Can you go from $\aleph_0$ to $\aleph_1$ with tetration or other higher order operators?

Question

The paradox of Hilbert's Hotel shows us that you can not get past the cardinality of the natural numbers ($\aleph_0$) by adding a finite number (one new guest), adding an infinite quantity (infinitely many new guests), or by multiplying by any infinite quantity (infinitely new buses with infinitely many guests each). Put simply, this shows that $\aleph_0 + n = \aleph_0 \cdot n = \aleph_0^n = \aleph_0$ for all $n$ in the natural numbers.

I know that you can get from $\aleph_0$ to $\aleph_1$ by taking $2^{\aleph_0}$, but what I want to know is if it is possible for tetration, or any similar operators that are extensions of multiplication/exponentiation/tetration? I am almost certain the answer is no, since I don't believe any of those operators can surpass the speed of exponentiation as they tend towards infinity, but I haven't seen it explicitly discussed anywhere and I would like to be sure.

Answer 1

First of all, $2^{\aleph_0}$ is not necessarily $\aleph_1$. The definition of $\aleph_1$ is the cardinality of the least uncountable ordinal, whereas $2^{\aleph_0}$ is the cardinality of the power set of $\Bbb N$. Whether or not the two are equal is a provably unprovable statement, known as "The Continuum Hypothesis".

Now that we moved this out of the way, we don't have a lot to work with when it comes to cardinal arithmetic:

1. We have addition.
2. We have multiplication.
3. We have exponentiation.
4. We have the successor operation.

As defined, $\aleph_1$ is the successor of $\aleph_0$. And as you've argued, taking $\aleph_0+n$ or $\aleph_0\cdot n$, or even $\aleph_0^n$ all still give us $\aleph_0$. By definition this means that $\aleph_0+\aleph_0$ or $\aleph_0\cdot\aleph_0$ are both also $\aleph_0$.

Exponentiation with infinite powers, however, jumps up: $$2^{\aleph_0}\leq\aleph_0^{\aleph_0}\leq(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$$

Think about repeated exponentiations and so on. Either you are still $\aleph_0$, or you had to jump to $2^{\aleph_0}$ because your power was infinite. There is no middle ground there.

If you want to reach exactly $\aleph_1$, then the only way to guarantee that is to use the successor operation. Since $2^{\aleph_0}>\aleph_1$ is something that is mathematically consistent with all the things we know about sets. If you just want to reach $2^{\aleph_0}$, then $\aleph_0^{\aleph_0}$ already takes you there.

Answer 2

First of all let's skip the requirement that we must arrive at exactly $\aleph_1$. To be able to do that one have to either accept (as an axiom) the continuum hypothesis or it's negation. Let's just settle for a bigger set than $\mathbb N$.

This will require power sets as in my answer here either directly or indirectly. So basically you need the operation $2^{\aleph_0}$.

When it comes to tetration it's not sure that it's possible to extend that arbitrarily. It's quite clear that we can define tetration $^na$ on sets where $n$ and $a$ are finite. For infinite $a$ and still finite $n$ all we would need to do is to make sure that $a^b$ is defined for arbitrary sets, I think the axiom of powerset would guarantee that it's possible - we create a set (of same cardinality as $b$) of disjoint sets (of same cardinality as $b$) and then there one could apply the axiom of power set. It then becomes clear that if $|a|>2$ we would have that $^2a \ge 2^a$ which means that $^2\aleph_0 > 2^\aleph_0$ which is larger than $\mathbb N$.

However I don't think that you can define $^na$ in a meaninful way if $n$ is infinite.