I wasn’t sure if David was seriously asking, but I spent 15 minutes reading Wikipedia and I wanted to give it a go.

Disclaimers: This is a reworking of things I learned on Wikipedia. This probably isn’t the best explanation out there. I assume experience with some basic set theory, including working a bit with cardinals. (I also assume the axiom of choice.)

1

If you start with finite cardinals, it is very difficult to get to infinity. “Getting” from a collection of sets to a new set entails using some kind of operations. Let’s pick two operations which generalize nicely to all of set theory:

• Addition: take a set of cardinals, union them all together, and take its size.

• One reason why we want to look at exponentiation is that it’s a really powerful force in set theory. $2^S$ is always bigger than $S$ , by Cantor’s diagonal argument. That’s pretty much the only way to make a set that’s guaranteed to be bigger than a given set.

  Exponentiation: take a cardinal, find the set of all of its subsets, and take its size.

If you start from finite cardinals, you will never get to infinity using these two constructions. That is:

• There is no finite set of finite cardinals which adds up to an infinite cardinal.
• There is no finite cardinal with an infinite power set.

Formally, using the notation $\aleph_0$ for the smallest infinite cardinal – the size of the set of natural numbers – we can say:

• There is no $<\aleph_0$ -set of $<\aleph_0$ -cardinals which adds up to $\geq\aleph_0$ .
• There is no $<\aleph_0$ -cardinal S with $2^S \geq \aleph_0$ .

We can generalize these properties to arbitrary cardinals, and give them names:

• $\lambda$ is a regular cardinal if there is no $<\lambda$ -set of $<\lambda$ -cardinals which adds up to $\lambda$ .
• $\lambda$ is a strong limit cardinal if there is no $<\lambda$ -cardinal S with $2^S \geq \lambda$ .

If a cardinal is regular and strong-limit, it feels kind of “inaccessible” – you can’t get to it from smaller cardinals. We define a cardinal to be strongly inaccessible if it is regular and strong-limit... almost. We actually also include the requirement that it be greater than $\aleph_0$ , to make things more interesting. You’ll see why.

2

Once we move past $\aleph_0$ , things get pretty interesting. In particular, it gets very difficult to find cardinals which satisfy both of our properties at once.

2a: Making a regular cardinal

It’s easy to find another regular cardinal. Just take $\aleph_1$ , the next largest cardinal after $\aleph_0$ . (Turns out there’s a next largest cardinal after $\aleph_0$ . Take my word for it.) Every cardinal smaller than $\aleph_1$ is countable. So to show $\aleph_1$ is regular, you need to show that the union of a countable set of countable sets is countable. This is just a zig-zag-path argument, like the one used to show that the rationals are countable.

So $\aleph_1$ is regular. But there’s no way it’s strong-limit... $2^{\aleph_0}$ is bigger than $\aleph_0$ , so it must be bigger than or equal to $\aleph_1$ . This means that, though $\aleph_1$ is not accessible through addition, it is totally accessible through exponentiation. Not inaccessible!

2b: Making a strong limit cardinal

Let’s try a different tack, and look for a strong limit cardinal. We want to find a cardinal (past $\aleph_0$ ) so big that it’s not the exponentiation of any smaller cardinal. There’s a neat trick to pull this off! Consider the series:

We’ll give these terms names, using the Hebrew letter “beth”:

(So $\beth_{n+1} = 2^{\beth_n}$ .)

Start by looking at partial sums. The funny thing about adding infinite cardinals is that larger cardinals absorb smaller ones in addition. Addition just becomes a “max” operation:

Every partial sum is just some $\beth_n$ , which is definitely not a strong limit cardinal – it’s straight-up defined as $2^{\beth_{n-1}}$ , the exponentiation of a smaller cardinal.

But the partial sums are not our goal here! We want the whole series, the sum of all (infinitely countably many) terms. Call this $\beth_\omega$ .

We can’t say $\beth_\omega$ is the “max” of the $\beth_n$ , since there’s no largest $\beth_n$ , but it is still the “least upper bound” of the $\beth_n$ .

Can we access $\beth_\omega$ via exponentiation? Let’s start with an arbitrary $S < \beth_\omega$ and see if we can possibly get $2^S = \beth_\omega$ . If a number is less than the least upper bound of a collection of numbers, it must be less than some specific number in the collection. Similarly, $S < \beth_\omega$ implies there must be some finite n with $S < \beth_n$ . But... that means $2^S$ can’t be any bigger than $2^{\beth_n}$ , which is $\beth_{n+1}$ , so $2^S$ can’t possibly be $\beth_\omega$ , which is larger than ALL the $\beth_n$ .

The significance of the trick is revealed! By taking the limit of a tower of exponentiations, we construct a cardinal which can’t be the exponentiation of anything. $\beth_\omega$ is a strong limit ordinal.

But there’s no way $\beth_\omega$ is regular. (Awww.) We constructed it with a countable sum of $\beth_n$ ’s, each of which is smaller than $\beth_\omega$ . Even though $\beth_\omega$ is not accessible through exponentiation, it is totally accessible through addition. Not inaccessible!

3

So our initial attempts to construct a strongly inaccessible cardinal have failed. Our attempt at regularity lacks strong-limit-ness, and our attempt at strong-limit-ness lacks regularity.

We shouldn’t be hard on ourselves. It turns out that it is impossible to prove the existence of a strongly inaccessible cardinal from the basic axioms of ZFC.

I don’t have time to dig deeply into why this is so, but I think I get the general gist of it. Basically, a strongly inaccessible cardinal $\lambda$ gives you a platform to construct a entire self-contained universe of set theory. Take the set of all sets “hereditarily smaller” than $\lambda$ , where “hereditarily smaller” means the set is smaller than $\lambda$ , its elements are smaller than $\lambda$ , their elements are smaller than $\lambda$ , etc. If you start only with sets inside this set, which we call $V_\lambda$ , it is very difficult to get out. None of the operations provided for by the ZFC axioms will ever get you out of $V_\lambda$ . (This requires that $\lambda$ be a strongly inaccessible cardinal!) $V_\lambda$ actually provably provides a complete model of ZFC.

But if you can construct a model of a set of axioms, you prove the consistency of those axioms. So if ZFC could construct a strongly inaccessible cardinal, it would in effect prove its own consistency. Gödel’s (slightly less famous) second incompleteness theorem demonstrates this is impossible, unless ZFC itself is inconsistent and just proves everything willy-nilly. So a consistent ZFC cannot prove the existence of a strongly inaccessible cardinal.

Interesting side note: The argument above works most of the way for $\aleph_0$ , which we know is strongly inaccessible in every sense except that it is countable. $V_{\aleph_0}$ is called the universe of “hereditarily finite sets”, and it does indeed provides a model for all of ZFC... except for the axiom of infinity. (ZFC has an axiom specifically to make sure there is an infinite set.) So I think that’s why $\aleph_0$ isn’t considered strongly inaccessible. It’s not big enough to make a universe for infinitary ZFC. This pattern, in which the hypothetical existence of a certain kind of cardinal is used to make a model for a certain set of axioms, is the basis for the study of “large cardinals”.