Can we have a cardinal exponent of any number?

Question

Can we have a cardinal such as ${1.5}^{\aleph_0}$ in the way that we have a powerset as $2^{\aleph_0}$, or is $2^{\aleph_0}$ just the notation we use, rather than actual exponentiation. Or is something else that I don’t know about yet going on here?

Answer 1

What is "actual exponentiation"? What is $1.5^{1.5}$ or $\sqrt2^{\sqrt2}$ if not some notation?

Cardinal exponentiation is "actual exponentiation" since the cardinal is an arithmetic system that includes addition, multiplication, and exponentiation.

However cardinals are not real numbers. Let me say that again. $$\huge\textbf{Cardinals are }\underline{\textbf{not}}\textbf{ real numbers.}$$

It is a common beginner's mistake to think that the real numbers are "the 'real' numbers" and anything else is just a bunch of extensions. But it's just one way of extending the natural numbers. The ordinals is another, and the cardinals is another. And they do not agree on anything except the arithmetic of the natural numbers.

Since $1.5$ is not a cardinal number (nor it should be), the notation $1.5^{\aleph_0}$ is entirely meaningless, unless you decide that $1.5$ is a new notation for the cardinality of some set, or decided to use $\aleph_0$ to denote a real number or a complex number, or some other object which would make sense of this notation. In either case, you would just cause the majority of mathematician to cry a little bit inside.

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Let me finish by saying that unlike real exponentiation which is defined as a continuation of rational exponentiation, which itself is defined in terms of radicals and iterated multiplication, cardinal exponentiation is defined far more directly in terms of number of functions between two sets.

It is this simplicity, perhaps, that leads you to think that this is "just notation". It's not. It's called elegance. :)