same size as some ordinal).