Why does $1^{\aleph_0}=1$ but $\aleph_0^{\aleph_0}\neq\aleph_0$?

Question

Note: $k$ is finite number.

1. $1\cdot 1=1$
2. $1^k=\underbrace{1\cdot1\dots1}_{k}=1$
3. $1^{\aleph_0}=\underbrace{1\cdot1\dots1}_{\aleph_0}=1$

---

4. $\aleph_0\cdot\aleph_0=\aleph_0$
5. $\aleph_0^k=\underbrace{\aleph_0\cdot\aleph_0\dots\aleph_0}_{k}=\aleph_0$
6. $\aleph_0^{\aleph_0}=\underbrace{\aleph_0\cdot\aleph_0\dots\aleph_0}_{\aleph_0}=\aleph\neq\aleph_0$

I don't understand why statement 3 give the same result as statement 1 and statement 2 because in the infinite (like statement 3) I will expect to be different behavior like the examples i gave in statements 4, 5 and 6.

I hope the question is clear.

Answer 1

While for the natural numbers it is true that exponentiation is defined as iterated multiplication, this is not true in general. Even in the real numbers, what does it mean to "multiply $x$ by itself $\sqrt\pi$ times"?

Cardinal exponentiation has a definition. We can prove that for finite exponents, it does in fact agree with repeated multiplication, and you could argue that the definition makes so much sense that it should be considered repeated multiplication. Mainly because there should be a difference between multiplying something to itself finitely many times, and infinitely many times.

The definition of $\kappa^\lambda$ is the cardinality of the set of all functions from a set $L$ into a set $K$ such that $|L|=\lambda$ and $|K|=\kappa$.

So $1^{\aleph_0}$ is the set of all functions from $\Bbb N$ into $\{0\}$. How many functions like this you can find? And remember that a function needs to be defined on all of $\Bbb N$. So yes, there is only one.

In contrast, $\aleph_0^{\aleph_0}$ is the set of all functions from $\Bbb N$ to itself, and we can prove that there are uncountably many of them. So it is certainly not $\aleph_0$ again. But this can also be said about $2^k$ vs. $2^{\aleph_0}$. It's $1$ that is the odd one out here, not $\aleph_0$.

---

What goes wrong? You might expect exponentiation to be continuous somehow.

But why? There is no reason for it to be continuous. Is $x^y$ continuous as a function of two real variables? Before you say yes, remember that it is in fact not continuous at $(0,0)$. Or take complex logarithms, for example, they are continuous, but you need to choose a branch and remove that from the domain for the function to work at all.

There's no reason to expect that everything is continuous. Why should all cardinal arithmetic be continuous?

Answer 2

In contrast to @AsafKaragila's great answer, if we let $$"2^{<\aleph_0}"$$ be the cardinality of the set of functions whose domain is $<\aleph_0$ and whose range is $2$, then continuity is back: $$2^{<\aleph_0}=\aleph_0=\lim_n 2^{<n}=\lim_n 2^n-1.$$