Take the 2-minute tour ×

While having measure theory this year the following came in my mind:

  • When we go from finite objects to infinite we "lose" a lot of properties. For example the summation isn't well defined anymore, when the sum doesn't converge, and operations which are abelian in the finite case are not abelian in the infinite case (think of conditional convergent series).
  • When we go from countable to uncountable we have things like that too. Measures are most time countable additive but not uncountable additive (at least the lebesgue measure). Sums of uncountable indexing sets only can converge when they are in fact countable (meaning nearly all are 0).

So my question is after Arthur Comments slightly rephrased

What happens when you go from the cardinality $\mathfrak{c}$ to a higher cardinality? Are there more intermediate steps, meaning having a cardinality strictly greater than $\mathfrak{c}$ but not "to big"?

share|improve this question
I'm not certain if this is a well-defined question (at least in the final box). Note that there is no bound on $\mathfrak{c}$ among the alephs (i.e., for any aleph $\aleph_\alpha$ it is consistent with $\sf{ZFC}$ that $\mathfrak{c} > \aleph_\alpha$). –  Arthur Fischer May 31 '13 at 9:59
@ArthurFischer Oh right continuum hypothesis. What are the cardinals which are greater than $\mathfrak{c}$ per definition? –  Dominic Michaelis May 31 '13 at 10:02
@MphLee: Inaccessible cardinals are always bigger than the continuum. –  Asaf Karagila May 31 '13 at 13:13
@Dominic: $\beth_2$ is provably larger than the continuum. Also any other cardinal larger than $\beth_2$, as well $(2^{\aleph_0})^+$. –  Asaf Karagila May 31 '13 at 13:15
@MphLee: There exists one, but its value is provably unprovable. Not to mention that some models have more ordinals than others (consider the following case: $M$ is a countable transitive model of $\sf ZFC$ which has an inaccessible $\kappa$, then $N=(V_\kappa)^M$ is a smaller countable transitive model. We can force over $M$ To have $\frak c$ greater than $\kappa$, in which case this is an ordinal which $N$ doesn't even know about, regardless to the fact that $M$ and $N$ agree completely about the real numbers and their cardinality. –  Asaf Karagila May 31 '13 at 21:58

2 Answers 2

up vote 5 down vote accepted

Allow me to share a small piece of my philosophical about the uncountable. It may not actually answer your question, but in a way it does.

Mathematics is created by humans. Humans are essentially finite beings, and they communicate in a finite matter. This is why we base things on finite intuition, which is the leading cause of confusion when people first deal with infinite sets.

Countable sets are sets that can be enumerated, and approximated by finite means. This is exactly the idea behind Cauchy's definition of the limit (and even more so if we only use rational numbers). We can list the rational numbers, for example, in a way that is assuring us that every number is met at a finite stage.

For this reason separability is a useful property. It allows us to approximate every element of the space using countable means. This is visible in metric space more than it is in non-metrizable spaces, but it's still true. That's the idea behind separability. So in some sense we may have an uncountable set, but we can approximate it using its structure and finite means.

But the uncountable, the truly uncountable, is not describable by such means. It's too big for us to assure that finite approximations can eventually swallow the set whole. We can't describe its initial segments in a finite fashion. We can't even assure that we can order the set nicely. This is why the axiom of choice is so important to us in the modern era of mathematics. We deal with abstract objects and we usually don't pose much limitations on their size (even though we usually care for objects quite small), and the axiom of choice gives us reasonable means to handle their sizes and structure.

But the uncountable is still uncountable. Note that even if we assume that $2^{\aleph_0}=\aleph_2$, without further assumptions such as Martin's Axiom, then we don't even know what's going on at the $\aleph_1$ level. All we know is that we can find a subset of the real numbers which has this cardinality. This is why the continuum hypothesis - in my opinion - is unprovable. The uncountable is indeed too large for us to comprehend (and manage in full) using our essentially finite means.

So what does happen beyond the continuum? A lot that we can't get a good grip on. The Lowenheim-Skolem theorem tells us that given a first-order structure which has an infinite model, it has an infinite model of every cardinality. This means that everything that happens at one infinite cardinality happens at the others, at least first-order-wise (e.g. there aren't many Archimedean fields, but there are real-closed fields). So we have a lot of fields, rings, measure algebras, Banach spaces, ordered sets, and so on.

What properties do they have? The answer often depends on set theoretic assumptions now. Much like many things at the countable level can fail without some axiom of choice (which assures the transition from finite to countable is smooth), many more can fail at the uncountable level. Assumptions such as $\lozenge_\kappa$, Martin's Axiom, cardinal arithmetics (e.g. $\kappa^{<\kappa}=\kappa$), and many many more set theoretical assumptions can - and will - tell you about the behavior of very large objects, but you will need to know about them first.

share|improve this answer
@Ittay: Thanks. –  Asaf Karagila May 31 '13 at 23:15

I hope I'm answering your question here.

First, the issue with conditionally convergent series is well known. In a sense, we can consider such series to diverge. The sum is highly sensitive to the order of summation and moreover the series can be re-arranged to give any sum at all. That is not the kind of behaviour you would expect of a convergent object. In the theory of integration there is a similar issue. The integral $\int_{i\infty }^\infty \sin(x)$ does not exist, even though, in some sense, it can be argued to be $0$. Moreover, by chopping the interval $(-\infty ,\infty)$ at suitable places, the integral can be 'computed' by splitting the integral as a sum of integrals to have any value you want. This is why often in integration theory a function whose absolute value is not integrable is not considered integrable.

Just to formalize things a bit, I'll concentrate on the following property, phrased in $\mathbb R$ and then generalized. Given a set of non-negative real numbers $S\subseteq [0,\infty ]$, its sum is defined to be the supremum of the set $\{s_1+\cdots s_k\mid k>0, s_i\in S\}$. This definitions agrees with the common definition of the sum of a series of non-negative elements. (If we allow $S\subseteq [-\infty , \infty ]$, then $S$ will have a sum iff it is countable and the elements give rise to an absolutely convergent series (historical remark: this is essentially how Bolzano defined infinite sums).) Now, it is easy to show that if the cardinality of $S$ is greater than $\aleph_0$, then the sum is $\infty $. This is a property of the non-negative reals $[0,\infty ]$ considered as a lattice with a compatible addition. This is the reason why in measure theory only countable additivity is entertained.

Now, in other lattices this cardinality restriction on sums need not hold. That is, consider a lattice $V$ with a notion of addition that is compatible (in whatever reasonable way) with the lattice structure (a common demand is that addition distributed over arbitrary joins and/or meets). For a subset $S\subseteq V$ one can defined its sum in $V$ as the supremum in $V$ of the same set of sums derived from $S$ as above. In an arbitrary lattice it is not the case the the sum will be $\infty $, i.e., the top element in the lattice, if the cardinality of $S$ is larger than $\aleph_0$. It is easy to construct artificial examples that will show that. I'm sure there are more practical situations where this happens, but I can't think of one right now.

This turned out to be a rather lengthy answer, so now I hope even more that it at least addressed what you were asking.

share|improve this answer
More of a parenthetical remark, but for any infinite cardinal $\kappa$ there will be lattices (or Boolean algebras) which are $\kappa$-complete (all subsets of size $< \kappa$ have a supremum), or some which are not. The continuum, $\mathfrak{c}$, is just one possible $\kappa$. –  Arthur Fischer May 31 '13 at 10:36

Your Answer


